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3 years ago

Use properties to find sum or product of 4 + (6+21)

Mathematics

1 answer:

Maurinko [17]3 years ago

4 0

You need to use pemdas so first you do inside the parenthesis 6+21=27 then you add 4 27+4=31 that is your answer

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What is the sum of the zeros of the function: y= x^2 - 5x + 6?

yawa3891 [41]

Answer:

Step-by-step explanation:

The zeros are the values of x for which y = 0

Quadratic formula

x = [5 ± √(5² – 4·1·6)] / [2·1]

= [5 ± √1] / 2

= [5 ± 1] /2

= 2, 3

sum of zeros = 5

7 0

2 years ago

Factor each completely.<br><br> 2x3 - 9x2 + 9x

EleoNora [17]

Answer:

x(2x−3)(x−3)

Step-by-step explanation:

2x^3−9x^2+9x

= x(2x−3)(x−3)

4 0

3 years ago

How do you solve ? <br>2(x−(3+2x)+9)=3x−8

PIT_PIT [208]

Hey there!!

Given equation :

... 2 ( x - ( 3 + 2x ) + 9 ) = 3x - 8

Using the distributive property.

... 2 ( x - 3 - 2x + 9 ) = 3x - 8

... 2 ( -x + 6 ) = 3x - 8

Using the distributive property.

... -2x + 12 = 3x - 8

Subtracting 12 on both sides.

... -2x = 3x - 8 - 12

... -2x = 3x - 20

Subtracting 3x on both sides.

... -2x - 3x = -20

... -5x = -20

Dividing by -5 on both sides.

... x = -20 / -5

... x = 4

<em>Hence, the answer is 4. </em>

Hope my answer helps!

6 0

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:

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

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

In conclusion, the area is

A=3 √4 units ^2