What is the value of x in the equation (3/4 + 5/8) x = 3/5

What is 5 1/6 divided by (2 2/3 multiplied by 1 3/8)

Karo-lina-s [1.5K]

So first you need your denominators to be the same so you can multiply. They share a common denominator of 24. 2/3 equals 16/24 and 2/8 equals 9/24. So now you need to turn your fractions into mixed numbers and multiply. Which equals 2112/576. That # is way to big so then you simplify and get 11/3. so now you have 5 and 1/6 divided by 11/3. So now just flip the second fraction multiply it by its reciprocal. 31/6 X 3/11=___. So your final answer is 1 and 9/22

3 0

3 years ago

Delia can buy 3 identical sweatshirts for a total of $30. How much would it cost if she were to buy 4 of the same sweatshirts at

emmainna [20.7K]

$30/ 3 sweatshirts = $10 /shirt

$ 10/shirt * 4 shirts = $40

It would cost $40

7 0

3 years ago

Round 0.79 to the nearest ones

Sauron [17]

Answer:

1

Step-by-step explanation:

0.79 is more than half, so rounds to 1.

4 0

3 years ago

Please help me with this question

7nadin3 [17]

9514 1404 393

Answer:

10s place

Step-by-step explanation:

When you line up the three digits of the divisor with the three most significant digits of the dividend, you see that the least significant digit of the divisor lines up with the dividend digit that is in the 10s place.

The first (most-significant) quotient digit will be in the 10s place.

_____

In the attached, the left-most non-zero quotient digit is 4, in the tens place.

3 0

3 years ago

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

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