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levacccp [35]
3 years ago
5

What is the answer to this ?

Mathematics
1 answer:
harina [27]3 years ago
5 0

Answer:

The answer is option 1.

Step-by-step explanation:

In order to solve x, you have to make x the subject, by multiplying each sides by 5, to get rid of 5 on the left side.

\frac{x}{5}  = 8

\frac{x}{5}  \times 5 = 8 \times 5

x = 40

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A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along
mina [271]

Answer:

The tip of the man shadow moves at the rate of \frac{20}{3} ft.sec

Step-by-step explanation:

Let's draw a figure that describes the given situation.

Let "x" be the distance between the man and the pole and "y" be distance between the pole and man's shadows tip point.

Here it forms two similar triangles.

Let's find the distance "y" using proportion.

From the figure, we can form a proportion.

\frac{y - x}{y} = \frac{6}{15}

Cross multiplying, we get

15(y -x) = 6y

15y - 15x = 6y

15y - 6y = 15x

9y = 15x

y = \frac{15x}{9\\} y = \frac{5x}{3}

We need to find rate of change of the shadow. So we need to differentiate y with respect to the time (t).

\frac{dy}{t} = \frac{5}{3} \frac{dx}{dt} ----(1)

We are given \frac{dx}{dt} = 4 ft/sec. Plug in the equation (1), we get

\frac{dy}{dt} = \frac{5}{3} *4 ft/sect\\= \frac{20}{3} ft/sec

Here the distance between the man and the pole 45 ft does not need because we asked to find the how fast the shadow of the man moves.

7 0
3 years ago
You may remember that the perimeter of a rectangle is P=2(W+L) where W is the width and L is the length.
s344n2d4d5 [400]

Step-by-step explanation:

So, x=width and the width=13+L. The formula for perimeter is 2(L)+2(W) and the total perimeter in this question is 62 feet. That means 2(L)+2(13+L)=62 feet. Solve for L and that's going to be your length. Add the length plus 13 and you'll have the width. Double check the equation by plugging the width and length back into the equation. (Does that help?)

6 0
2 years ago
Carrie has 2 meters of ribbon. She cuts off pieces of ribbon that are 5/10 meter, 1/10 meter, and 7/10 meter. How long is the re
vekshin1

Answer:

\frac{7}{10}

Step-by-step explanation:

Carrie has 2 meters of ribbon. She cuts off pieces of ribbon that are 5/10 meter, 1/10 meter, and 7/10 meter

Lets add all the cut of pieces and subtract it from 2 meters

\frac{5}{10} +\frac{1}{10}+\frac{7}{10}=\frac{13}{10}

Now we subtract 13/10 from 2 meters

2 - \frac{13}{10}

To subtract , make the denominator same

\frac{2*10}{1*10} - \frac{13}{10}=\frac{20}{10} - \frac{13}{10}=\frac{7}{10}

7/10 meter is the remaining piece of ribbon

8 0
3 years ago
Which reflection rule, if any, can be used to prove that rectangle A(-8, -3), B(-2,-3), C (-2,-6), D(-8,-6) and rectangle A'(8.-
Zina [86]

Answer:

d

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Plz help!! A jar contained 6 red marbles, 10 green marbles, and 8 blue marbles. A marble was selected at random, the color was r
Inessa [10]

Answer:

To calculate the relative frequency, first we need to know what exactly is and how to calculate it.

Relative frequency is the ratio between the absolute frequency (how many repetitions have a specific outcome) and the total outcomes. Also, this type of frequency is used to show the representation that some data have over the whole distribution.

So, in this case, we need to just divide 13, which belongs to red marble's results, to 60 which is the total outcomes, as it's presented:

         13redmarble                                                                                                                       Fr = -------------------------

       60 totalmarbles  

Normally, relative frequency is shown as a percentage multiplying this result by 100. Therefore, 22% is the approximate percentage of the relative frequency, which means that 22% is the representation of red marbles outcomes of the whole distribution, or we can say it as a probability: there's 22% of chances when someone extract a marble, it will be red.

3 0
2 years ago
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