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3 years ago

14

Hurry its timed ill mark brainliest

1 answer:

Taya2010 [7]3 years ago

7 0

Answer:

The height of the pyramid is $6\ cm$

Step-by-step explanation:

we know that

The volume of the pyramid is equal to

$V=\frac{1}{3}Bh$

where B is the area of the base of the pyramid

h is the height of the pyramid

we have

$V=12\ cm^{3}$

$B=(3)(2)=6\ cm^{2}$ ----> the area of the base is the area of a rectangle

substitute the values in the formula and solve for h

$12=\frac{1}{3}(6)h$

$h=12/2=6\ cm$

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HELP HOW DO I SOLVE THE FOLLOWING? PLEASE HELP

levacccp [35]

$f(x) = \sqrt[3]{2x - 1} \\f(14) = \sqrt[3]{2(14) - 1} \\f(14) = \sqrt[3]{28 - 1} \\f(14) = \sqrt[3]{27} \\f(14) = 3$

$f(x) = \sqrt[3]{2x - 1} \\f(-62) = \sqrt[3]{2(-62) - 1} \\f(-62) = \sqrt[3]{-124 - 1} \\f(-62) = \sqrt[3]{-125} \\f(-62) = -5$

$f(x) = \sqrt[3]{2x - 1} \\f(20) = \sqrt[3]{2(20) - 1} \\f(20) = \sqrt[3]{40 - 1} \\f(20) = \sqrt[3]{39}$

4 0

3 years ago

What are the domain and range?

DaniilM [7]

Answer:

sooo i think it would be

-infinity > x > infinity

edit:

it will also be

-infinity > y > 64

i think since y ends at 64 (hightest height above the ground)

Step-by-step explanation:

umm because the graph goes forever? not really sure though since in my college the topics are a little different and they may not teach in yours

6 0

2 years ago

GUYS PLEASE HELP DUE SOONNN

jek_recluse [69]

Answer: -1 1/2 to -1: -1 1/3 and -5/4

-1 to -1/2: -2/3, -3/4, -6/7, -5/6

-1/2 to 0: -1/3 and -3/8

0 to 1/2: 1/5, 1/10, 3/12, 3/8, 3/7, 1/3

1/2 to 1: 4/5, 6/10, 7/8, 7/9, 3/4, 8/10

1 to 1 1/2: 1 5/12, 9/8

Step-by-step explanation:

4 0

3 years ago

The sum of the first two terms of an infinite GP is 6 and each term is 5 times the sum of the succeeding termsthen the second te

kozerog [31]

Answer:

$\dfrac{6}{7}$ .

Step-by-step explanation:

It is given that the sum of first two terms of an infinite GP is 6.

nth term of a GP is

$a_n=ar^{n-1}$

$a+ar=6$ ...(1)

Each term is 5 times the sum of the succeeding terms.

$a_n=5(a_{n+1}+a_{n+2}+...+\infty)$

$ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)$

$ar^{n-1}=5ar^n(1+r+r^2+...+\infty)$

Divide both sides by a.

$r^{n-1}=5r^n(\dfrac{1}{1-r})$ $[\text{Sum of infinite GP}=\dfrac{a}{1-r}]$

$\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}$

$\dfrac{1}{r}=\dfrac{5}{1-r}$

$1-r=5r$

$1=6r$

$\dfrac{1}{6}=r$

The common ratio is 1/6.

Put r=1/6 in (1).

$a+a(\dfrac{1}{6})=6$

$6a+a=36$

$7a=36$

$a=\dfrac{36}{7}$

Second term $ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}$

Therefore, the second term is $\dfrac{6}{7}$ .

4 0

3 years ago

There is 10 pounds of dog food at the dog shelter currently. Each dog bowl gets 1/4 pound of dog food. How many dogs can they fe

Neporo4naja [7]

Answer: 40

Step-by-step explanation: Multiply by the reciprocal (flip the fraction 1/4), and the new equation is 10*4, which is 40

3 0

3 years ago

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