Would appreciate some help

Mathematics

1 answer:

agasfer [191]1 year ago

4 0

Answer:

e) 36x² - 32y³ = (4)(9x² - 8y³)

f) 45a⁴ - 54a³ = (9a³)(5a - 6)

Step-by-step explanation:

e) 36x² - 32y³ = (4)(9x² - 8y³) becuase 4*9 = 36 and 4*8 = 32

f) 45a⁴ - 54a³ = (9a³)(5a - 6) becuase 9*5 = 45 and 9*6 = 54 and when a is multiplied by a³ you get a⁴

(the exponent are added)

PLEASE RATE!! I hope this helps!!
If you have any questions comment below

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NemiM [27]

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Step-by-step explanation:

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2 years ago

P = a^b + b^a. Given a,b,p are prime numbers. find the value of p

VARVARA [1.3K]

2³ + 3² = 2*2*2+3*3= 8+9=17
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2 years ago

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:

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

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