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2 years ago

Ummm how do you do this two questions?

Mathematics

1 answer:

guajiro [1.7K]2 years ago

8 0

Answer:

Here are the formulae you need. Just plug in the values you have and you will get your answers.

Area of a triangle: 1/2*base*height

Volume of a rectangular prism: length*width*height

Volume of a cylinder: pi*r^2*height

Step-by-step explanation:

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Can someone please help with this i don’t know how to do it 85points

Dmitry [639]

Answer:

Step-by-step explanation:

You can find the area by cutting it across in the middle into a trapezoid and a rectangle.

The trapezoid will have:

height = 10-4 = 6ft

top = 9-6 = 3ft

bottom = 8ft

So its area = 1/2*(3+8)*6 = 33 ft^2

The rectangle will have:

length = 10ft

width = 6ft

So its area = 10*6 = 60 ft^2

Total area = 60+33 = 93 ft^2

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2 years ago

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Write two monomials with a product of -84a^6b^10

rjkz [21]

(-2a^5b^4) * (42ab^6) = -84a^6b^10

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3 years ago

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A store owner orders 25 calculator that cost $38.00 each. The store owner can sell each calculator for $50. The store owner sold

USPshnik [31]

The owners profit is 252 dollars

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3 years ago

Divide: 83.5 ÷ 6.25 The first step in solving 83.5 ÷ 6.25 is to multiply the divisor and dividend by . The new division problem

kherson [118]

Answer:

The first step in solving 83.5 ÷ 6.25 is to multiply the divisor and dividend by

[✔ 100 ]

sorry idk the answer for the second one :(

Step-by-step explanation:

7 0

3 years ago

An equation of an ellipse is given. y2 = 1 − 3x2 (a) Find the vertices, foci, and eccentricity of the ellipse. vertex (x, y) = (

oksian1 [2.3K]

Answer:

Step-by-step explanation:

Given

$y^2=1-3x^2$

$3x^2+y^2=1$

$\frac{x^2}{(\frac{1}{\sqrt{3}})^2}+\frac{y^2}{1}=1$

therefore it is a vertical ellipse

thus a=1

$b=\frac{1}{\sqrt{3}}$

eccentricity of Ellipse

$e^2=1-\frac{b^2}{a^2}$

$e^2=1-\frac{1}{(\sqrt{3})^2}$

$e^2=1-\frac{1}{3}$

$e^2=\frac{2}{3}$

$e=\sqrt{\frac{2}{3}}$

Focii are $(0,ae)$ and $(0,-ae)$

$ae=1\times \sqrt{\frac{2}{3}}$

thus focii are $(0,\sqrt{\frac{2}{3}})$ & $(0,-\sqrt{\frac{2}{3}})$

(b) Length of major axis $=2a=2\times 1$

length of minor axis $=2b=2\times \sqrt{\frac{2}{3}}=2\cdot \sqrt{\frac{2}{3}}$

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3 years ago