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3 years ago

10

I dont under stand these?

1 answer:

Alexeev081 [22]3 years ago

8 0

The first one means what is the smallest number that goes into 12 which would be 4. The second one mean what does not go into 4 and 6 so that would be 8 because 6 is not a multiple of 8.The third one is either 1 or 2 not sure. And the 4th one is little complicated you might want to look up video on prime factorization that can really help you .

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I can't figure out how to solve for x

Romashka [77]

The answer i got was x=2 :)

Step 1: Simplify both sides of the equation.
x−2 over 3−(x+3 over 5)=−1
Simplify: (Show steps)
2 over 15x+−19 over 15=−1
Step 2: Add 19/15 to both sides.
2 over 15x+−19 over 15+19 over 15=−1+19 over 15
2 over 15x=4 over 15
Step 3: Multiply both sides by 15/2.
(15 over 2)*(2 over 15x)=(15 over 2)*(4 over 15)

3 0

3 years ago

10. The value of a car, in dollars, x years after its purchase is represented by the function f(x) = 8000(0.55) x . What is the

Scrat [10]

F(1)=8000x0.55x1=4400
f(2)=8000x0.55x2=8800
(8800-4400)/2=2200

7 0

3 years ago

Those are not the powers question from transformation of product and sum trigonometry

OlgaM077 [116]

Answer:

see explanation

Step-by-step explanation:

Using the sum to product identity

cosx + cosy = 2cos( $\frac{x+y}{2}$ )cos( $\frac{x-y}{2}$ )

Consider left side

(cos5A +cos3A) + (cos15A + cos7A)

= 2cos( $\frac{5A+3A}{2}$ )cos( $\frac{5A-3A}{2}$ ) + 2cos( $\frac{15A+7A}{2}$ ) cos( $\frac{15A-7A}{2}$ )

= 2cos( $\frac{8A}{2}$ ) cos( $\frac{2A}{2}$ ) + 2cos( $\frac{22A}{2}$ )cos( $\frac{8A}{2}$ )

= 2cos4AcosA + 2cos11A cos4A ← factor out 2cos4A from both terms

= 2cos4A( cos11A + cosA) ← repeat the process

= 2cos4A( 2cos( $\frac{11A+A}{2}$ )cos( $\frac{11A-A}{2}$ )

= 2cos4A(2cos( $\frac{12A}{2}$ )cos( $\frac{10A}{2}$ )

= 2cos4A(2cos6A cos5A)

= 4cos4Acos5Acos6A

= right side , thus verified

8 0

2 years ago

Juliette [100K]

The answer is y+4=x thats the answer on E d g e n u i t y

8 0

3 years ago

A student at St. F. X. decided to become his own employer by using his car as a taxi for the summer. It costs the student $693.

polet [3.4K]

Given :

Insurance cost , $693.00 .

Gas cost , $452.00 .

Average charge per fare , $7.00 .

Tuition fee , $3280.00 .

To Find :

The fare he have to get to pay his fee .

Solution :

Let , number of fair required is n .

Therefore , total profit is :

$Profit = earning - expenditure \\Profit = 7n - 693-452\\Profit = 7n-1145$

Now , to pay his fee this profit must be equal to his fee because he don't have other expenditure .

So ,

$7n-1145=3280\\\\7n=4425\\\\n=\dfrac{4425}{7}\\\\n=632.1$

But number of rides cannot be in decimal so minimum rides he must get is

632+1 = 633 .

Hence, this is the required solution .

8 0

3 years ago