EASY POINTS!!
Question 4 of 10
True or false? The graph represents a function
A. True
B. False
SUBMIT
Answers
Answer: True
Step-by-step explanation:
It passes the vertical line test, no two points are on the same x value
Correct option is A .True . The given graph is of a function by vertical line test
What is a vertical line test for function?The vertical line test is a test that checks whether a given graph is of a function or not. It states that every vertical line passing through the given graph should intersect the graph at most one point .( which comes from the definition of function that that a function takes every point in the input set to a unique image in the codomain set)
For the given graph whenever we will draw a vertical line passing through any point on the domain then it will cut the graph at one point only this can be seen in the attached figure.
Therefore, its true that the given graph represents a function.
∴Correct option is A. True.
Also, Learn more about the vertical line test from the link below:
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Related Questions
The length of a rectangle is 7 cm less than four times its width. The area of the rectangle is 36 square cm
Answers
Answer:
W = 4 cm and L = 9 cm
Step-by-step explanation:
I don't see a question, but will assume the problem wants the length(L) and width(W) of the described rectangle.
Let L and W stand for Length and Width.
Area of a rectangle is given by L*W
We are told that L*W = 36 cm^2
We are also told that L = 4W-7 ["length of a rectangle is 7 cm less than four times its width"]
Substituting the second into the first equation:
L*W = 36 cm^2
(4W-7)*W = 36 cm^2 [L = 4W-7]
4W^2-7W - 36 cm^2 = 0
(W-4)(4W+9) = 0
The roots are: 4 and -(9/4)
We'll use the positive value: W = 4
Since L = 4W-7:
L = 4(4)-7
L = 16-7
L = 9 cm
∠A and \angle B∠B are vertical angles. If m\angle A=(7x-6)^{\circ}∠A=(7x−6)
∘
and m\angle B=(8x-27)^{\circ}∠B=(8x−27)
∘
, then find the measure of \angle B∠B
Answers
keeping in mind that vertical angles are always congruent.
[tex]\stackrel{\measuredangle A}{7x-6}~~ = ~~\stackrel{\measuredangle B}{8x-27}\implies -6=x-27\implies 21=x~\hfill \underset{\measuredangle B}{\stackrel{8(21)~~ - ~~27}{141}}[/tex]
What are range, index of qualitative variation (IQV), interquartile range (IQR), standard deviation, and variance
Answers
Answer:
To find the interquartile range (IQR), first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.
Step-by-step explanation:
7. What value of c will make x2 – 20x + c
a perfect square trinomial?
Answers
Need help with this problem
Answers
Answer:
A)
Step-by-step explanation:
Sum of all angles of triangle = 180
53 + 68 + x = 180
121 + x = 180
x = 180 - 121
x = 59°
Answer:
180 - 53 - 68 = 59
The third angle is 59°.
Step-by-step explanation:
NEED HELP ASAP!!!! Will give brainiest
Answers
The temperature is dropping at a rate of five degrees per hour.
Let d represent the number of degrees the temperature drops.
Let t represent the number of hours that pass.
Which is the dependent variable?
Answers
Answer:
The number of degrees the temperature drops°
Step-by-step explanation:
hope this helps
and hope this is the answer you was looking for
pls mark brainliest
Samir recorded the grade-level and instrument of everyone in the middle school School of Rock below. Seventh Grade Students Instrument # of Students Guitar 6 Bass 4 Drums 6 Keyboard 7 Eighth Grade Students Instrument # of Students Guitar 9 Bass 9 Drums 9 Keyboard 10 Based on these results, express the probability that a seventh grader chosen at random will play an instrument other than drums as a fraction in simplest form.
Answers
Using it's concept, it is found that there is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.
What is a probability?A probability is given by the number of desired outcomes divided by the number of total outcomes.
In this problem:
There is a total of 6 + 4 + 6 + 7 = 23 seventh graders.Of those, 23 - 6 = 17 play instruments that are not the drum.Hence:
[tex]p = \frac{17}{23}[/tex]
There is a [tex]\frac{17}{23}[/tex] probability that a seventh grader chosen at random will play an instrument other than drums.
More can be learned about probabilities at https://brainly.com/question/14398287
I need this for school, please help!!
Answers
What are the solutions of the equation (x + 2)2 + 12(x + 2) – 14 = 0? Use u substitution and the quadratic formula to
solve.
-8+5√2
O x=-6252
O x=-4+5√2
x=-2 +5√2
Answers
Answer:
-8+5√2
Step-by-step explanation:
(x+2)^2+12(x+2)–14=0
(x+2)^2=(x+2)(x+2)=x^2+4+4x
12(x+2)=12x+24
x^2+4+4x+12x+24-14=0
x^2+4x+12x+4+24-14=0
x^2+16x+14=0
quadratic formula
x = {-b +- square root of (b^2 – 4ac)} ÷ {2a}
a= 1
b = 16
c = 14
x = {-16 +- square root of (16^2 – 4*1*14)} ÷ {2*1}
x = {-16 +- square root of (256 – 56)} ÷ {2*1}
x = ((-16 +- square root of (200)) ÷ (2)
x = ((-16 +- 10√2)) ÷ (2)
x= -8+-5√2
8. Daisies and tulips are planted in a
garden. There are 11 fewer tulips
planted than daisies.
a. Write an expression that represents
the number of tulips in terms of the
number of daisies. Define any
variables used.
b. If 18 daisies are planted, how many
tulips are planted?
Answers
Answer:
a) [tex]d-11 = t[/tex]
b) 7
Step-by-step explanation:
Let [tex]d[/tex] = daisies
Let [tex]t[/tex] = tulips
a) 11 fewer tulips than daisies: [tex]d-11 = t[/tex]
b) Substitute 18 into [tex]d[/tex] and solve for [tex]t[/tex].
[tex]18-11= t[/tex]
[tex]7 = t[/tex]
What is 40 x 40 x 40 please help fast ASAP
Answers
Which is
64 times 1000
Which is
64000
A right triangle includes one algae that measures 14º. what is the measure of the third angle
A 14º C 90º
B 76º D 104º
Answers
Answer: B. 76
Step-by-step explanation: A right triangle is 180 degrees. It has an angle of 90 since it is a right triangle. 90 + 14 = 104. 180 - 104 = 76
Answer:
B. 76 degrees
Step-by-step explanation:
EVERY triangle's angles add up to 180 degrees. We already know that since it's a right triangle, one of the angles equals 90 degrees (that's a right angle) and they give us the second angle measurement, 14 degrees. If we add those two angle measures together and subtract them from 180, we should get the measure of the third angle as our answer.
14 + 90 = 104
180 - 104 = 76
Therefore, the third and final angle in this right triangle equals 76, so your answer is B. I hope this helps! Have a lovely day!! :)
Hello Calculus!
Find the value
[tex]\\ \rm\Rrightarrow {\displaystyle{\int\limits_3^5}}(e^{3x}+7cosx-3tan^3x)dx[/tex]
Note:-
Answer with proper explanation required and all steps to be mentioned .
Answers
Answer:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]
General Formulas and Concepts:
Calculus
Differentiation
DerivativesDerivative NotationDerivative Property [Multiplied Constant]:
[tex]\displaystyle (cu)' = cu'[/tex]
Derivative Property [Addition/Subtraction]:
[tex]\displaystyle (u + v)' = u' + v'[/tex]
Derivative Rule [Basic Power Rule]:
f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Integration
IntegralsIntegration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
Integration Method: U-Substitution + U-Solve
Step-by-step explanation:
Step 1: Define
Identify given.
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx[/tex]
Step 2: Integrate Pt. 1
[Integral] Rewrite [Integration Rule - Addition/Subtraction]:[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + \int\limits^5_3 {7 \cos x} \, dx - \int\limits^5_3 {3 \tan^3 x} \, dx[/tex][Integrals] Rewrite [Integration Property - Multiplied Constant]:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^3 x} \, dx[/tex][3rd Integral] Rewrite:
[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \int\limits^5_3 {e^{3x}} \, dx + 7 \int\limits^5_3 {\cos x} \, dx - 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx[/tex]
Step 3: Integrate Pt. 2
Identify variables for u-substitution and u-solve.
1st Integral
Set u:[tex]\displaystyle u = 3x[/tex][u] Differentiate [Derivative Properties and Rules]:
[tex]\displaystyle du = 3 \, dx[/tex][Bounds] Swap:
[tex]\displaystyle \left \{ {{x = 5 \rightarrow u = 3(5) = 15} \atop {x = 3 \rightarrow u = 3(3) = 9}} \right.[/tex]
3rd Integral
Set v:[tex]\displaystyle v = \sec x[/tex][v] Differentiate [Trigonometric Differentiation]:
[tex]\displaystyle dv = \sec x \tan x \, dx[/tex][dv] Rewrite:
[tex]\displaystyle dx = \frac{1}{\sec x \tan x} \, dv[/tex]
Step 4: Integrate Pt. 3
Let's focus on the 3rd integral first.
Apply Integration Method [U-Solve]:[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \int\limits^{x = 5}_{x = 3} {\frac{v^2 - 1}{v}} \, dv[/tex][Integral] Rewrite [Integration Property - Addition/Subtraction]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \int\limits^{x = 5}_{x = 3} {v} \, dv - \int\limits^{x = 5}_{x = 3} {\frac{1}{v}} \, dv \Bigg)[/tex][Integrals] Apply Integration Rules [Reverse Power Rule and Logarithmic Integration]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{v^2}{2} \bigg| \limits^{x = 5}_{x = 3} - \ln | v | \bigg| \limits^{x = 5}_{x = 3} \Bigg)[/tex][v] Back-Substitute:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 x}{2} \bigg| \limits^{5}_{3} - \ln | \sec x | \bigg| \limits^{5}_{3} \Bigg)[/tex]Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
[tex]\displaystyle 3 \int\limits^5_3 {\tan^2 x \bigg( \sec^2 x - 1 \bigg)} \, dx = 3 \Bigg( \frac{\sec^2 5 - \sec^2 3}{2}- \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \Bigg)[/tex]
Step 5: Integrate Pt. 4
Focus on the other 2 integrals and solve using integration techniques listed above.
1st Integral:
[tex]\displaystyle\begin{aligned}\int\limits^5_3 {e^{3x}} \, dx & = \frac{1}{3} \int\limits^5_3 {3e^{3x}} \, dx \\& = \frac{1}{3} \int\limits^{15}_9 {e^{u}} \, du \\& = \frac{1}{3} e^u \bigg| \limits^{15}_9 \\& = \frac{1}{3} \bigg( e^{15} - e^9 \bigg)\end{aligned}[/tex]
2nd Integral:
[tex]\displaystyle\begin{aligned}7 \int\limits^5_3 {\cos x} \, dx & = 7 \sin x \bigg| \limits^5_3 \\& = 7 \bigg( \sin 5 - \sin 3 \bigg)\end{aligned}[/tex]
Step 6: Integrate Pt. 5
[Integrals] Substitute in integrals:[tex]\displaystyle \int\limits^5_3 {\bigg( e^{3x} + 7 \cos x - 3 \tan^3 x \bigg)} \, dx = \frac{e^{15} - e^9}{3} + 7 \bigg( \sin 5 - \sin 3 \bigg) - 3 \bigg( \frac{\sec^2 5 - \sec^2 3}{2} - \ln \bigg| \frac{\cos 3}{\cos 5} \bigg| \bigg)[/tex]
∴ we have evaluated the integral.
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---
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Answer:
1086950.36760
Formula's used:
[tex]\rightarrow \sf \int sin(ax+b)=-\dfrac{1}{a} cos(ax+b)+c[/tex]
[tex]\rightarrow \sf \int cos(ax+b)=\dfrac{1}{a} sin(ax+b)+c[/tex]
[tex]\rightarrow \sf \int \dfrac{1}{ax+b} =\dfrac{1}{a} ln|ax+b|+c[/tex]
[tex]\rightarrow \sf \int e^{ax+b}=\dfrac{1}{a} e^{ax+b} + c[/tex]
[tex]\rightarrow \bold{ ln|a| - ln|b| = ln|\frac{a}{b} | }[/tex]
Explanation:
[tex]\dashrightarrow \sf \int \left(e^{3x}+7cos\left(x\right)-3tan^3\left(x\right)\right)[/tex]
apply sum rule: [tex]\bold{\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx}[/tex]
[tex]\dashrightarrow \sf \int \:e^{3x}dx+\int \:7\cos \left(x\right)dx-\int \:3\tan ^3\left(x\right)dx[/tex]
Integrate simple followings first, using formula's given above
[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-\int 3tan^3x[/tex]
Breakdown the component
[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int tan^2x(tanx)[/tex]
[ tan²x = sec²x - 1 ][tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\int (sec^2x-1)(tanx)[/tex]
===========================================================
for integration of [tex]\bold{\int (sec^2x-1)(tanx)}[/tex]
apply substitution ... u[tex]\dashrightarrow \int \dfrac{-1+u^2}{u}[/tex]
[tex]\dashrightarrow \sf \int \:-\dfrac{1}{u}+udu[/tex]
[tex]\dashrightarrow \sf - \int \dfrac{1}{u}du+\int \:udu[/tex]
[tex]\dashrightarrow -\ln \left|u\right|+\dfrac{u^2}{2}[/tex]
substitute back u = sec(x)[tex]\dashrightarrow \sf-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}[/tex]
================================================= insert back
[tex]\dashrightarrow \sf \dfrac{1}{3}e^{3x}+7\sin \left(x\right)-3\left(-\ln \left|\sec \left(x\right)\right|+\dfrac{\sec ^2\left(x\right)}{2}\right)[/tex] outcome after integrating
Now apply the given limits
[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{3(5)}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{3(3)}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]
simplify
[tex]\sf \hookrightarrow \sf \dfrac{1}{3}e^{15}+7\sin \left(5\right)-3\left(-\ln \left|\sec \left(5\right)\right|+\dfrac{\sec ^2\left(5\right)}{2}\right) - (\sf \dfrac{1}{3}e^{9}+7\sin \left(3\right)-3\left(-\ln \left|\sec \left(3\right)\right|+\dfrac{\sec ^2\left(3\right)}{2}\right))[/tex]
and group the variables
[tex]\sf \hookrightarrow \dfrac{e^{15}-e^9}{3}-\dfrac{3}{2\cos ^2\left(5\right)}+\dfrac{3}{2\cos ^2\left(3\right)}+7\sin \left(5\right)-7\sin \left(3\right)+3\ln \left(\dfrac{1}{\cos \left(5\right)}\right)-3\ln \left(-\dfrac{1}{\cos \left(3\right)}\right)[/tex]
value:
[tex]\sf \hookrightarrow 1086950.36760[/tex]
15 × [-5] +15 × [-3] = solution
Answers
[tex]Heyo![/tex]
SaddySokka is here to help!!
Let's do this step-by-step explanation!
[tex](15)(-5)+(15)(-3)[/tex]
[tex]=-75+(15)(-3)[/tex]
[tex]=-75+-45[/tex]
[tex]=-120[/tex]
Answer:
[tex]-120[/tex]
Hopefully, this helps you!!
Have a great day!!
SaddySokka~
Answer:
-120
Step-by-step explanation:
15×[-5]+15×[-3]
Use BODMAS
-75+-45
-120
Mr. Fuller wants to put fencing around his rectangular-shaped yard. the width of the yard is 55 feet and the length is 75 feet. how many feet of fencing does Mr. Fuller
Answers
The total length of the fencing is the perimeter of the rectangular yard which is P = 260 feet
What is the Perimeter of a Rectangle?
The perimeter P of a rectangle is given by the formula, P=2 ( L + W) , where L is the length and W is the width of the rectangle.
Perimeter P = 2 ( Length + Width )
Given data ,
Let the perimeter of the rectangular yard be = P
Now , the equation will be
The width of the rectangular yard W = 55 feet
The length of the rectangular yard L = 75 feet
And , the required fencing = Perimeter of rectangular yard
Perimeter of rectangular yard P = 2 ( L + W )
Substituting the values in the equation , we get
Perimeter of rectangular yard P = 2 ( 55 + 75 )
Perimeter of rectangular yard P = 2 ( 130 )
Perimeter of rectangular yard P = 260 feet
Therefore , the value of P is 260 feet
Hence , the perimeter of yard is 260 feet
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Which expression is equal to 0.75×0.09
Answers
ASAP PLS
Write an equation to represent the following scenario: Ms. Cloutier’s wedding photographer requires a $1000 deposit, and then $250 for every hour she is working.
Answers
x being the number of hours she’s working
Which pair of expressions has equivalent values?
1^13 and 1^15
6^1and 9^1
7^8and 8^7
9- and 4^3
Answers
Answer:
1^13 and 1^15
Step-by-step explanation:
1 raised to anything is still just 1
so, 1^13 = 1 and 1^15 =1
The scale factor for a model is 8 cm m Model : 11.23cm actual: 40.2 m
Answers
Find the circumference of the circle of 13 inches use 3.14 for pi and round to the nearest whole number
Answers
Answer:
ohh just use this formula
this is for Area- A= π r^2
this is for Circumference- C= 2 π r
Step-by-step explanation:
The circumference of the circle of the radius of 13 inches is, 41 inches
What is circumference ?Circumference is the distance around the perimeter of a circular object. It is defined as the length of the circle that is found by multiplying the diameter of the circle by π (pi), which is approximately equal to 3.14.
The formula for the circumference of a circle is given by: C = 2πr,
Given that,
The diameter of the circle is 13 inches,
The radius of the circle can be calculated as follows:
r = d/2
= 13 inches / 2
= 6.5 inches
Using the formula for the circumference, we can calculate the circumference as follows:
C = 2πr
= 2 × 3.14 × 6.5 inches
= 40.76 inches
Rounding to the nearest whole number, the circumference of the circle is 41 inches.
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Wind is
• air moving from areas of high pressure to areas of low pressure.
• air moving from areas of low pressure to areas of high pressure.
air moving from areas of high temperature to areas of low temperature.
Answers
Wind is caused by differences in the atmospheric pressure. When a difference in atmospheric pressure exists, air moves from the higher to the lower pressure area, resulting in winds of various speeds. On a rotating planet, air will also be deflected by the Coriolis effect, except exactly on the equator.
Hope this helped!
Which value is not a solution of the inequality x-4 symbol 2
Answers
The inequality x -4 > 2 uses a greater than symbol
All numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2
How to determine the value not in the solution?The inequality is given as:
x -4 > 2
Add 4 to both sides of the inequality
x - 4 + 4 > 2 + 4
Evaluate the sum
x > 6
The above means that only numbers greater than 6 are in the solution of the inequality.
Since the options are not given, I will give a general solution that all numbers lesser or equal to 6 are not a solution of the inequality x -4 > 2
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Prove that: a + b + c / a^-1+ b^-1+ c^-1 = abc
Answers
Answer:
It's right
Step-by-step explanation:
(dk how to show prove but thank?
A kitchen can be broken into 2 rectangles. One rectangle has a base of 7 feet and height of 5 feet. The second rectangle has a base of 2 feet and height of 2 feet. One package of tile will cover 3 square feet. How many packages of tile will she need? 8 13 15 39
Answers
Answer:
its 13 or B
Step-by-step explanation:
Number 21 please help me solve it thank youu
Answers
Answer:
64 pi
Step-by-step explanation:
32 is diameter
diameter is circumference
2 circles
so 2*32=64
help me please need help
Answers
Answer:
Step-by-step explanation:
1. x -> opposite side of 48°
o → hypotenuse
b → adjacent side of 48°
[tex]\sf Sin \ 48^\circ = \dfrac{opposite \ side }{hypotenuse}\\\\\\0.7431 = \dfrac{15}{o}\\\\\\0.74 * o = 15\\\\\\ o = \dfrac{15}{0.74}\\\\\\[/tex]
o = 20.27
[tex]\sf cos \ 48^\circ = \dfrac{adjacent \ side }{hypotenuse}\\\\\\0.67 =\dfrac{b}{o}\\\\\\0.67=\dfrac{b}{20.27}[/tex]
b = 0.67*20.27
b = 13.58
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
2) i → opposite side of 25°
n → adjacent side of 25°
[tex]\sf Sin \ 25 =\dfrac{i}{t}\\\\\\0.42=\dfrac{i}{30}\\\\\\0.42*30=i[/tex]
i = 12.6
[tex]\sf Cos \ 30^\circ =\dfrac{n}{t}\\\\0.91=\dfrac{n}{30}\\\\\\0.91*30 = n[/tex]
n = 27.3
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
3) a → opposite side of 70°
e → adjacent side of 70°
[tex]Sin \ 70^\circ =\dfrac{a}{l}\\\\0.94 =\dfrac{a}{25}\\\\0.94*25=a[/tex]
a = 23.5
[tex]\sf Cos \ 70^\circ =\dfrac{e}{l}\\\\0.34=\dfrac{e}{25}\\\\0.34*25=e[/tex]
e = 8.5
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
4)
[tex]\sf Sin \ 52^\circ = \dfrac{x}{75}\\\\0.79*75=x\\[/tex]
x = 59.25
[tex]\sf Cos \ 52^\circ = \dfrac{z}{75}\\\\0.62*75 =z[/tex]
z = 46.5
Arrange the expressions below in order from least to greatest. place the least at the top and greatest at the bottom. ( 72 ÷ 8 ) − 2 × 3 1 72 ÷ ( 8 − 2 ) × 3 1 72 ÷ ( 8 − 2 ) × ( 3 1 ) 72 ÷ 8 − 2 × ( 3 1 )
Answers
The expressions in order from least to greatest 72 / 8 - 2 x (3 + 1), (72 / 8) - 2 x 3 + 1, 72 / (8 - 2) x 3 + 1 and 72 / (8 - 2) x (3 + 1).
What is BODMAS?BODMAS stands for B - Bracket, O - order of Power, D - Division, M - Multiplication, A - Addition, and S - Subtraction.
To Arrange the expressions below in order from least to greatest. place the least at the top and the greatest at the bottom
72 / 8 - 2 x (3 + 1) equals 1
(72 / 8) - 2 x 3 + 1 equals 4
72 / (8 - 2) x 3 + 1 equals 37
72 / (8 - 2) x (3 + 1) equals 48
Thus, The expressions in order from least to greatest 72 / 8 - 2 x (3 + 1), (72 / 8) - 2 x 3 + 1, 72 / (8 - 2) x 3 + 1 and 72 / (8 - 2) x (3 + 1).
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Tom wants the scale model to be 9 inches tall.
How wide should the scale model be?
A. 1.7 inches
B. 5.4 inches
O c. 15 inches
OD. 20 inches
Answers
Answer:
C 15
Step-by-step explanation:
9 = 12 3/4
20 3/4 = 15
Identify the surface area of the cylinder to the nearest tenth. Use 3.14 for π.
Answers
Answer:
967.6
Step-by-step explanation:
967.6
967.12 in
Step-by-step explanation:
the formula for the area (surface area) of a cylinder is: A=2πrh+2πr2
to solve we need to determine the values
R= radius = half the diameter = 14/2 =7
D= diameter =14inches
H= height= 15 inches
plug in
A=2πrh+2πr2 = A=2π([tex]\frac{d}{2}[/tex])h+2π([tex]\frac{d}{2}[/tex])^2
= 2[tex]\pi[/tex]([tex]\frac{14}{2}[/tex])(15) + 2[tex]\pi[/tex]([tex]\frac{14}{2}[/tex])^2
= 2[tex]\pi[/tex](7)(15) + 2[tex]\pi[/tex](7)^2
=[tex]\pi[/tex]((2x7x15)+(2x7^2))
=[tex]\pi[/tex](210+98)
=[tex]308\pi[/tex]
=967.12 in