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Kipish [7]
3 years ago
7

Find the slope and y-intercept, then write the equation of the line WILL GIVE BRAINLIEST URGENT

Mathematics
1 answer:
anyanavicka [17]3 years ago
7 0

Answer:

Y-intercept= (0,-2)

Provide a clearer graph so I can find the slope

Step-by-step explanation:

I will edit my answer after you provide a clearer graph because I cannot see clearly.

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Which of these numbers is irrational?
Sidana [21]
<span>Which of these numbers is irrational?
a. square root of 3
b. 0.25
c. 1/5
d. square root of 9

answer
</span><span>a. square root of 3
</span><span>--------------------------------
What is (-7)^5 in expanded form?
a. -7 • -7 • -7 • -7 • -7 
b. -7 • 5
c. (-7) • 7 • (-7) • 7 • (-7) • -7
d. 5 • 5 • 5 • 5 • 5 • 5 • 5

answer

</span><span>a. -7 • -7 • -7 • -7 • -7 </span>
6 0
3 years ago
(60y – 18) = 2(8-16)​
Alekssandra [29.7K]

Answer:

3/10

Step-by-step explanation:

(60y-18)=2(8-16)

60y-18=16-16

60y-18=0

60y=18 /:60

y= 18/60=3/10

4 0
2 years ago
Sal is making bracelets for a fundraiser he uses 6 inches of yarn per bracelet how many bracelets can sal make with 5 feet of ya
natali 33 [55]
There are 12 inches in 1 foot, and each bracelet takes 6 inches
12 divided by 6 = 2
there can be 2 bracelets per foot of yarn
5 x 2 = 10 bracelets can be made
let me know if you have any further questions
:)
3 0
3 years ago
Read 2 more answers
Soda cans move along a conveyor belt at a constant rate. When each can reaches the
PilotLPTM [1.2K]

Answer:

yeah so you just do that

Step-by-step explanation:

3 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
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