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2 years ago

The wind blew away 26 of y out r muffins. That was 2/3 of all of them! With how many did you start?

Mathematics

1 answer:

lorasvet [3.4K]2 years ago

7 0

26 / 2/3 = 39

you started with 39 muffins

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37% of what number is 99.9?

siniylev [52]

Answer:

270

Step-by-step explanation:

37 % of a number is the same as divide by 100 and multiply by 37. To get this number we do the opposite: divide 99.9 by 37 which gives us 27 and then multiply by 100 which will give us 270.

if we check this: 270 /100 *37 = 99,9

3 0

3 years ago

Work out the percentage change to 2 decimal places when a price of £57.99 is decreased to £49.99.

IRISSAK [1]

Answer:

Percentage change in price = 13.80%

Step-by-step explanation:

Given that:

Original price = £57.99

Decreased price = £49.99

Difference = Original price - Decreased price

Difference = 57.99 - 49.99

Difference = £8.00

Percent change = $\frac{Difference}{Original\ price}100$

Percent change = $\frac{8.00}{57.99}*100$

Percent change in price = 13.80%

Hence,

Percentage change = 13.80%

5 0

3 years ago

Nicholas wrote the steps below to simplify the fraction 20/30 find his error and correct it

Dmitrij [34]

Unfortunately, Nicholas' steps in simplifying the fraction is not reflected in the question. However, the general idea of the simplifying fraction is to find the greatest common factor (GCF) of both the numerator and the denominator. In this case, the GCF is 10. Then, divide both numbers by the GCF. The answer for this item should be 2/3.

7 0

3 years ago

Read 2 more answers

PLEASE help I’ll give brainlest

dimulka [17.4K]

Answer:

the perimeter is 87

Step-by-step explanation:

6 0

2 years ago

Find the average value of f(x+y)=sin(x+y) over

Troyanec [42]

For each given region $D$ , the average value of $f$ over $D$ is the integral of $f$ over $D$ divided by the area of $D$ . In both cases, $D$ is a rectangle so the area is trivial.

a. The area of $D$ is $\dfrac\pi3\cdot\dfrac{5\pi}3=\dfrac{5\pi^2}9$ . The integral of $f$ over $D$ is

$\displaystyle\int_0^{5\pi/3}\int_0^{\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$\displaystyle=\int_0^{5\pi/3}\left(\cos y-\cos\left(y+\dfrac\pi3\right)\right)\,\mathrm dy$

$=\sin\dfrac{5\pi}3-\sin\left(\dfrac{5\pi}3+\dfrac\pi3\right) - \sin0+\sin\dfrac\pi3=0$

So the average value of $f$ over this region is 0.

b. The area of $D$ is $\dfrac{2\pi}3\cdot\dfrac{7\pi}6=\dfrac{7\pi^2}9$ . The integral of $f$ is

$\displaystyle\int_0^{7\pi/6}\int_0^{2\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^{7\pi/6}\left(\cos y-\cos\left(y+\frac{2\pi}3\right)\right)\,\mathrm dy$

$=\sin\dfrac{7\pi}6-\sin\left(\dfrac{7\pi}6+\dfrac{2\pi}3\right) - \sin0+\sin\dfrac{2\pi}3=\dfrac{\sqrt3}2$

The average vale of $f$ over this region is then $\dfrac{\frac{\sqrt3}2}{\frac{7\pi^2}9}=\dfrac{9\sqrt3}{14\pi^2}$ .

4 0

2 years ago