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3 years ago

Classify each representation shown below as either linear or exponential.

Mathematics

1 answer:

hichkok12 [17]3 years ago

4 0

1. Exponential

2. Linear

3. Exponential

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What is the answer to this?

Dennis_Churaev [7]

Answer:

Step-by-step explanation:

The angles (5x - 5)° and (6x + 31)° are supplementary angles.

Supplementary Angles add up to 180°.

Therefore we have the equation:

5x - 5 + 6x + 31 = 180 <em>combine like terms</em>

(5x + 6x) + (-5 + 31) = 180

11x + 26 = 180 <em>subtract 26 from both sides</em>

11x = 154 <em>divide both sides by 11</em>

x = 14

The angles (5x - 5)° and z° are vertical angles.

Vertical angles are congruent.

Therefore we have the equation:

z = 5x - 5 <em>put x = 14 to the equation</em>

z = 5(14) - 5

z = 70 - 5

z = 65

4 0

3 years ago

Drag and drop the equation to correctly complete the sentence.

Klio2033 [76]

Answer:

y= 2x - 11

Step-by-step explanation:

(-7 +1 )/ (2-5) = 2

y= 2x +b

-1 = 2* 5 + b

solve for b and you get -11 so the equation of the line is :

y = 2x -11

4 0

3 years ago

Another one, soon it'll be over (i hope)

maks197457 [2]

Answer:

24. 8 x 3.....................

6 0

3 years ago

Read 2 more answers

Which expression is equal to (6x−4)(3x−1)?

Natali5045456 [20]

Answer:

$(6x-4)(3x-1) = 18x^2 - 18x + 4$ is the required simplification.

Step-by-step explanation:

Here, the given expression is (6x−4)(3x−1).

By DISTRIBUTIVE PROPERTY of numbers:

A ( B + C ) = AB + AC

Simplifying the given expression, we get:

$(6x-4)(3x-1) = 6x(3x-1)-4(3x-1)\\= 6x(3x) -1(6x) -4(3x) +4\\=18x^2 - 6x - 12x + 4\\18x^2 - 18x + 4$

or, $(6x-4)(3x-1) = 18x^2 - 18x + 4$

Hence the given expression is simplified.

3 0

3 years ago

A vertical right circular cylindrical tank has height h=8 feet high and diameter d=6 feet. It is full of kerosene weighing 50 po

Mariana [72]

Answer:

The work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

Step-by-step explanation:

The work is defined by:

$W=\int dFdx$ (1)

The force here will be the product between the volume and the kerosene weighing, so we have :

$dF=\pi R^{2}dy*50$

This force will be in-lbs.

Where R is the radius (3 feet)

Then using (1), we have:

$W=\int \pi R^{2}dy*50(8-y)$

Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.

$W=\int_{0}^{8} \pi 3^{2}dy*50(8-y)$

$W=450\pi \int_{0}^{8}dy(8-y)$

$W=450\pi(\int_{0}^{8} 8dy-\int_{0}^{8} ydy)$

$W=450\pi(8y|_{0}^{8} -\frac{y^{2}}{2}|_{0}^{8})$

$W=450\pi(8*8 -\frac{8^{2}}{2})$

$W=450\pi(64 -\frac{64}{2})$

Therefore, the work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

I hope it helps you!

6 0

3 years ago