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mariarad [96]
3 years ago
11

What is the ratio of 140:160

Mathematics
1 answer:
levacccp [35]3 years ago
4 0

Answer:

it reduces to 7:8

Step-by-step explanation:

just put it like a fraction like 140/160. it simplifies to 7/8

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The measure of L in triangle LMN is 33°. The measure of M is 79°. What is the measure of N?
FromTheMoon [43]

Answer:

<h2>N = 68°</h2>

Step-by-step explanation:

Since ∆LMM is a triangle all it's interior angles sum up to 180°

To find M add up all the angles and N and equate it to 180° to find N

That's

L + M + N = 180

33 + N + 79 = 180

N + 112 = 180

N = 180 - 112

We have the final answer as

<h3>N = 68°</h3>

Hope this helps you

3 0
4 years ago
There are 128 Year 7 students. Three quarters of them own a pet. A survey shows that 37 students have cats, 1/3 of the students
VikaD [51]
Hmmm, I think it’s 17. Three quarters of 128 is 96. 96 minus 37 is 59. And then 1/3 of 96 is 32, so 59 minus 32 is 17. Then there remains 17 students! So 17 students have horses. I hope this helped!
6 0
3 years ago
What is the vaule of 12(3x+5x),when x=7
german
Fist plug in 7 in to the X values like so 12 (3 (7) +5 (7)) then solve in side the parenthesis 3×7 =21 and 5×7=35 then add them 35+21=56 and then you are left with 12(56) or 12×56= 672

8 0
3 years ago
If you can mow 1000 square feet per minute, how long to mow a circle with diameter of 800 feet?
maria [59]

Answer:

You can calculate the area of the circle

We know the formula is: \pi*r^2

And if diammeter is 800, then the radio is a half: 400

We replace:

\pi*r^2 --> \pi * 400^2 \approx 502654.8  feet

So now we can divide:

502654/1000 = 502,654 minutes

3 0
3 years ago
Find the deriative dy/dx for y=x^2-2x/x^3+3
vaieri [72.5K]

Answer:

\frac{dy}{dx}=(\frac{(2x-2)(x^3+3)-(x^2-2x)(3x^2)}{(x^3+3)^2})

Step-by-step explanation:

So we want to find the derivative of the rational equation:

y=\frac{x^2-2x}{x^3+3}

First, recall the quotient rule:

(\frac{f}{g})'=\frac{f'g-fg'}{g^2}

Let f be x^2-2x and let g be x^3+3.

Calculate the derivatives of each:

f=x^2-2x\\f'=2x-2

g=x^3+3\\g=3x^2

So:

\frac{dy}{dx}=(\frac{x^2-2x}{x^3+3})'

Use the above format:

\frac{dy}{dx}=\frac{f'g-fg'}{g^2}\\\frac{dy}{dx}=(\frac{(2x-2)(x^3+3)-(x^2-2x)(3x^2)}{(x^3+3)^2})

And that's our answer :)

(If you want to, you can also expand. However, no terms will be canceled.)

8 0
3 years ago
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