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PSYCHO15rus [73]
3 years ago
12

1. Find the area of each of the composite figures: a) 6 m 8 m 10 m

Mathematics
2 answers:
Dafna1 [17]3 years ago
7 0

Answer:

Could u include a picture of the figures as well?

Step-by-step explanation:

Brrunno [24]3 years ago
5 0
Include picture . I can help but send pic of the composite figure.
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2 fuel tanks with demensions shown have the same volume. what is the value of h?
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A right prism has a base that is an equilateral triangle. The height of the prism is equal to the height of the base. If the vol
Romashka [77]

<u>Answer:</u>

Lengths of the sides (there are 3 sides in an equilateral triangle) of the base of prism is <u>5.72cm </u>each.

<u>Step-by-step explanation:</u>

As the base of triangular prism is equilateral:

∴ Area of base = (√ 3 / 4)× (side)²

According to question:

height of prism = height (side) of base

Volume of prism = 81 cm³,

∴ Volume of prism = Area of base x height

⇒  81 cm³ = (√ 3 / 4)× (side)² × height       ∵ side = height

⇒   81 cm³ = (√ 3 / 4)× (side)³

⇒  81 × (4/√3) cm³ = side³    

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6 0
2 years ago
Write the standard form of the line that passes through (-1,-3) and (2,1).
kobusy [5.1K]

Answer:

Step-by-step explanation:

\mathrm{Line\:passing\:through\:}\left(-1,\:-3\right)\mathrm{,\:}\left(2,\:1\right)\\\\\mathrm{Find\:the\:line\:}\\\mathbf{y=mx+b}\mathrm{\:passing\:through\:}\left(-1,\:-3\right)\mathrm{,\:}\left(2,\:1\right)\\\\\mathrm{Compute\:the\:slope\:}\left(-1,\:-3\right),\:\left(2,\:1\right):\quad \\m=\frac{4}{3}\\\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}\\\\\left(x_1,\:y_1\right)=\left(-1,\:-3\right),\:\left(x_2,\:y_2\right)\\\left\\

\\m=\frac{1-\left(-3\right)}{2-\left(-1\right)}\\\\Refine\\\\m=\frac{4}{3}\\\\\mathrm{Compute\:the\:}y\mathrm{\:intercept}:\quad b=-\frac{5}{3}\\\\y=\frac{4}{3}x-\frac{5}{3}\\

Convert equation to standard form

y - \frac{4}{3} x + \frac{5}{3} = 0\\\\Multiply \:through\: by\: 3;\\\\3\times (y - \frac{4}{3} x + \frac{5}{3} )=0\\\\3y -4x+5 = 0\\Arrange\:n\:form\:of \:;\\Ax +By = C\\\\-4x +3y =-5

5 0
3 years ago
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