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Dmitriy789 [7]
3 years ago
15

A glass paperweight has a composite shape: a square pyramid fitting exacty on top of an 8 centimeter cube. The pyramid has a hei

ght of 3 cm. Each triangular face has a height of 5 centimeters.
what is the volume of the paperweight?
what is the total surface area of the paperweight?​
Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
8 0

Answer:

Part 1) The volume of the paperweight is 576\ cm^{3}

Part 2) The total surface area of the paperweight is 400\ cm^{2}

Step-by-step explanation:

Part 1) what is the volume of the paperweight?

we know that

The volume of the paperweight is equal to the volume of the square pyramid plus the volume of the cube

step 1

Find the volume of the pyramid

The volume of the pyramid is equal to

V=\frac{1}{3}BH

where

B is the area of the square base

H is the height of the pyramid

B=8^{2}=64\ cm^{2}

H=3\ cm

substitute

V=\frac{1}{3}(64)(3)=64\ cm^{3}

step 2

Find the volume of the cube

The volume of the cube is equal to

V=b^{3}

V=8^{3}=512\ cm^{3}

step 3

Find the volume of the paperweight

64\ cm^{3}+512\ cm^{3}=576\ cm^{3}

Part 2) what is the total surface area of the paperweight?​

we know that

The total surface area of the paperweight is equal to the surface area of 5 faces of the cube plus the lateral area of the pyramid

step 1

Find the surface area of 5 faces of the cube

SA=5b^{2}

SA=5(8^{2})=320\ cm^{2}

step 2

Find the lateral area of the pyramid

LA=4[\frac{1}{2}bh]

LA=4[\frac{1}{2}(8)(5)]=80\ cm^{2}

step 3

Find the total surface area of the paperweight

320\ cm^{2}+80\ cm^{2}=400\ cm^{2}

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Answer:

\displaystyle \tan A=\frac{4}{3}

Step-by-step explanation:

<u>Funciones Trigonométricas</u>

La identidad principal en trigonometría es:

sen^2A+cos^2A=1

Si sabemos que A es un ángulo agudo (que mide menos de 90°), su seno y coseno son positivos.

Dado que Sen A = 4/5, calculamos el coseno:

cos^2A=1-sen^2A

Sustituyendo:

\displaystyle cos^2A=1-\left(\frac{4}{5}\right)^2

\displaystyle cos^2A=1-\frac{16}{25}

\displaystyle cos^2A=\frac{25-16}{25}

\displaystyle cos^2A=\frac{9}{25}

Tomando raíz cuadrada:

\displaystyle cos\ A=\sqrt{\frac{9}{25}}=\frac{3}{5}

La tangente se define como:

\displaystyle \tan A=\frac{sen\ A}{cos\ A}

Substituyendo:

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\displaystyle \tan A=\frac{4}{3}

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Answer:

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Step-by-step explanation:

3x + y = 6         -------------(i)

5x - 2y = 10      -------------(ii)

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Substitute x = 2 in (i)

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