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3 years ago

A pyramid has a height of 18 centimeters and a base with an area of 26 square centimeters. Find the volume. please Help!

Mathematics

1 answer:

juin [17]3 years ago

4 0

Answer:

156cm cubed

Step-by-step explanation:

i took the test...

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Find the value of Z. [Inscribed Angle]

jekas [21]

Answer:

87°

Step-by-step explanation:

If a quadrilateral is inscribed in a circle, then opposite angles are supplementary. That means that z + 93 = 180 and z = 87°

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3 years ago

Read 2 more answers

MARKING AS BRAINLIEST! LAST ATTEMPT! ) show ur work

Ivenika [448]

$- 3 \sqrt{180 {x}^{5}y } = - 3 \sqrt{36 {x}^{4} \times 5xy } = - 18 {x}^{2} \sqrt{5xy}$

$5 \sqrt{108xy {z}^{2} } = 5 \sqrt{36 {z}^{2} \times 3xy } = 30z \sqrt{3xy}$

$3x \sqrt{x {y}^{10} {z}^{7} } = 3x \sqrt{ {y}^{10} {z}^{6} \times xz } = 3x {y}^{5} {z}^{3} \sqrt{xz}$

$\sqrt{ {x}^{3} } = \sqrt{ {x}^{2} \times x } = x \sqrt{x}$

$\sqrt{x {y}^{3}z } = y \sqrt{xyz}$

$\sqrt{18 {x}^{2}y } = \sqrt{9 {x}^{2} \times 2y } = 3x \sqrt{2y}$

$\sqrt{32 {x}^{4} {y}^{2} } = \sqrt{16 ({ {x}^{2} })^{2} {y}^{2} \times 2 } = 4 {x}^{2} y \sqrt{2}$

$2 \sqrt{16 {x}^{7} {y}^{3} {z}^{4} } = 2 \sqrt{ {4}^{2} {x}^{6} {y}^{2} ( { {z}^{2} )}^{2} \times xy } = 8 {x}^{3} y {z}^{2} \sqrt{xy}$

${x}^{2} \sqrt{ {x}^{3} {y}^{2} } = {x}^{2} \sqrt{ {x}^{2} {y}^{2} \times x } = {x}^{3} y \sqrt{x}$

Answer: ......

Ok done. Thank to me :>

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2 years ago

What is the length of JH ?

fredd [130]

Answer: 34

Step-by-step explanation:

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2 years ago

Jim reads 1 page for every 3 pages that randy reads if randy read 6 then how many does jim read?

Ksivusya [100]

Answer:

Jim reads 2 pages

Step-by-step explanation:

Jim reads 1 page for every 3 pages that randy reads

3 + 3 = 6

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3 years ago

Find the coefficient of x^4 in the expansion of (1 + 2x + 3x^3)^10?

Lesechka [4]

$(1+2x+3x^3)^{10}$

The only possible configurations of terms that contribute to generating the $x^4$ term in the expansion are given by the multinomial theorem to be

$\left(\dbinom{10}{6,4,0}(1)^6(2x)^4(3x^3)^0+\dbinom{10}{8,1,1}(1)^8(2x)^1(3x^3)^1\right)x^4$

where

$\dbinom{10}{k_1,\ldots,k_m}=\dfrac{10!}{k_1!\cdots k_m!}$

Simplifying a bit gives a coefficient of

$\dbinom{10}{6,4,0}(2)^4+\dbinom{10}{8,1,1}(2)(3)$
$16\dbinom{10}{6,4,0}+6\dbinom{10}{8,1,1}$
$16\dfrac{10!}{6!4!0!}+6\dfrac{10!}{8!1!1!}$
$16(5\times3\times2\times7)+6(10\times9)=3900$

3 0

3 years ago