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3 years ago

Solve the equation: 1.) y/6 + 12 = (-12) 2.) y/3 - 4 = (-20)

Mathematics

1 answer:

vivado [14]3 years ago

4 0

I am sorry but what level is this

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Two opposite sides of the square are increased by 8 inches and the other two sides are decreased by 5 inches. the area of the ne

Iteru [2.4K]

Answer:

The length of a side of the square would be 8.

Step-by-step explanation:

7 0

2 years ago

Simplify and then classify by degree and number of terms 2x+3x^2(4x-5)

Delicious77 [7]

Answer:

2x + 3x^2 (4x - 5)

2x + 12x^3 - 15x^2

12x^3 - 15x^2 + 2x

Degree - 3

Terms - 3 (Trinomial)

<h3>Lets explore more - </h3>

The degree of a polynomial is the highest power of its variable
Degree represents the no. of zeros of the polynomial

Linear  - Degree 1
Quadratic - Degree 2
Cubic - Degree 3
Biquadratic - Degree 4

Terms - Classification of polynomial On the basis of terms

monomial - polynomial which has only 1 term
Binomial - Polynomial which has 2 terms
Trinomial - Polynomial which has 3 terms

5 0

3 years ago

Read 2 more answers

Solve the equation as shown 2(x+6) = 22

lutik1710 [3]

Answer:

x = 5

Step-by-step explanation:

2(x+6) = 22

2x+12 = 22

2x = 10

x = 5

6 0

3 years ago

Read 2 more answers

How do I fill out the table help me answer it but try to explain.

Elodia [21]

I believe it would be 6 ounces

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3 years ago

The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati

makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

$y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\$

And the second one would be

$y = 2-x \,\,\,\,\,, 1 \leq x \leq 2$ .

If you rotate the first region around the "y" axis you get that

$A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51$

And if you rotate the second region around the "y" axis you get that

$A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188$

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first region around the x axis you get that

$A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708$

And if you rotate the second region around the x axis you get that

$A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472$

And the sum would be 1.5708+1.0472 = 2.618

7 0

3 years ago

Solve the equation: 1.) y/6 + 12 = (-12) 2.) ﻿ ﻿ y/3 - 4 = (-20)

Solve the equation: 1.) y/6 + 12 = (-12) 2.) y/3 - 4 = (-20)