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3 years ago

Write the prime factorization of these numbers 18

Mathematics

1 answer:

Fofino [41]3 years ago

3 0

18 = 2·9 = 2·3²

answer: 2·3²

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Can someone please help?

mel-nik [20]

Answer:

C. y = 2/3 x - 2.

Step-by-step explanation:

The general form of the slope-intercept equation of a line is y = mx + c where m = the slope and (0. c) is the y-intercept.

We are given that the slope = 2/3 and the y intercept is at (0,-2) so our equation is y = 2/3 x - 2.

4 0

3 years ago

Read 2 more answers

A jewelry store marks up the price of bracelets they purchase at $30.00 each by 75%. What is the markup and new price of each br

horrorfan [7]

The new price is 52.50
30x2=60
30/4=7.5
7.5x3=22.5
22.50+30= 52.50

3 0

3 years ago

The seventh grade wants to break last year’s record of 78 coats collected for the annual clothing drive. They have already colle

Dimas [21]

Answer:

c + 13c ≥ 78c

78c - 13c =

65c

8 0

2 years ago

1) After a dilation, (-60, 15) is the image of (-12, 3). What are the coordinates of the image of (-2,-7) after the same dilatio

BARSIC [14]

Answer:

a) k = 5; (-10, -35)

Step-by-step explanation:

Given:

Co-ordinates:

Pre-Image = (-12,3)

After dilation

Image = (-60,15)

The dilation about the origin can be given as :

Pre-Image $(x,y)\rightarrow Image(kx,ky)$

where $k$ represents the scalar factor.

We can find value of $k$ for the given co-ordinates by finding the ratio of $x$ or $y$ co-ordinates of the image and pre-image.

$k=\frac{Image}{Pre-Image}$

For the given co-ordinates.

Pre-Image = (-12,3)

Image = (-60,15)

The value of $k=\frac{-60}{-12}=5$

or $k=\frac{15}{3}=5$

As we get $k=5$ for both ratios i.e of $x$ and $y$ co-ordinates, so we can say the image has been dilated by a factor of 5 about the origin.

To find the image of (-2,-7), after same dilation, we will multiply the co-ordinates with the scalar factor.

Pre-Image $(-2,-7)\rightarrow$ Image $((-2\times5),(-7\times 5))$

Pre-Image $(-2,-7)\rightarrow$ Image $(-10,-35)$ (Answer)

7 0

3 years ago

If tan a = 9／40<br>use trigonometric identities to find the values of sin a and cos a.

Sindrei [870]

Answer:

see explanation

Step-by-step explanation:

Given

tan a = $\frac{9}{40}$ = $\frac{opposite}{adjacent}$

We require the hypotenuse h

Using Pythagoras' identity

h² = 9² + 40² = 81 + 1600 = 1681 ( take square root of both sides )

h = $\sqrt{1681}$ = 41 , thus

sin a = $\frac{opposite}{hypotenuse}$ = $\frac{9}{41}$

cos a = $\frac{adjacent}{hypotenuse}$ = $\frac{40}{41}$

4 0

3 years ago