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Sunny_sXe [5.5K]
2 years ago
7

Someone please help me with number 10 and 16 I’m lost I tried to figure it out on my own but couldn’t

Mathematics
1 answer:
Lelu [443]2 years ago
5 0

Answer:

10. \frac{g}{3}  - 7 = 15

g - 7 = 15 \times 3

g - 7 =4 5

g = 45  +  7

g = 52

16. - 3 +  \frac{p}{7}  =  - 5

- 3  + p =  - 5 \times 7

- 3 + p =  - 35

p =  - 35 + 3

p =  - 32

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Please anyone help me with this?!​
andriy [413]

Answer:

It would be 55.88 but if you round it it would be 55.9 then round it again it would be 56%.

7 0
3 years ago
The width of a rectangle is 3 feet less than it’s length. The perimeter of the rectangle is 98 feet. Find the dimensions of the
horsena [70]

Answer:

The length is 26 feet and the width is 23 feet

Step-by-step explanation:

Let l represent the length of the rectangle.

The width can be represented by l - 3, since it is 3 feet less than the length

Set up an equation:

l + l + (l - 3) + (l - 3) = 98

Add like terms and solve for l:

l + l + (l - 3) + (l - 3) = 98

4l - 6 = 98

4l = 104

l = 26

So, the length is 26 feet.

Since the width is l - 3, we can plug this in for l to find the width:

l - 3

26 - 3

= 23

So, the length is 26 feet and the width is 23 feet

4 0
2 years ago
A car’s stopping distance in feet is modeled by the equation d(v)=2.15vsquared over 58.4f, where v is the initial velocity of th
frutty [35]
According to your description, you can simply plug in all the numbers:

d(47) = 2.15 * 45^2 / (58.4*0.34) = 219.27 m
3 0
3 years ago
Read 2 more answers
Find the point-slope form. Then use that to find the slope-intercept form
umka2103 [35]
\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ 5 &,& 10~) 
%  (c,d)
&&(~ 8 &,& 16~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{16-10}{8-5}\implies \cfrac{6}{3}\implies 2
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-10=2(x-5)\implies y-10=2x-10
\\\\\\
y-=2x-10+10\implies y=2x
4 0
3 years ago
A horse walks around a circular track while its trainer stands in the center. The trainer is 14 feet from the horse at all times
max2010maxim [7]

The horse traveled 439.6 feet after walking around the track 5 times

<u><em>Solution:</em></u>

Given that, horse walks around a circular track while its trainer stands in the center

The trainer is 14 feet from the horse at all times

Therefore, radius of circular track = 14 feet

The circumference of circle is the distance traveled by horse for 1 lap

<em><u>The circumference of circle is given as:</u></em>

C = 2 \pi r

Where, "r" is the radius and \pi is a constant equal to 3.14

C = 2 \times 3.14 \times 14\\\\C = 87.92

Thus the distance traveled by horse for one time in circular track is 87.92 feet

<em><u>About how far had the horse traveled after walking around the track 5 times? </u></em>

Multiply the circumference by 5

distance = 5 \times 87.92\\\\distance = 439.6

Thus the horse traveled 439.6 feet after walking around the track 5 times

5 0
3 years ago
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