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olga nikolaevna [1]
3 years ago
13

Parallelogram DEFG is graphed on the coordinate plane. It has coordinates D(4, 1), E(10, -3), F(p, -4), and G(-5, q).

Mathematics
1 answer:
labwork [276]3 years ago
5 0
It would be E im pretty sure be you simplify the p and q
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garri49 [273]

Hey there!!

Simple equation :

mᵃ × mᵇ = m ᵃ⁺ᵇ

We have :

5⁻⁷ * 5¹² * 5⁻²

... 5⁻⁷⁺¹²⁻²

... 5¹²⁻⁹

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Hope my answer helps!!

5 0
3 years ago
I WILL CHOOSE THE BEST ANSWER PLSS HELP ME
Elden [556K]

Answer:

-3, -1, 0, +2

Step-by-step explanation:

8 0
2 years ago
What is the simplified form of (2 X 102)2 in standard notation? Do not use commas in your answer. (HAS TO BE A NUMBER)
Oxana [17]

Answer:

40000

Step-by-step explanation:

(2 \times  {10}^{2} )^{2}  \\  =  {2}^{2}  \times  {10}^{2 \times 2} \\  = 4 \times  {10}^{4}  \\  = 40000

7 0
4 years ago
Name the property<br> cx1=c<br><br> 83+(52+17)=(83+52)+17<br> 22+b+18=b+22+18
jek_recluse [69]

Answer:

distributive property :)

6 0
3 years ago
For the real-valued functions f(x)=2x+1 and g(x)=sqrt(x-1), find the composition f o g and specify its domain using interval not
Anettt [7]

Answer:

fog = 2√(x-1) + 1

Domain = [1, \infty)

Step-by-step explanation:

Given the functions  f(x)=2x+1 and g(x)=sqrt(x-1), we are to find the composite function fog

fog = f(g(x))

f(g(x)) = f(√(x-1))

f(√(x+1)) means that we are to replace variable x in f(x) with the function √(x-1)

f(√(x-1)) = 2(√(x-1))+1

f(√(x+1)) = 2√(x-1) + 1

fog = 2√(x-1) + 1

<em>For the function to exist on any real valued function, then the function inside square root i.e x-1 must be greater than or equal to zero (x-1≥0)</em>

If x-1≥0

x≥0+1

x≥1

This means the range of variable x must be values of x greater than or equal to 1.

Domain = [1, \infty)

8 0
3 years ago
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