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3 years ago

Need help on factoring this

Mathematics

1 answer:

vovangra [49]3 years ago

3 0

First find what a times c is, then find the factors of that number that add up to the value of b. Then expand the original equation. After that, you will create two binomials by factoring out a term.

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Find the diameter of a cone that had a volume of 56.52 cubic inches and a height of 6 inches

bulgar [2K]

Answer: 5.98 inches

Step-by-step explanation:

1. You have that the formula for calculate the volume of a cone is:

$V=\frac{r^{2}h\pi}{3}$

Where V is the volume, r is the radius and h is the height.

2. By definition, the diameter is twice the radius. Therefore, you need to solve for the radius:

$V=\frac{r^{2}h\pi}{3} \\3V=r^{2}h\pi\\r=\sqrt{\frac{3V}{h\pi}}$

3. Substitute values:

$r=\sqrt{\frac{3(56.52in^{3})}{(6in)\pi}}=2.99in$

4. The diameter is:

$D=2r\\D=2(2.99in)\\D=5.98in$

5 0

3 years ago

Read 2 more answers

I need help with this

olga2289 [7]

Answer:

x = 24.52

Step-by-step explanation:

Since this is a right triangle, use Pythagorean Theorem to solve for the hypotenuse.

Pythagorean Theorem: a² + b² = c²

24² + 5² = x²

576 + 25 = x²

601 = x²

√601 = x

4 0

3 years ago

Coco's garden has 888 tomato plants, 444 pepper plants, 333 squash plants, and 666 pumpkin plants.

raketka [301]

8 + 4 + 3 + 6 = 21There are 21 plants altogether.21 is 3 times 7.For every 21 plants there are 6 pumpkin plants, sofor every 7 plants, there are 2 pumpkin plants.

7 0

4 years ago

If the solution to an equation is x = -3, what could the original equation be?

seropon [69]

Answer: on edge c. X + 8=5

Step-by-step explanation:

8 0

4 years ago

If a, b, c are in A.P. show that<br>a (b + c)/bc,b(c + a) /ca, c(a-b )/bc<br>are in A.P.<br>

vfiekz [6]

Answer:

Step-by-step explanation:

$\frac{a(b+c)}{bc} ,\frac{b(c+a)}{ca} ,\frac{c(a+b)}{ab} ~are~in~A.P.\\if~\frac{ab+ca}{bc} ,\frac{bc+ab}{ca} ,\frac{ca+bc}{ab} ~are~in~A.P.\\add~1~to~each~term\\if~\frac{ab+ca}{bc} +1,\frac{bc+ab}{ca} +1,\frac{ca+bc}{ab} +1~are~in~A.P.\\if~\frac{ab+ca+bc}{bc} ,\frac{bc+ab+ca}{ca} ,\frac{ca+bc+ab\\}{ab} ~are~in~A.P.\\\\divide~each~by~ab+bc+ca\\if~\frac{1}{bc} ,\frac{1}{ca} ,\frac{1}{ab} ~are ~in~A.P.\\if~\frac{a}{abc} ,\frac{b}{abc} ,\frac{c}{abc} ~are~in~A.P.\\if~a,b,c~are~in~A.P.\\which~is~true.$

3 0

3 years ago