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3 years ago

Find the missing dimension. Use the scale factor 1: 15. Door Model Height 10 in Actual

Mathematics

1 answer:

anygoal [31]3 years ago

5 0

Answer:

The answer is "150 unit"

Step-by-step explanation:

scale factor = 1:15

height = 10

dimension =?

let dimension = x

$\Rightarrow \frac{10}{x} = \frac{1}{15} \\\\\Rightarrow 150 = x\\\\\Rightarrow x= 150 \ units$

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Solve for g and h please

Ivenika [448]

Step-by-step explanation:

since it's an Equilateral triangle

then:

g=15

h=96°

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3 years ago

Simplified form of the expression (x^3+3x^2+6)-(9x^2-5x+7)

JulijaS [17]

$(x^3 + 3x^2 + 6) - (9x^2 - 5x + 7)\\\\x^3 + 3x^2 + 6 - 9x^2 + 5x - 7\\x^3 - 6x^2 + 5x -1$

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4 years ago

Pls help with math fast !!

Nutka1998 [239]

Answer:

1 would: You dont have to pay anything because 80% of 750 is $150 which you also have to deduct $150 so you dont pay anything.

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3 years ago

I need a lot of help, I’m really confused on what we are learning currently.

Virty [35]

7m+2=7n−5

Swap sides so that all variable terms are on the left hand side.

7n−5=7m+2

Add 5 to both sides.

7n=7m+2+5

Add 2 and 5 to get 7.

7n=7m+7

Divide both sides by 7.

7m+7

Dividing by 7 undoes the multiplication by 7.

7m+7

Divide 7+7m by 7.

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3 0

3 years ago

Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A

iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

$\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA$

7 0

3 years ago