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2 years ago

What is the volume of this composite solid? 16 m 5 m 12 m 9 m

Mathematics

1 answer:

son4ous [18]2 years ago

3 0

Answer:

864 cubic meters

Step-by-step explanation:

Rectangular prism = 5x9x12=540

Triangular prism = 1/2(9)(6)(12)=324

540+324=864 cubic meters

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A: What is 7*<img src="https://tex.z-dn.net/?f=10%5E%7B15%7D" id="TexFormula1" title="10^{15}" alt="10^{15}" align="absmiddl

pickupchik [31]

Step-by-step explanation:

7*10^10=70,000,000

(70milion)

7*10^15=7000,000,000,000,000

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2 years ago

Which expression shows the correct way to find the distance between (23,17) and (43,14)? ( TTM)

Lesechka [4]

Answer:

d= $\sqrt{x} (43-23)^2+(14-17)^2$

Answer= 20.224

Step-by-step explanation:

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2 years ago

Read 2 more answers

What is the area of this parallelogram? 104 m² 88 m² 56 m² 32 m²

iragen [17]

Answer: To find area of a rectangle, multiply its base (length) by its height (width). Plug in the base, 7 meters, and the height, 8 meters. So the area of Rectangle DFBE is 56 square meters.

88m

HOPE THIS HELPS

6 0

3 years ago

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

3 0

3 years ago

ILL MARK U BRAINLIEST!!

dmitriy555 [2]

Answer:

Kidneys,nephrons

Step-by-step explanation:

6 0

2 years ago