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sineoko [7]
3 years ago
11

Can you please help me

Mathematics
1 answer:
inessss [21]3 years ago
5 0
2)3x+2y-(2x+2)
3)3x+2y-x(2+2)
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PLEASE HELPPPPPP<br> The perimeter is 50 ft and the length is 15 ft.<br> What's the width?
SVEN [57.7K]

Answer:

10

Step-by-step explanation:

I assume it's a rectangle, therefore:

P = 50

a = 15 ft

b = ?

P = 2a + 2b

50 = 2 * 15 + 2b

50 = 30 + 2b

2b = 50 - 30

2b = 20

b = 10

8 0
3 years ago
I need to simplify this expression:<br> 4×4+4(5-2)+7
atroni [7]

Answer:

35

Step-by-step explanation:

4x4+4(5-2)+7

=4x4+4(3)+7

=16+12+7

=28+7

=35

7 0
3 years ago
Twice the sum of a number and 8 is the same aa the difference of 3 and the number. Find the equation that could be used to find
Nezavi [6.7K]

Answer:

f your number is x, then the difference between that number and 8 would be x-8. Twice that is 2(x-8)

 

Three times the sum of the number and 3 would be 3(x+3). So, you get the equation:

 

2(x-8)=3(x+3)

2x-16=3x+9

x=-25

5 0
3 years ago
MATHMATECIS HELP. HELP APPRECIATED. I NEED AN EXPLANATION STEP BY STEP PLEASE
VARVARA [1.3K]

Answer:

i think B

Step-by-step explanation:

because if we think logically...the price have to more than 10-28.All answer except B are illogical for me i think

hope its help

8 0
2 years ago
Read 2 more answers
F(t)=t^2+19t+60<br> What are the zeros of the function and what is the vertex of the parabola?
SVEN [57.7K]

Answer: -4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)

Step-by-step explanation:

Given

Function is F(t)=t^2+19+60

Zeroes of the function are

t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15

Using completing the square method

y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)

4 0
3 years ago
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