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astraxan [27]
3 years ago
10

HELP WILL GIVE BRAINIEST TO BEST ANSWER I NEED HELP ASAP

Mathematics
1 answer:
Ad libitum [116K]3 years ago
6 0
Wait lemme get my worksheet I just had this same one
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A right triangle is formed by joining three
MatroZZZ [7]

Answer:

D:  x^{2} = 576 + 7^{2}

Step-by-step explanation:

Well you want to find what x is so you need the square root

We already have the area for one of the squares (576) which can help us find what x is.

And the bottom small square is 7 and since it's a sqaure all the sides are congruent/ same. Which means u get 7^{2}.

And you need to add them all up to get what x is

and is x^{2} since both sides of that triangle will have the same lengths, and when u have both of those lengths its easier to find the area

Hope this helped!

4 0
4 years ago
What are these expressions simplified
Serjik [45]

Answer:

1) 14i+11t

2) 11x+8y

3) -7a+2c

Step-by-step explanation:

7 0
4 years ago
Read 2 more answers
Find the value of the expression (x2−4)3x when x = 5.
Degger [83]

Answer:

315

Step-by-step explanation:

You plug 5 in for x

(5^2-4)3(5) = 315

7 0
2 years ago
Read 2 more answers
What is the graph to the rational function?
Aneli [31]

We are given

y=\frac{x^2-4x+3}{x^2-9}

Vertical asymptotes:

Firstly, we will factor numerator and denominator

we get

y=\frac{(x-1)(x-3)}{(x-3)(x+3)}

We can see that (x-3) is common in both numerator and denominator

so, we will only set x+3 to 0

and then we can find vertical asymptote

x+3=0

x=-3

Hole:

We can see that (x-3) is common in both numerator and denominator

so, hole will be at x-3=0

x-3=0

x=3

Horizontal asymptote:

We can see that degree of numerator is 2

degree of denominator is also 2

for finding horizontal asymptote, we find ratio of leading coefficients of numerator and denominator

and we get

y=\frac{1}{1}

y=1

so, option-D............Answer

7 0
3 years ago
Solve each equation. Show your work please. Part 3​
irina1246 [14]

Answer:

Please see the attached pictures for the full solution.

4 0
3 years ago
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