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Vesnalui [34]
3 years ago
13

Can someone help me with this one

Mathematics
1 answer:
rewona [7]3 years ago
5 0
4 over 5 or y=4/5x+0
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Explain how to do this problem in steps and include answers as well
vlabodo [156]
8(4x - 5)-12=20x-10

32x - 40-12=20x-10

32x -52=20x-10

32x -52 -20x=-10

12x -52=-10

12x =-10 +52

12x =42

x= \frac{42}{12}

x =  \frac{7}{2}

x=3 \frac{1}{2}  ≈ 3.5
5 0
4 years ago
Read 2 more answers
A store sold 655 bicycles last year. This year the store sold 582 bicycles. What is the percent of change in the number of bicyc
Triss [41]

Answer:

12.5%

Step-by-step explanation:

Given data

Number of bicycles sold last year= 655

Number of bicycles sold this year= 582

%percent change= Iniitial -FInal/initial *100

%percent change= 655-582/582*100

%percent change= 73/582 *100

%percent change=0.125 *100

%percent change=12.5%

Hence the percent change is 12.5%

7 0
3 years ago
-7 x - 2y = -13 <br> x - 2y = 11
VashaNatasha [74]
Well the first one would convert to
-2y= 7x-13 and the second one would be -2y=-x+11 and you solve from there (lmk if you need the steps) but your final answer would be. Y= -7/2x+ 13/2 and y=1/2x - 11/2
8 0
3 years ago
Please help with my geometry
nasty-shy [4]

Answer:

45

Step-by-step explanation:

90+45+45=180,k

8 0
3 years ago
Read 2 more answers
A rock is dropped from a height of 100 feet. Calculate the time between when the rock was dropped and when it landed. If we choo
andre [41]

The time it takes between when the rock was dropped and when it landed is 1.33 seconds.

<h3>What is the time it takes for the rock to land?</h3>

The time it takes the rock to land on the ground from where it is dropped can be determined by using the function given.

The height of the function when it hits the ground will be zero, so the function can be written as:

\mathbf{h(t) =36t^2 -64}

\mathbf{0 =36t^2 -64}

64 = 36t²

\mathbf{t^2 = \dfrac{64}{36}}

t² =1.778

t = \sqrt{1.778}

t = 1.333 seconds

Learn more about calculating the time of a function here:

brainly.com/question/26898697

#SPJ1

6 0
2 years ago
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