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rosijanka [135]

4 years ago

11

A triangle has a perimeter of 28 cm. If the three sides of the triangle are n, 3n − 4, and 5n − 13, what is the length of each s

ide?
Separate multiple entries with a comma.

1 answer:

jeka57 [31]4 years ago

8 0

28=9n-17 45=9n 5=n One side is 5 One side is 11 One side is 12

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Help im stuck haveing such a bad time understanding this

asambeis [7]

The answer would be 12 to the -10 because with multiplication you you add the -4 and -6

6 0

4 years ago

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Lindsay got her haircut the cost of a haircut for $16 if she wants to tip for hairstylist 20% how much does she leave for the ti

gtnhenbr [62]

Answer:

$3.20

Step-by-step explanation:

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5 0

3 years ago

Which is bigger 0.125 or 0.15

meriva

0.15 is bigger, this is because the 0.05 in 0.15 has more value than 0.02 in 0.125, even though it has a third digit.

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3 years ago

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At Farmer Jones Market, bananas cost $.80 per pound. At Farmer Garcia's

e-lub [12.9K]

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4 0

3 years ago

Evaluate using <br> Definite integrals

swat32

Since $[0,4]=[0,1]\cup(1,4]$ , we can rewrite the integral as

$\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt$

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

$\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt$

Both integrals are quite immediate: you only need to use the power rule

$\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}$

to get

$\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4$

Now we only need to evaluate the antiderivatives:

$\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15$

So, the final answer is 15.

4 0

4 years ago