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egoroff_w [7]
3 years ago
15

Which is the domain of the function f(x) = Negative five-sixths (three-fifths) x?

Mathematics
1 answer:
andrew-mc [135]3 years ago
6 0
Maybe 3/5 i’m not sure
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The price of a sweater was reduced from $25 to &19. By what percentage was the sweater reduced by?
JulijaS [17]

Answer:

24%

Step-by-step explanation:

Difference in price of sweater

=$25-$19

=$6

6/25×100%=24%

8 0
3 years ago
A jet ski rental company charges a $100 rental fee, plus $20 per hour. Write and dole an equation to determine how many hours yo
adell [148]

3 hours because if u have 175 dollers and u spend 100 so now u have 75. 75 -20 is 55. then 55 - 20 is 35. then 35 - 20 is 15 so then you have 15 dollers to spare. so 3

4 0
3 years ago
What is the value of y?
nika2105 [10]

Answer:

y =24

Step-by-step explanation:

The exterior angle is the sum of the opposite interior angles

y+18  + 3y -41  = y+49

Combine like terms

4y -23 = y+49

Subtract y from each side

4y -y -23 = y+49-y

3y -23 = 49

Add 23 to each side

3y -23+23 = 49 +23

3y = 72

Divide by 3

3y/3 = 72/3

y =24

4 0
3 years ago
Trigonometry help!! - double angle formulae
ivolga24 [154]

Answer:

The two rules we need to use are:

Sin(a + b) = sin(a)*cos(b) + sin(b)*cos(a)

cos(a + b) = cos(a)*cos(b) - sin(a)*sin(b)

And we also know that:

sin^2(a) + cos^2(a) = 1

To solve the relations, we start with the left side and try to construct the right side.

a) Sin(3*A) = sin (2*A + A) = sin(2*A)*cos(A) + sin(A)*cos(2*A)

sin(A + A)*cos(A) + sin(A)*cos(A + A)

(sin(A)*cos(A) + sin(A)*cos(A))*cos(A) + sin(A)*(cos(A)*cos(A) - sin(A)*sin(A))

sin(A)*cos^2(A) + sin(A)*cos^2(A) + sin(A)*cos^2(A) - sin^3(A)

3*sin(A)*cos^2(A) - sin(A)*sin^2(A)

sin(A)*(3*cos^2(A) - sin^2(A))

Now we can add and subtract 4*sin^3(A)

sin(A)*(3*cos^2(A) - sin^2(A)) + 4*sin^3(A) -  4*sin^3(A)

sin(A)*(3*cos^2(A) + 3*sin^2(A)) - 4*sin^3(A)

sin(A)*3*(cos^2(A) + sin^2(A)) - 4*sin^3(A)

3*sin(A) - 4*sin^3(A)

b) Here we do the same as before:

cos(3*A) = 4*cos^3(A) - 3*cos(A)

We start with:

Cos(2*A + A) =  cos(2*A)*cos(A) - sin(2*A)*sin(A)

= cos(A + A)*cos(A) - sin(A + A)*sin(A)

= (cos(A)*cos(A) - sin(A)*sin(A))*cos(A) - ( sin(A)*cos(A) + sin(A)*cos(A))*sin(A)

= (cos^2(A) - sin^2(A))*cos(A) - sin^2(A)*cos(A) - sin^2(A)*cos(A)

= cos^3(A) - 3*sin^2(A)*cos(A)

=  cos(A)*(cos^2(A) - 3*sin^2(A))

now we subtract and add 4*cos^3(A)

= cos(A)*(cos^2(A) - 3*sin^2(A)) + 4*cos^3(A) - 4*cos^3(A)

= cos(A)*(-3*cos^2(A) - 3*sin^2(A)) + 4*cos^3(A)

= cos(A)*(-3)*(cos^2(A) + sin^2(A)) + 4*cos^3(A)

= -3*cos(A) + 4*cos^3(A)

8 0
3 years ago
joseph needs to run 3 miles during his workout for the soccer team. he begins by running 1/2 half and he takes 3 breaks during p
Bingel [31]

At the end of practice, Joseph will need to run another 1.75 miles or 1 3/4 miles.

To determine this we need to subtract the amounts he already has run during practice from the goal of 3 miles.

3 miles - 1/2 miles - 1/4 miles - 1/4 miles - 1/4 miles = 1.75 miles

3 0
3 years ago
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