What makes this equation true?

Mathematics

1 answer:

DIA [1.3K]2 years ago

5 0

Answer:

$B.\cfrac{1}{2}$

<h3>Given equation</h3>

$\cfrac{12y-1}{2} =\cfrac{9y+8}{5}$

<h3>Solve it for y</h3>

1. Cross - multiply:

$5(12y-1)=2(9y+8)$

2. Open parenthesis:

$60y-5=18y+16$

3. Collect like terms:

$60y-18y=16+5$

4. Simplify:

$42y=21$

5. Divide both sides by 42:

$y=21/42=1/2$

Matching answer choice is B

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The region r is enclosed by the curves y=4-4x2 and y=0

Misha Larkins [42]

The region r is enclosed by the curves y₁ = 4 - 4x² and y₂ = 0 is 16/3 or 5.333 square units.

<h3>What is an area bounded by the curve?</h3>

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The region r is enclosed by the curves y₁ = 4 - 4x² and y₂ = 0

The intersection points will be

y₁ = y₂

4 - 4x² = 0

x = ±1

Then the area bounded by the curves will be

$\rm Area = \int _{-1}^1 (y_1- y_2) dx\\\\Area = \int _{-1}^1 (4 - 4x^2) dx\\\\Area = \left [ 4x - \dfrac{4x^3}{3} \right ]_{-1}^1\\\\Area = 4 \left ( 1 + 1 \right ) - \dfrac{4}{3} \left ( 1^3 - (-1)^3 \right )\\\\Area = 8 - \dfrac{8}{3}\\\\Area = \dfrac{16}{3} = 5.333 \$