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3 years ago

Which expression is equivalent to (5y+4)−2(7−2y)?

Mathematics

2 answers:

Gemiola [76]3 years ago

4 0

Answer:

9y-10

Step-by-step explanation:

mafiozo [28]3 years ago

3 0

Answer:

Step-by-step explanation:

5y +y-2y

=6y-2y

=4y

hoped it helped :)

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Molodets [167]

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Step-by-step explanation:

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3 years ago

The area of a rectangle is 50 ft, and the length of the rectangle is 5 ft less than three times the width. Find the dimensions o

IgorC [24]

Answer:

L - 10 . W - 5

Step-by-step explanation:

Length = 3*width - 5

area = length * width

50 = (3w-5)* w

50 = 3w^2-5w

0 = 3w^2 - 5w -50

0= (3w+10)(w-5)

w = 5 , w can't equal -10/3 because width can't be negative

width is 5

Length = 3 * 5 - 5

Length = 10 , width = 5

5 0

3 years ago

Read 2 more answers

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2