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faltersainse [42]
3 years ago
7

Help me please everyone always passes by and doesn’t help me :( help me in the way u can please

Mathematics
1 answer:
tiny-mole [99]3 years ago
6 0

Answer:

C': (0,3)

D': (Stays the Same)

E': (-1,6)

Step-by-step explanation:

When reflecting across x = 2, you would draw a horizontal line on the x-axis on the number 2. From there, you reflect over the horizontal line. (It helps if you count the original distance from the horizontal line and then count the same amount on the other side of AWAY from the line!)

I hope this helps! I'm really not good at explaining...Good Luck!

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Camille put $1,250 in a new account at her bank. - The bank pays 2.25% interest compounded annually on this account. - Camille m
Galina-37 [17]

Answer:

FV= $1,397.1

Step-by-step explanation:

Giving the following information:

Initial investment (PV)= $1,250

Interest rate (i)= 2.25% compounded annually

Number of periods (n)= 5 years

<u>To calculate the future value, we need to use the following formula:</u>

FV= PV*(1+i)^n

FV= 1,250*(1.0225^5)

FV= $1,397.1

5 0
3 years ago
I need help with this geometry question
Alex17521 [72]

Answer:

x=9; JL= 23

Step-by-step explanation:

To find x:

If JL and KM are equivalent, then JL = KM. Substitute the equations and you get 2x+5 = 7x-40.

Put the similar terms on the same side by subtracting 2x from both sides and adding 40 to both sides. You will get 45=5x.

Get x alone by dividing both sides by 5 and you will get 9=x.

To find JL:

Substitute the x in 2x+5 (JL length equation) with 9. Your new equation will be 2(9) + 5.

2•9 = 18.

18+5=23

8 0
3 years ago
Which function does not have a vertical asymptote? A) y=(x) /(1-x²) . B) y=(5x) /(1-2x²) . C) y=(5x-1) /(3+x^2) . D) (5x) /(x+x²
uysha [10]

A function has a vertical asymptote x=a at point a, where the denominator becomes equal to 0.

A. The denominator of the function f(x)=\dfrac{x}{1-x^2} turns into 0 at x=1 or x=-1. Then x=1 and x=-1 are two vertical asymptotes of this function.

B. The denominator of the function f(x)=\dfrac{5x}{1-2x^2} turns into 0 at x=\sqrt{\frac{1}{2}} or x=-\sqrt{\frac{1}{2}}  Then x=\sqrt{\frac{1}{2}}  and x=-\sqrt{\frac{1}{2}}  are two vertical asymptotes of this function.

C. The denominator of the function f(x)=\dfrac{5x-1}{3+x^2} never turns into 0, then this function hasn't any asymptotes.

D. The denominator of the function f(x)=\dfrac{x}{x+x^2} turns into 0 at x=0. Then x=0 is vertical asymptote of this function.

Answer: correct choice is C.

5 0
3 years ago
A man and a woman share a prize
VladimirAG [237]

Step-by-step explanation:

What type of line the picture represent

3 0
3 years ago
Find the two values that x can have if x2 – 12 = 37​
Sauron [17]

Answer:

x1= -7 , x2 = 7

Step-by-step explanation:

x^2 = 37 +12

x^2 = 49

x = -7 or 7

4 0
4 years ago
Read 2 more answers
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