Answers
Answer:
108 square yards
Step-by-step explanation:
l = 6 yd
w = 4 yd
h = 3 yd
Surface area of rectangular prism = 2*(lw +wh + hl)
= 2*(6*4 + 4*3 + 3*6)
= 2* ( 24 + 12 + 18)
= 2 *54
= 108 square yards
Related Questions
The volume of a right cylinder is 108π and its height is 12.
What is the length of the cylinder's radius?
Answers
Answer:
volume of a cylinder = π r^2 *h
108π=π*r^2*h
108=r^2*h
12=r^2
3.464=r
radius =3.464
Step-by-step explanation:
Brainliest would be appreciated
Ask for questions!
Answer:
The length of the cylinder's radius is 3
Step-by-step explanation:
A cylinders radius is found by the formula, r= √ v/πh
So, 108π/πh.
Because there is pi in both the numerator and denominator, we can cancel them out. This leaves the equation to be : r = √ 108/12.
r = √ 108/12
108/12 is 9.
r = √9
r = 3
Find the values that will make the equations true.
options:
a = 49
a = 27.5
a = 7
a = 24.5
Answers
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
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:
A football field is 120 yards long by 53yr wide. If a player runs diagonally from one corner to the opposite corner, how far will they travel
Answers
Answer:
They will travel about 131.18 yards.
Step-by-step explanation:
340 to the nearest 100
Answers
Answer:
300
Step-by-step explanation:
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
What is 40 x 40 x 40 please help fast ASAP
Answers
Which is
64 times 1000
Which is
64000
enter the value of c when the expression 21.2x+c is equivalent to 5.3(4x-2.6). PLEASE GIVE ME AN EXPLANATION AS WELL. I DON’T WANT JUST AN ANSWER.
Answers
21.2x+c = 5.3(4x-2.6)
To substitute h the is out, you could write this if it feels better for you:
21x+c = y and 5.3(4x-2.6) = y
Hope this helps!
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
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]
HURRY PLEASE! I NEED HELP. PLEASE GIVE ME THE ANSWER
Use the table to describe the function.
A 2-column table with 6 rows. Column 1 is labeled x with entries negative 1,000, negative 0.1, negative 0.0001, 0.0001, 0.1, 1,000. Column 2 is labeled f (x) with entries negative 1.9999982, 1.78, 1.799998 times 10 Superscript 6 Baseline, 1.799998 times 10 Superscript 6 Baseline, 1.78, negative 1.9999982.
The end behavior of the function is that as
x →±∞, y approaches
.
The Limit of f (x) as x approaches infinity
.
The function has an asymptote at
.
Answers
Answer:
-2
-2
y=-2
Step-by-step explanation:
Answer:
-2
-2
y=-2
Step-by-step explanation:
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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The quantity z varies directly with w and inversely with x. When w = 12 and x = 4, z = -42. Find z when w = 11 and x = 7.
PLEASE HELP ME!!!!!!!
Answers
Given the joint variation and the constant of variation, the value of z when w = 11 and x = 7 is -22
How to solve variation?z =( k × w) / x
where,
z = -42 w = 12x = 4k = constant of variation-42 = (k × 12) / 4
cross product
4 × -42 = 12k
-168 = 12k
k = -168 /12
k = -14
Find z when w = 11 and x = 7z =( k × w) / x
z = (-14 × 11) / 7
= -154 / 7
z = -22
Therefore, the value of z when w = 11 and x = 7 is -22
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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
7. What value of c will make x2 – 20x + c
a perfect square trinomial?
Answers
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:
i doont understand??
Answers
Give him brainliest <3
Which expression is equal to 0.75×0.09
Answers
Which expression shows the factored form of 15x + 4
Answers
Answer:
Step-by-step explanation:
The only factor you can use is H. This doesn't factor into anything nice (like the common factor would be a decimal which is usually not allowed when doing this kind of question.
Answer: The question is the answer. 15x + 4 or H
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
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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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
I need this for school, please help!!
Answers
The scale factor for a model is 8 cm m Model : 11.23cm actual: 40.2 m
Answers
When one variable precedes the other in time and the correlation between them is high, which statement can be made with confidence?
A) The variable that comes earliest in time causes the other variable
B) There is a high level of association between the two variables
C)The two variables have an association with an outcome variable
D)There are no extraneous variables at work in the situation
Answers
Using correlation coefficients, it is found that the correct option is given by:
B) There is a high level of association between the two variables.
What is a correlation coefficient?It is an index that measures correlation between two variables, assuming values between -1 and 1.If it is positive, the relation is positive, that is, they are direct proportional. If it is negative, they are inverse proportional.If the absolute value of the correlation coefficient is greater than 0.6, the relationship is strong.In this problem, it is said that the correlation between them is high, that is, strong, hence there is a high level of association, which means that option B is correct.
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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?
Hi, I think the domain is 5 am I right?
Answers
Starts at -1 and goes to 3
3+1 =4
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:
The ratio 125 : x is equivalent to the ratio x^2 : 125. What is the value of x?
Answers
Answer:
x = 25
Step-by-step explanation:
The proportion can be written using fractions and solved for x in the usual way.
[tex]\dfrac{125}{x}=\dfrac{x^2}{125}\\\\125^2=x^3\qquad\text{multiply by 125x}\\\\x=\sqrt[3]{125^2}=\sqrt[3]{5^6}=5^2\qquad\text{take the cube root}\\\\\boxed{x=25}[/tex]