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max2010maxim [7]
4 years ago
9

Can you help me with this question.

Mathematics
2 answers:
never [62]4 years ago
6 0

Answer:

\large\boxed{\dfrac{1}{3}}

\large\boxed{78\ mi}

Step-by-step explanation:

Q1.\\\text{The scale of similar is the ratio of lengths of corresponding sides}\\\text{Scale:}\ \dfrac{6cm}{18cm}=\dfrac{1}{3},\ \dfrac{4cm}{12cm}=\dfrac{1}{3}

Q2.\\1\ in\to12\ mi\\\\6\dfrac{1}{2}\ in\to6\dfrac{1}{2}\cdot12\ mi=\dfrac{6\cdot2+1}{2}\cdot12\ mi=\dfrac{13}{2}\cdot12\ mi=(13)(6)\ mi=78\ mi

HACTEHA [7]4 years ago
3 0

For the scale factor, use the dimensions from similar sides:

Using 18 from the original image you would use 6 from the scaled image.

18 /6 = 3

This means the scaled image is 3 times smaller than the original. Because the scaled image is smaller divide 1 by the scale factor to get 1/3.

Q2:

The scale factor is 1 inch = 12 miles.

Multiply the total inches by 12:

6.5 x 12 = 78 miles.

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What is 7 29/80 in a decimal
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Answer:

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hope it helped

 

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
PLEASE HELP FAST URGENT
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Answer:

I think it is c

Step-by-step explanation:

6 0
3 years ago
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Acute △ABC with angles α, β, and γ is inscribed in a circle. Tangents to the circle at points A, B, and C intersect in points M,
SOVA2 [1]

Answer:

The measures of angles of triangle MNP are

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

step 1

Find the measure of arcs AB, BC and AC

we know that

The inscribed angle is half that of the arc it comprises.

so

\gamma=\frac{1}{2}[arc\ AB] ----> arc\ AB=2\gamma\\\alpha=\frac{1}{2}[arc\ BC] ----> arc\ BC=2\alpha\\\beta=\frac{1}{2}[arc\ AC] ----> arc\ AC=2\beta

step 2

Find the measure of angle M

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

M=\frac{1}{2}[arc\ AB+arc\ BC-arc\ AC]

substitute

M=\frac{1}{2}[2\gamma+2\alpha-2\beta]\\M=[\gamma+\alpha-\beta]

step 3   

Find the measure of angle N

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

N=\frac{1}{2}[arc\ AC+arc\ BC-arc\ AB]

substitute

N=\frac{1}{2}[2\beta+2\alpha-2\gamma]\\N=[\beta+\alpha-\gamma]

step 4    

Find the measure of angle P

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

P=\frac{1}{2}[arc\ AC+arc\ AB-arc\ BC]

substitute

P=\frac{1}{2}[2\beta+2\gamma-2\alpha]\\P=[\beta+\gamma-\alpha]

5 0
4 years ago
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