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2 years ago

monique mad 5/8 pound of trail mix and divided the mix into 4 equal portions wha is the weight of each portion

Mathematics

1 answer:

agasfer [191]2 years ago

4 0

So you take 5/8 and multiply it by 4’s reciprocal which is 1/4 and you get 5/32 pounds for each portion

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Hamid has gained weight. He now weights 88kg, which is 10% higher than his normal weight. What is Hamids normal weight

zubka84 [21]

Let his normal weight = 100%

He is 10% more so he is 110%

Divide his weight now by 110% to find his normal weight.

88/1.1 = 80

His normal weight is 80 kg

3 0

2 years ago

Can anyone help not understanding? And can you explain how you got it. Thanks for all your help!

Allushta [10]

No, the height wouldn't be 70 feet, it would be 80 feet.

You would get this answer by using the equation h= A/b (h meaning height)
A = area ( or 20,000)
b= base ( or 250)

Then you divide A by b:
20,000/ 250 =
80

I hoped this helped! :)

8 0

3 years ago

Can you please help me out if you know this answer<br>

koban [17]

Step-by-step explanation:

A strike in bowling meant all 10 pins knocked down.

In this case, there is only 1 strike

4 0

2 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

____ [38]

Answer:

$\int_{-\infty}^0 5 e^{60x} dx = \frac{1}{12}[e^0 -0]= \frac{1}{12}$

Step-by-step explanation:

Assuming this integral:

$\int_{-\infty}^0 5 e^{60x} dx$

We can do this as the first step:

$5 \int_{-\infty}^0 e^{60x} dx$

Now we can solve the integral and we got:

$5 \frac{e^{60x}}{60} \Big|_{-\infty}^0$

$\int_{-\infty}^0 5 e^{60x} dx = \frac{e^{60x}}{12}\Big|_{-\infty}^0 = \frac{1}{12} [e^{60*0} -e^{-\infty}]$

$\int_{-\infty}^0 5 e^{60x} dx = \frac{1}{12}[e^0 -0]= \frac{1}{12}$

So then we see that the integral on this case converges amd the values is 1/12 on this case.

6 0

3 years ago

Simplify x^2-9 over x^2-3x

Alla [95]

Answer:

$\frac{x+3}{x}$ or x+3/x

Step-by-step explanation:

Using D.O.T.S (difference of two squares) formula :

$a^{2} -b^{2} = (a+b)(a-b)$

$x^{2} -9= (x+3)(x-3)$

Factorise the second bit:

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

Now write out this fraction:

$\frac{(x+3)(x-3)}{x(x-3)}$

Now we can cancel out repeated terms (x-3):

$\frac{(x+3)}{x}$
This is our final answer:

$\frac{x+3}{x}$

5 0

2 years ago

Read 2 more answers