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3 years ago

6

A photograph is reduced by a scale factor of 2/3. If the original length of the photograph was 6 inches, what is the length of t

he reduced photo?
A. 2 inches
B. 3 inches
C. 4 inches
D. 5 1/3 inches

1 answer:

AlladinOne [14]3 years ago

3 0

The answer is c, 4 inches
6 x (2/3) = 4

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Determine side x. Round to the nearest unit. 20 9 12 5

tatyana61 [14]

Answer:

12

Step-by-step explanation:

First, you will have to use trigonometry. Since the side of the triangleis the opposite of the angle 53 and the hypotenuse, you will use sin.

Sin 53 degrees = opposite side / hypotenuse side.

The opposite side is x and the hypotenuse side equals 15.

sin53= x/15

15 ( sin 53 ) =x

x= 11.97953

Round

x=12

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3 years ago

Assured normally costs £42 if you receive a 15% discount how much change will you get from £50

vesna_86 [32]

5% of £42 = £2.10
10% of £42 = £4.20
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3 years ago

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Use radical notation to rewrite the expression 144 1/2

andrew-mc [135]

$144^{\frac{1}{2}} = \sqrt{144}$

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3 years ago

The table below shows the heights of several books. Jean stacks a dictionary on top of her novel. How high is the stack of the t

aev [14]

<u><em>Answer:</em></u>

The height of the stack = $5\frac{1}{4}$ in

<u><em>Explanation:</em></u>

We are given that Jean stacks a dictionary on top of a novel.

This means that the total height of the stack will be equal to the height of the dictionary added to the height of the novel.

<u>From the givens, we have:</u>

height of dictionary = $4\frac{1}{2}$ in

height of novel = $\frac{3}{4}$ in

<u>Now, adding the two height, we will have:</u>

Total height of stack = height of novel + height of dictionary

Total height of stack = $\frac{3}{4}+4 \frac{1}{2} = \frac{3}{4}+ \frac{18}{4}= \frac{21}{4}=5 \frac{1}{4}$ in

Hope this helps :)

6 0

3 years ago

If 1st and 4tg terms of G.p are 500 and 32 respectively it's second term is ?

bixtya [17]

Answer:

$T_{2} = 200$

Step-by-step explanation:

Given

Geometry Progression

$T_1 = 500$

$T_4 = 32$

Required

Calculate the second term

First, we need to write out the formula to calculate the nth term of a GP

$T_n = ar^{n-1}$

For first term: Tn = 500 and n = 1

$500 = ar^{1-1}$

$500 = ar^{0}$

$500 = a$

$a = 500$

For fought term: Tn = 32 and n = 4

$32 = ar^{4-1}$

$32 = ar^3$

Substitute 500 for a

$32 = 500 * r^3$

Make r^3 the subject

$r^3 = \frac{32}{500}$

$r^3 = 0.064$

Take cube roots

$\sqrt[3]{r^3} = \sqrt[3]{0.064}$

$r = \sqrt[3]{0.064}$

$r = 0.4$

Using: $T_n = ar^{n-1}$

$n = 2$ $r = 0.4$ and $a = 500$

$T_{2} = 500 * 0.4^{2-1}$

$T_{2} = 500 * 0.4^1$

$T_{2} = 500 * 0.4$

$T_{2} = 200$

<em>Hence, the second term is 200</em>

5 0

3 years ago