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3 years ago

11

How to graph 2x + 6y = -8

1 answer:

weqwewe [10]3 years ago

7 0

this is how this should help.

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Dilate point a in relation to the origin by a dilation factor of 5.?

IRINA_888 [86]

Answer:

origin dilation if 5. is 5

7 0

2 years ago

What is the product of and (6+5i) and (3-7i) ?<br><br> -17-27i<br> -17+27i<br> 53+27i<br> 53-27i

USPshnik [31]

We are going to make the product step by step.
We have:
(6 + 5i) * (3-7i)
Multiplying:
((6) * (3)) + ((6) * (- 7i)) + ((5i) * (3)) + ((5i) * (- 7i))
Rewriting:
18 - 42i + 15i - 35i ^ 2
18 - 42i + 15i - 35 * (- 1)
18 - 42i + 15i + 35
53 - 27i
Answer:
the product of and (6 + 5i) and (3-7i) is:
53-27i

5 0

3 years ago

How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x

Fofino [41]

Answer:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

Step-by-step explanation:

We are given the function:

$f(x)=-4x^5-8x^3+12x$

And we want to finds its zeros.

Therefore:

$0=-4x^5-8x^3+12x$

Firstly, we can divide everything by -4:

$0=x^5+2x^3-3x$

Factor out an x:

$0=x(x^4+2x^2-3)$

This is in quadratic form. For simplicity, we can let:

$u=x^2$

Then by substitution:

$0=x(u^2+2u-3)$

Factor:

$0=x(u+3)(u-1)$

Substitute back:

$0=x(x^2+3)(x^2-1)$

By the Zero Product Property:

$x=0\text{ and } x^2+3=0\text{ and } x^2-1=0$

Solving for each case:

$x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}$

Therefore, our real and complex zeros are:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

4 0

2 years ago

The diameter of a small gear is 16 cm. This is a 2.5 cm more than 1/4 of the diameter of a large gear. What is the diameter of t

eimsori [14]

The large gear has a diameter of 54 CM.

6 0

3 years ago

Read 2 more answers

When four times a number n is decreased by 3 it’s at most 21

pashok25 [27]

The number is = 6
……………………………….

8 0

3 years ago