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3 years ago

Please help me and my friend need the answer

Mathematics

1 answer:

Iteru [2.4K]3 years ago

6 0

Answer:

20m-60 = 280

Step-by-step explanation:

1. Start off defineing a variable, let’s call it m for months.

2. We know that she saves 20$ every month, so we can just divide the total to find the number of months. But, she withdrew $60. We can add those back to get the total she saved. So we have

280 + 60 = 340

Then we can divide by 20:

340/20 = 17.

So the equation would be

20m - 60 = 280.

I hope this helps!

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Solve x2 – 3x - 108 = 0.

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X=12, -9 I hope this somewhat helps you :)

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3 years ago

There were 60 runners to start the race. In the first half of the race,1/3 of them dropped out. In the second half of the race,1

Verizon [17]

In the question, it is already given that the total number of runners in the race is 60 and out of them 1/3 dropped out in the first half. In the second half 1/4 of the remaining runners dropped out.
Now
Total number of runners in the race = 60
Number of runners that dropped out in the first half = 1/3 * 60
   = 20
Number of runners remaining = 60 - 20
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Number of runners dropping out in the second half = 40 * 1/4
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Then the number of runners that finished the race = 40 - 10
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2 years ago

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3 years ago

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The five circles making up this archery target have diameters of length $2,4,6,8,$ and $10$. What is the total red area?

Sliva [168]

Answer:

$A=15\pi\ units^2$

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

we know that

It is given that the diameter of 5 circles making up the archery is 2,4,6,8, and 10.

To determine the total red area, we use the formula for area of the circle

$A=\pi r^{2}$

step 1

Find the Area of the 1st red circle

$r=2/2=1\ unit$ ---> the radius is half the diameter

$A_1=\pi (1)^{2}=\pi\ units^2$

step 2

Find the Area of the 2nd white circle

$r=4/2=2\ units$ ---> the radius is half the diameter

$A_2=\pi (2)^{2}=4\pi\ units^2$

step 3

Find the Area of the 3rd red circle

$r=6/2=3\ units$ ---> the radius is half the diameter

$A_3=\pi (3)^{2}=9\pi\ units^2$

step 4

Find the Area of the 4th white circle

$r=8/2=4\ units$ ---> the radius is half the diameter

$A_4=\pi (4)^{2}=16\pi\ units^2$

step 5

Find the Area of the 5th red circle

$r=10/2=5\ units$ ---> the radius is half the diameter

$A_5=\pi (5)^{2}=25\pi\ units^2$

The total red area is given by

$A=A_5-A_4+A_3-A_2+A_1$

substitute

$A=25\pi-16\pi+9\pi-4\pi+\pi$

$A=15\pi\ units^2$

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3 years ago

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$