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3 years ago

Which scales are equivalent to the scale 1cm to 5km? More than one is the right answer

Mathematics

2 answers:

Oduvanchick [21]3 years ago

8 0

The answer would be A because each side is being multiplied by 3

bearhunter [10]3 years ago

6 0

Answer:

Option A and D.

Step-by-step explanation:

The scale is given 1 centimeter = 5 kilometers

We have to find the equivalent scales from the given options.

A. 3 cm to 15 km

∵ 1 cm to 5 km

∴ 3 cm to 3 × 5 = 15 km

This option is right.

B. 1 mm to 150 km

∵ 1 mm = 0.1 cm

∴ 0.1 cm = 150 km

This is not the correct option

C 5 Cm to 1 km

∵ 1 cm to 5 km

∴ 5 cm = 5 × 5 = 25 km

This option is incorrect.

D. 5 mm to 2.5 km

1 cm = 10 mm

5 mm = 0.5 cm

∵ 0.5 cm = 2.5 km

∴ 1 cm = 2.5 ÷ 0.5 = 5 km

This is the correct option

E. 1 mm to 500 km

10 mm = 1 cm

1 cm to 5000 km

This is not correct

Therefore, Option A and D is equivalent to the scale.

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Step-by-step explanation:

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Step-by-step explanation:

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Read 2 more answers

What is the best approximation of the projection of (5,-1) onto (2,6)?

Hatshy [7]

Answer:

Hence, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\sqrt{10} }{5}$ , and the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

Step-by-step explanation:

We have given two points $(5, -1)$ and $(2, 6)$ .

Let, $\vec a=5\hat {i}-\hat {j}$ and $\vec b= 2\hat {i}+6\hat{j}$ .

and we have calculate the projection of $\vec a$ onto $\vec b$ .

Now,

For the calculation of projection, first we need to calculate the dot product of $\vec a$ and $\vec b$ .

$\vec a.\vec b=(5\hat {i}-\hat{j}).(2\hat{i}+6\hat{j})$

$=10-6$

$=4$

then, we have to calculate the magnitude of $\vec b$ .

$\mid {\vec {b}}\mid$ = $\sqrt{2^{2}+6^{2} }$ = $\sqrt{40}$ = $2\sqrt{10}$ .

Now, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\vec a.\vec b}{\mid b\mid}$

= $\frac{4}{2\sqrt{10} }$ $\frac{2}{\sqrt{10} } \times\frac{\sqrt{10} }{\sqrt{10} } =\frac{2\sqrt{10} }{10} = \frac{\sqrt{10} }{5}$

and the vector projection of $\vec a$ onto $\vec b$ = $\frac{\vec a. \vec b}{\mid\vec b \mid^{2} } . \vec b$

= $\frac{4}{40} . (2\hat i+ 6\hat j)$

= $\frac{1}{5} \hat i+\frac{3}{5} \hat j$

Hence, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\sqrt{10} }{5}$ , and the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

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3 years ago