what is the answer to 1/3 + 3/4 + 2/5 =

Answer:

x = - 11

Step-by-step explanation:

[tex]\frac{-x+10}{3}[/tex] = 7 ( multiply both sides by 3 to clear the fraction )

- x + 10 = 21 ( subtract 10 from both sides )

- x = 11 ( multiply both sides by - 1 )

x = - 11

Answer:

B

Step-by-step explanation:

It made more machines per hour than the other ones

Answer:

B or 5 and 1/2

Step-by-step explanation:

Answer:

The answer is B. 5 1/2

Step-by-step explanation:

8 5/8 - 3 1/8 = ?

The first thing to check for, is that the fractions have the same denominator!

They do: 8

8 5/8 - 3 1/8 = 5 4/8 reduce

5 1/2.  

5 1/2 is your answer!

The measure of the angle ∠DEF is 85 degrees and the measure of the angle ∠CFE is 57 degrees.

It is a polygon that has four sides. The sum of the internal angle is 360 degrees.

In a cyclic quadrilateral, the sum of opposite angles is 180°.

∠DCE + ∠DEF = 180°

             ∠DEF = 180° - ∠DCF

             ∠DEF = 180° - 95°

             ∠DEF = 85°

Similarly for the other two angles, we have

∠CDE + ∠CFE = 180°

             ∠CFE = 180° - ∠CDE

             ∠CFE = 180° - 123°

             ∠CFE = 57°

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a. Area of the exposed water of the prism pool: 33.5 ft²

b. The area of exposed water for the cone pool = 102.1 ft²

c. Hemisphere pool = πr² = π(4²) = 50.3 ft²

Volume = (2/3)πr

Volume = 1/3πr²h

Volume = Base area × height

a. Area of the exposed water = area of the square top = base area of the prism

Find base area using, Volume = Base area × height. Thus:

134 = Base area × 4

Base area = 134/4 = 33.5 ft²

Area of the exposed water for the prism pool = 33.5 ft²

b. Find the radius of the cone using, volume = 1/3πr²h.

134 = 1/3πr²(4)

(3)(134) = πr²(4)

402/4π = r²

32 = r²

r = 5.7 ft

The area of exposed water for the inverted cone pool = πr² = π(5.7)² = 102.1 ft²

c. The radius is the depth of the hemisphere pool

The area of the exposed water for the hemisphere pool = πr² = π(4²) = 50.3 ft²

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Answer:

car and van 602

Step-by-step explanation:

brain liest me please need

There are 427 vans and 435 cars were washed.

An equation is an expression that shows the relationship between two or more numbers and variables.

A mathematical equation is a statement with two equal sides and an equal sign in between. An equation is, for instance, 4 + 6 = 10. Both 4 + 6 and 10 can be seen on the left and right sides of the equal sign, respectively.

We are given that school car wash charged $5 for a car and $6 for a van.

The total of 86 cars and vans were washed on the weekend and they earned $475.

The equation are;

x + y = 86

5x + 6y = 475

Now we have;

5x + 6y = 475

5(86- y) + 6y = 475

y = 435

Therefore,

435 + x= 8

x = - 427

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Step-by-step explanation:

[tex]\sf x = \sqrt{a^{sin^{-1} \ t}}\\\\\\Derivative \ rule:\boxed{\dfrac{d(\sqrt{x})}{dx}=\dfrac{1}{2}*x^{\frac{-1}{2}}=\dfrac{1}{2\sqrt{x}}}[/tex]

[tex]\sf \dfrac{d(\sqrt{a^{sin^{-1} \ t}}}{dt}=\dfrac{1}{2\sqrt{a^{sin^{-1} \ t}}}*\dfrac{d(a^{sin^{-1} \ t})}{dt}\\\\\\Derivative \ rule: \boxed{\dfrac{d(a^{x})}{dx}=log \ a *a^{x}}[/tex]

                      [tex]\sf = \dfrac{1}{2\sqrt{a^{sin^{-1}} \ t}}*a^{sin^{-1} \ t}* log \ a *\dfrac{d(Sin^{-1} \ t)}{dt}\\\\[/tex]

[tex]Derivative \ rule:\boxed{\dfrac{d(sin^{-1} \ x}{dx}=\dfrac{1}{\sqrt{1-x^2}}}[/tex]

                       [tex]\sf = \dfrac{1}{2\sqrt{a^{sin^{-1} \ t}}}*a^{Sin^{-1} \ t}*log \ a*\dfrac{1}{\sqrt{1-x^2}}}}}\\\\ = \dfrac{a^{Sin^{-1} \ t}*log \ a}{2\sqrt{a^{sin^{-1} \ t}}*\sqrt{1-x^2}}[/tex]

[tex]\boxed{ \dfrac{a^{sin^{-1} \ t}}{\sqrt{a^{sin^{-1} \ t}}}=\dfrac{\sqrt{a^{sin^{-1} \ t}}*\sqrt{a^{sin^{-1} \ t}}}{\sqrt{a^{sin^{-1} \ t}}} = \sqrt{a^{sin^{-1} \ t}}}[/tex]

                         [tex]\sf = \dfrac{a^{sin^{-1} \ t}*log \ a}{2\sqrt{1-x^2}}[/tex]

[tex]\sf \dfrac{dy}{dt}=\dfrac{d(a^{cos^{-1} \ t})}{dt}[/tex]

     [tex]= \dfrac{1}{2\sqrt{a^{cos^{-1} \ t}}}*a^{cos^{-1} \ t}*log \ a *\dfrac{-1}{\sqrt{1-x^2}}}\\\\\\=\dfrac{(-1)*a^{cos^{-1} \ t}*log \ a}{2*\sqrt{a^{cos^{-1} \ t}}*\sqrt{1-x^2}}[/tex]

    [tex]\sf = \dfrac{(-1)*\sqrt{a^{Cos^{-1} \ t}}* log \ a }{2\sqrt{1-x^2}}\\\\[/tex]

                     [tex]\sf \bf \dfrac{dy}{dx}=\dfrac{dy}{dt} \div \dfrac{dx}{dt}\\[/tex]

                            [tex]\sf \bf = \dfrac{(-1)*\sqrt{a^{cos^{-1} \ t}}*log \ a}{2*\sqrt{1-x^2}} \ \div \dfrac{\sqrt{a^{sin^{-1} \ t}} *log \ a}{2*\sqrt{1-x^2}}\\\\\\=\dfrac{(-1)*\sqrt{a^{cos^{-1} \ t}}*log \ a}{2*\sqrt{1-x^2}} \ * \dfrac{2*\sqrt{1-x^2}}{\sqrt{a^{sin^{-1} \ t}} *log \ a}\\\\= \dfrac{(-1)* \sqrt{a^{cos^{-1} \ t}} }{\sqrt{a^{sin^{-1} \ t}}}\\\\= \dfrac{-y}{x}[/tex]  

[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Let's solve ~

[tex]\qquad \sf  \dashrightarrow \: x = \sqrt{ {a}^{sin {}^{ - 1}t } } [/tex]

here, let's differentiate it with respect to t ~

[tex]\sf   \dashrightarrow \: \dfrac{dx}{dt} = \dfrac{1}{2 \sqrt{a {}^{sin {}^{ - 1}t } } } \times a {}^{sin {}^{ - 1}t } \sdot ln(a) \times \dfrac{1}{ \sqrt{1 - {x}^{2} } }[/tex]

[tex]\sf  \dashrightarrow \: \dfrac{dx}{dt} = \dfrac{ \sqrt{ {a}^{sin {}^{ - 1}t } } \sdot ln(a)}{2 \sqrt{1 - {x}^{2} } } [/tex]

[tex]\sf \dashrightarrow \: \cfrac{dt}{dx} = \dfrac{2 \sqrt{1 - {x}^{2} } }{ \sqrt{a {}^{sin {}^{ - 1} t} \sdot ln(a)} }[/tex]

Smililarly,

[tex]\sf  \dashrightarrow \: \dfrac{dy}{dt} = \dfrac{1}{2 \sqrt{a {}^{cos{}^{ - 1}t } } } \times a {}^{cos {}^{ - 1}t } \sdot ln(a) \times \dfrac{ - 1}{ \sqrt{1 - {x}^{2} } }[/tex]

[tex]\sf  \dashrightarrow \: \dfrac{dy}{dt} = - \dfrac{ \sqrt{ {a}^{cos {}^{ - 1}t } } \sdot ln(a)}{2 \sqrt{1 - {x}^{2} } }[/tex]

Now : Lets get Required result ~

[tex]\sf \dashrightarrow \: \dfrac{dy}{dx} = \dfrac{dy }{dt} \times \dfrac{dt}{dx} [/tex]

[tex]\sf \dashrightarrow \cfrac{dy}{dx} = - \dfrac{\sqrt{ {a}^{cos {}^{ - 1}t } \sdot \cancel{ ln(a)}}}{ \cancel{2 \sqrt{1 - {x}^{2}}}} \sdot \dfrac{ \cancel{2 \sqrt{1 - {x}^{2}} } }{ \sqrt{a {}^{sin {}^{ - 1} t} }\sdot \cancel{ln(a)}}[/tex]

[tex]\sf \dashrightarrow \cfrac{dy}{dx} = - \dfrac{\sqrt{ {a}^{cos {}^{ - 1}t } }}{ \sqrt{a {}^{sin {}^{ - 1} t} }}[/tex]

[tex]\sf \dashrightarrow \cfrac{dy}{dx} = - \dfrac{y}{x} [/tex]

[ since y = [tex]\sf{\sqrt{a^{cos^{-1}t}} } [/tex] and x = [tex]\sf{\sqrt{a^{sin^{-1}t}} } [/tex] ]

Answer:

FACT CHECK THE ANSWERS!! they are based of mild research and could be wrong!!))

both questions come back as true from the best of my ability and understanding

Step-by-step explanation:
1) In metaphysics and philosophy of language, the correspondence theory of truth states that the truth or falsity of a statement is determined only by how it relates to the world and whether it accurately describes (i.e., corresponds with) that world.

2) Some philosophers view the concept of truth as basic, and unable to be explained in any terms that are more easily understood than the concept of truth itself. Most commonly, truth is viewed as the correspondence of language or thought to a mind-independent world. This is called the correspondence theory of truth.

both exlanations are snippets from Wikipedia

Answer:

x=14

Step-by-step explanation:

3x+1 and 43 are both vertical angles so they are equivant.

3x+1=43

3x=42

x=14

The area of the considered regular hexagon which has got 14.7 inches of apothem and a perimeter of 101.8 inches is 748.2 sq. inches.

Apothem for a regular polygon is a line segment which originates from the center of the regular polygon and touches the mid of one of the sides of the regular polygon. It is perpendicular to the regular polygon's side it touches.

Regular polygons have all side same and that apothem bisects the side in two parts, (provable by symmetry).

Consider the diagram attached below.

The area of the regular hexagon considered = 6 times (area of triangle ABC) (because of symmetry).

Also, we have:

Area of triangle ABC = 2 times (Area of triangle ABD).

Thus, we get:
Area of the considered hexagon = 6×2×(Area of triangle ABD)

Area of the considered hexagon = 12×(Area of triangle ABD)

Perimeter of a closed figure = sum of its sides' lengths.

There are 6 equal sides in a regular hexagon (due to it being regular).

Thus, if each side is of 'a' inch length, then:

Perimeter = 6×a inches

[tex]101.8 = 6a\\\\\text{Dividing both the sides by 6, to get 'a' on one side}\\\\a = \dfrac{101.8}{6} \approx 16.967 \: \rm inches[/tex]

This is bisected by the apothem.

Thus, we get:
Length of the line segment BD = |BD| = a/2 ≈ 8.483 inches

Since it is given that the length of the apothem = |AD| = 14.7 inches, therefore, we get:

[tex]\text{Area of ABD} = \dfrac{1}{2} \times \rm base \times height \approx \dfrac{14.7 \times 8.483}{2} \approx 62.35 \: \rm in^2[/tex]

Thus, we get:

Area of the considered hexagon = 12×(Area of triangle ABD)

Area of the considered hexagon [tex]\approx 12 \times 62.35 = 748.2 \: \rm in^2[/tex]

Thus, the area of the considered regular hexagon which has got 14.7 inches of apothem and a perimeter of 101.8 inches is 748.2 sq. inches.

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Answer:

748.23

Step-by-step explanation:

On edge

The maximum height is the highest level of height an object can reach. . The greatest height of the hangar is 32feet

The maximum height is the highest level of height an object can reach. Given the height in feet of the curved roof of an aircraft hangar can be modeled by y=-0.02x^2+1.6x

The velocity of the aircraft is zero at the maximum height. Therefore:

dy/dx = -0.04x + 1.6  = 0

Determine the value of x

0.04x = 1.6
x = 1.6/0.04
x = 40

Substitute x = 40 into the function to get the greatest height

y=-0.02x^2+1.6x

y=-0.02(40)^2+1.6(40)

y = -32 + 64

y = 32ft

Hence the greatest height of the hangar is 32feet

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The question is an illustration of combination, and there are 1820 ways to select the 12 students

The distribution of the students is given as:

Senior = 7

Junior = 5

Sophomore = 4

The total number of students is:

Total = 7 + 5 + 4

Evaluate

Total = 16

To select 12 students from the 16 students, we make use of the following combination formula

Ways = 16C12

Evaluate the expression using a calculator

Ways = 1820

Hence, there are 1820 ways to select the 12 students

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Using the normal distribution, it is found that the replacement time that separates the top 18% from the bottom 82% is of 11.23 years.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem, the mean and the standard deviation are, respectively, given by [tex]\mu = 9.4, \sigma = 2[/tex].

The desired value is the 82nd percentile, which is X when Z = 0.915, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.915 = \frac{X - 9.4}{2}[/tex]

X - 9.4 = 0.915(2)

X = 11.23

The replacement time that separates the top 18% from the bottom 82% is of 11.23 years.

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Answer:

n = 10

Step-by-step explanation:

12n ÷ 8 = 15

12n = 15 × 8

12n = 120

n = 120/ 12

n = 10

Answer:

C

Step-by-step explanation:

using Pythagoras' identity in the right triangle.

the square on the hypotenuse is equal to the sum of the squares on the other 2 sides.

let x be the hypotenuse , then

x² = 7² + 8² = 49 + 64 = 113 ( take square root of both sides )

x = [tex]\sqrt{113}[/tex]

Answer:

the interest owed is 1684.8

Step-by-step explanation:

Formula for simple interest =

I = Prt, where P is amount borrowed, r for interest rate and t for time.

Since the first three years is 1.6% interest we can write:

I = 15600 x 0.016 x 3

= 748.8

Then for the following 3 years the interest rate is 2%:

I = 15600 x 0.02 x 3

= 936

Adding the values gives us 1684.8

Answer:

Step-by-step explanation:

Formula

P = s1 + s2 + s3

Givens

s1 = 7x - 4

s2 = x + 3

s3 = 3y + 2

Solution

P = 7x - 4 + x+3 + 3y + 2

P = 8x + 3y + 3 + 2 - 4

Answer

P = 8x + 3y + 1

Note: I can't read the choices well enough to tell which answer it is. To me, it looks like A

Answer:

1. angle addition postulate

2. angle addition postulate

3. given

4. substitution

5. given

6. substitution

7. subtraction property of equality

Answer:

4968

Step-by-step explanation:

4500 x (1-4%) x (0.15 + 1) = 4500 x 0.96 x 1.15

(4500 x 0.96) x 1.15 = 4320 x 1.15

4320 x 1.15 = 4968

Hope this helps!

Answer:

f=2×12k⁴-6

Step-by-step explanation:

root(f+6/2)=12k²

take the root away f+6/2=12k²

f+6/2=12k⁴

f=2×12k⁴-6

Answer:

105 square feet

Step-by-step explanation:

First of all, find the area of the rectangle (length x width). 7*12 = 84. Now, find the area of the triangle [(base x height)/2]. 6*7 = 42. 42/2 = 21.

Add both of these areas together: 84 + 21 = 105. The area is 105 square feet.

Answer:

105 square feet

Step-by-step explanation:

There are two shapes in this figure, a rectangle and a triangle. We have to find the area of the two shapes separately.

l x w is the rectangle area formula.

12 x 7 = 84.

The area of the rectangle is 84 square feet

[tex]\frac{l * w}{2}[/tex] is the triangle area formula

[tex]\frac{7* 6}{2}[/tex] = 21

84 + 21 = 105 square feet.

Answer:

your second choice

Step-by-step explanation:

Answer:

B. 6 < s < 12.8

Step-by-step explanation:

I just took the test

[tex]\large \mathbb\purple {✒✨ANSWER ✨⚘ }[/tex]

⊱─━━━━━━━━━⊱༻●༺⊰━━━━━━━━━─⊰

➝ the absolute value of a number is its distance from 0 on the number line.

⊱─━━━━━━━━━⊱༻●༺⊰━━━━━━━━━─⊰

Answer:

c. 11.7

Step-by-step explanation:

Distance

Answer: $95

Step-by-step explanation:

Te new price will be 90 + 5x, if the price increases "x" times

The number of wind chimes sold per month will become 140 - 7x

[tex]\begin{aligned}&\text {Revenue, } \mathrm{R}(\mathrm{x})=(90+5 \mathrm{x})(140-7 \mathrm{x}) \\&R^{\prime}(x)=5(140-7 x)-7(90+5 x) \\&R^{\prime}(x)=700-35 x-630-35 x \\&R^{\prime}(x)=70-70 x=0 \\&70 x=70 \\&x=1\end{aligned}[/tex]

Therefore, if the price becomes (90 + 5(1)) = $95 per wind chime, then the revenue will be maximum

i am from afghanistan and you all gus

The trigonometric relation is solved and distance of the plane from the runway is D = 18,818.52 feet

Trigonometry is the study of the relationships between the angles and the lengths of the sides of triangles

The six trigonometric functions are sin , cos , tan , cosec , sec and cot

Let the angle be θ , such that

sin θ = opposite / hypotenuse

cos θ = adjacent / hypotenuse

tan θ = opposite / adjacent

tan θ = sin θ / cos θ

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

Given data ,

An airplane is flying 4,000 feet above the ground

A person is standing on the runway looking up at the plane. And the angle of elevation is 12°

So , from the trigonometric relation , we get

tan 12° = 4000 / D

On simplifying , we get

D = 4000 /  0.21255656167

D = 18,818.52 feet

Hence , the distance is D = 18,818.52 feet

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Answer:

6g quarts

Step-by-step explanation:

three is a factor and multiple of 6

Answer:

167.283525618[tex]in^{3}[/tex] or 53.248[tex]\pi[/tex][tex]in^{3}[/tex]

Step-by-step explanation:

first divide the diameter by 2 to get the radius and you'll get 1.6 and since the formula for finding the volume of a cylinder is [tex]\pi r^{2} h[/tex] you then square 1.6 and get 2.56 multiply that by the height and get 13.312 then multiply by pie to get 41.8208814046 and then multiply that by the number of jars you have which is four to get the total 167.283525618

or 53.248[tex]\pi[/tex] the problem did not specify how it wanted the answer to be

the answer is cubed because the volume is 3 dimentional