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sergejj [24]
3 years ago
12

Your car insurance comes due annually and generally costs about $1,500. You decide that you would like to set aside a monthly am

ount, beginning in January, to be prepared for when this bill comes at the end of the year. How much should you set aside each month?
Mathematics
1 answer:
Volgvan3 years ago
5 0

Answer:

$125

Step-by-step explanation:

Since there are 12 months in a year, divide the total bill by the total number of months.

1500/12 = $125

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There are 9 female performers in a dance recital. The ratio of men to women is 2:3. How many men are in the dance recital?
Firdavs [7]

Answer:

6 men

Step-by-step explanation:

4 0
3 years ago
Find the sum of 1/6and 7/12 in lowest terms.
disa [49]

Answer:3/4

Step-by-step explanation:

Multiply 1/6 by 2 to get 2/12 and then add that to 7/12 and get 9/12

If you simplify that you end up with 3/4

6 0
3 years ago
Read 2 more answers
Given X is midpoint of RS , MX =XS Prove Mx =RX
xz_007 [3.2K]

Answer:

Transitive Property

Step-by-step explanation:

Since X is the midpoint, it bisects the points RS. If MX=XS, and RS is the same as XS, then they are the same. This also follows the Transitive propety  (if a=b and b=c, then a=c)

3 0
3 years ago
Which of the following are identities? Check all that apply
Natasha2012 [34]

Answer:

A, C

Step-by-step explanation:

Actually, those questions require us to develop those equations to derive into trigonometrical equations so that we can unveil them or not. Doing it only two alternatives, the other ones will not result in Trigonometrical Identities.

Examining

A) True

\frac{1-tan^{2}x}{2tanx} =\frac{1}{tan2x} \\ \frac{1-tan^{2}x}{2tanx} =\frac{1}{\frac{2tanx}{1-tan^{2}x}}\\ tan2x=\frac{1-tan^{2}x}{2tanx}

Double angle tan2\alpha =\frac{1 -tan^{2}\alpha }{2tan\alpha}

B) False,

No further development towards a Trig Identity

C) True

Double Angle Sine Formula sin2\alpha =2sin\alpha *cos\alpha

sin(8x)=2sin(4x)cos(4x)\\2sin(4x)cos(4x)=2sin(4x)cos(4x)

D) False No further development towards a Trig Identity

[sin(x)-cos(x)]^{2} =1+sin(2x)\\ sin^{2} (x)-2sin(x)cos(x)+cos^{2}x=1+2sinxcosx\\ \\sin^{2} (x)+cos^{2}x=1+4sin(x)cos(x)

7 0
3 years ago
Read 2 more answers
(2x-6)(x +11)<br> What is the answer?
ycow [4]

Answer:

2x^2+16x-66

Step-by-step explanation:

Use FOIL (First, Outside, Inside, Last)

(2x-6)(x+11)

F- 2x*x=2x^2

O-2x*11=22x

I--6*x=-6x

L--6*11=-66

2x^2+22x-6x-66

Combine like terms

2x^2+16x-66

5 0
3 years ago
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