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Paul [167]
3 years ago
8

Can you guys help me pleaseeeee!

Mathematics
1 answer:
Vladimir79 [104]3 years ago
3 0

Answer: A) True

Step-by-step explanation:

GF and KJ are marked with one dash indicating the are congruent. Same for FH and JL but this time it's marked with 2 dashes. Angles F and J are congruent because they are both right angles. So, we have a side congruent an angle congruent and a side congruent. We can use SAS to say they are congruent.

(Brainliest?)

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Can someone help me please
Valentin [98]

Answer:

It is B (5x10^11)

Step-by-step explanation:

hope this helps! pls award/stars/heart

3 0
3 years ago
T= u+v/2 for u<br><br> •u= 2t+v<br> •u= 2t/v<br> •u= 2(t-v)<br> •u= 2t-v
ratelena [41]

Answer:

u=2t-v

Step-by-step explanation:

t=(u+v)÷2 |×2

2t=u+v

-u=v-2t |:(-1)

u=-v+2t

u=2t-v

3 0
3 years ago
Read 2 more answers
What is the value of y?<br><br> Enter your answer, as an exact value, in the box.
OLga [1]
We know the bottom triangle is a 45, 45, 90 triangle, so the hypotenuse is √2 times the value of the legs:

(√2)(√2)
=√4
=2

Now, we can use this to solve for y. The top triangle is a 30, 60, 90 triangle. The side we found above is the side across from the 30 degree angle. The side opposite the 60 degree angle is √3 times the side across the 30 degree angle. Therefore, we can solve for y by multiplying 2 by √3

y=2√3
7 0
4 years ago
The volume of water and a rectangle swimming pool can be modeled by the function v(x)=x^3+13x-210. If the depth of the pool is g
S_A_V [24]

Answer:

Required,

L=\frac{3+\sqrt{37}}{2}-\frac{144}{x-3}

W=\frac{3-\sqrt{37}}{2}-\frac{144}{x-3}

Step-by-step explanation:

Given volume and depth respectively,

V(x)=x^3+13x-210 and x-3

To find length and width of the rectanglular swiming pool we know,

Volume=length\timesheight\timesdepth.

Let depth=D=x-3, length=L, width=W, then

V=DLW

x^3+13x-210= LW(x-3)

LW=\frac{x^3+13x-210}{x-3}

After divide we will get x^2-3x-22 with remainder -144.

Thus,

x^3+13x-210=(x^2-3x-22)(x-3)-144=(x-3)LW

Now to find root of,

x^2-3x-144=\frac{3\pm\sqrt{9+88}}{2}=\frac{3\pm \sqrt{37}}{2}

Thus,

L=\frac{3+\sqrt{37}}{2}-\frac{144}{x-3}

W=\frac{3-\sqrt{37}}{2}-\frac{144}{x-3}

5 0
3 years ago
X2+11x-26=0<br> X2 - 25 = 0
Oksanka [162]
X² + 11x - 26 = 0
x² - 25 = 0

(x + 13)(x - 2) = 0
(x + 5)(x - 5) = 0

x = -13
x = 2
x = -5
x = 5
8 0
3 years ago
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