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inna [77]
3 years ago
7

3/4 cup of sugar is needed to make six muffins. How much sugar is in each muffin?

Mathematics
1 answer:
tia_tia [17]3 years ago
4 0
1/8 cup of sugar in each mufin
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A park has a large circle painted in the middle of the playground area. The circle is divided into 444 equal sections, and each
balu736 [363]

Answer:

The area of each section of the circle is A=25\pi\ m^{2}

Step-by-step explanation:

we know that

Each section represent a quarter of circle

The area of a quarter of circle is equal to

A=\frac{1}{4}\pi r^{2}

we have

r=10\ m

substitute

A=\frac{1}{4}\pi (10)^{2}

A=25\pi\ m^{2}

7 0
3 years ago
What is 14/4y + 4/2=3y-3
OverLord2011 [107]

Answer:

-10

Step-by-step explanation:

Solve for y by simplifying both sides of the equation, then isolating the variable

4 0
3 years ago
Which value of x from the set {5,6,7,8} makes this equation true?<br><br> 3x = 24
N76 [4]
The answer is 8 because when you replace x by 8 it equals 24
3 0
3 years ago
Write a story problem to go with the multiplication problem 3 x 7/8
Mariulka [41]

Answer:

Elias Nebiyu gives seven breads for his eight friends and he continues this give away for three days.

Step-by-step explanation:

3x 7/8 result is attached above.

4 0
2 years ago
Brandon is on one side of a river that is 50 m wide and wants to reach a point 300 m downstream on the opposite side as quickly
AlexFokin [52]
Let P be Brandon's starting point and Q be the point directly across the river from P. 
<span>Now let R be the point where Brandon swims to on the opposite shore, and let </span>
<span>QR = x. Then he will swim a distance of sqrt(50^2 + x^2) meters and then run </span>
<span>a distance of (300 - x) meters. Since time = distance/speed, the time of travel T is </span>

<span>T = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x). Now differentiate with respect to x: </span>

<span>dT/dx = (1/4)*(2500 + x^2)^(-1/2) *(2x) - (1/6). Now to find the critical points set </span>
<span>dT/dx = 0, which will be the case when </span>

<span>(x/2) / sqrt(2500 + x^2) = 1/6 ----> </span>

<span>3x = sqrt(2500 + x^2) ----> </span>

<span>9x^2 = 2500 + x^2 ----> 8x^2 = 2500 ---> x^2 = 625/2 ---> x = (25/2)*sqrt(2) m, </span>

<span>which is about 17.7 m downstream from Q. </span>

<span>Now d/dx(dT/dx) = 1250/(2500 + x^2) > 0 for x = 17.7, so by the second derivative </span>
<span>test the time of travel, T, is minimized at x = (25/2)*sqrt(2) m. So to find the </span>
<span>minimum travel time just plug this value of x into to equation for T: </span>

<span>T(x) = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x) ----> </span>

<span>T((25/2)*sqrt(2)) = (1/2)*(sqrt(2500 + (625/2)) + (1/6)*(300 - (25/2)*sqrt(2)) = 73.57 s.</span><span>
</span><span>
</span><span>
</span><span>
</span><span>mind blown</span>
8 0
3 years ago
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