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3 years ago

(a) (-5z)^3 (b) 5x^4(x^2)

Mathematics

1 answer:

yuradex [85]3 years ago

6 0

A) ( - 5 )^3 * z^3 = -125z^3;
b) 5x^( 4 + 2 ) = 5x^6;

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Find an equation for the line that passes through the points (-5, -5) and (3, 5)

Luda [366]

Answer:

5x-4y+5=0.

Step-by-step explanation:

1) if the point A(-5;-5) and B(3;5), the formula is:

$\frac{x-X_A}{X_B-X_A} =\frac{y-Y_A}{Y_B-Y_A}$

2) according to the formula above:
$\frac{x+5}{3+5} =\frac{y+5}{5+5}; \ = > \ \frac{x+5}{4}=\frac{y+5}{5};$

or 5x-4y+5=0.

8 0

2 years ago

What is the vertex of the absolute value function defined by ƒ(x) = |x + 2| + 4?

Marrrta [24]

Answer:

C. (-2,4)

Step-by-step explanation:

We have been given a function $f(x)=|x+2|+4$ and we are asked to find the vertex of our absolute value function.

The rules for the translation of a function are as follows:

$f|x-h|=\text{Graph shifted to right by h units}$

$f|x+h|=\text{ Graph shifted to left by h units}$

$f|x|+h=\text{ Graph shifted upward by h units}$

$f|x|-h=\text{ Graph shifted down by h units}$

Upon comparing our absolute function with above transformations we can see that our function is shifted to two units right of the origin(0,0) so x coordinate of our absolute function will be -2.

Our function is shifted upward from origin by 4 units, therefore, y-coordinate of our absolute value function will be 4.

$f(x)=|x--2|+4$

Therefore, the vertex of our absolute value function will be on point (-2,4) and option C is the correct choice.

4 0

3 years ago

Read 2 more answers

I need help with no. 11 & 12. Pls show your work.

Hitman42 [59]

But you're already done....

5 0

3 years ago

2(a³+b²+11)+1(3+a³+b+11)

Ann [662]

Answer:

$2(a^3+b^2+11)+1(3+a^3+b+11)=3a^3+2b^2+b+36$

Step-by-step explanation:

I assume that you need simplification of the given expression.

The given expression is:

$2(a^3+b^2+11)+1(3+a^3+b+11)$

Using distributive property and multiplying 2 inside the parenthesis and 1 inside the other parenthesis. This gives,

$2(a^3+b^2+11)=2\times a^3+2\times b^2+2\times 11\\2(a^3+b^2+11)=2a^3+2b^2+22\\\\1(3+a^3+b+11)=1\times 3+1\times a^3+1\times b+1\times 11\\1(3+a^3+b+11)=3+a^3+b+11=a^3+b+14$