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cupoosta [38]
3 years ago
14

Which choices are equivalent to the expression below check all that apply √-25 please help.

Mathematics
2 answers:
nadezda [96]3 years ago
7 0
B and D
Hope this helps
klio [65]3 years ago
6 0

Answer: A and B

Ape x

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A new plasma screen tv costs $2,970. A family makes a down payment of $720 and pays off the balance in 18 equal monthly payments
ratelena [41]
The monthly payment would be 125. 2,970-720=2,250 2,250/18=125 a month 
4 0
3 years ago
Find the exact value of sec theta if csc theta = -4/3 and the terminal side of theta lies in quadrant III
GREYUIT [131]

Answer:

-4 sqrt(7)/7

Step-by-step explanation:

csc theta = -4/3

csc theta = hypotenuse / opposite side

hypotenuse = 4  

opposite  = 3

Using the pythagorean theorem

a^2 + b^2 = c^2

3^2 + b^2 = 4^2

9+b^2 = 16

b^2 = 16-9

b^2 = 7

Taking the square root

sqrt(b^2) = sqrt(7)

b = sqrt(7)

We are in the third quadrant so only tan  and cot are positive

that means the x and y values are "negative"  so a = -3 and b = - sqrt(7)

sec theta = hypotenuse / adjacent

                = 4/ - sqrt(7)

rationalizing  

                   -4 sqrt(7)/ sqrt(7)* sqrt(7)

                = -4 sqrt(7)/7

           

3 0
3 years ago
In a certain Algebra 2 class of 29 students, 7 of them play basketball and 14 of them
Mariulka [41]

Answer:

<em>Two possible answers below</em>

Step-by-step explanation:

<u>Probability and Sets</u>

We are given two sets: Students that play basketball and students that play baseball.

It's given there are 29 students in certain Algebra 2 class, 10 of which don't play any of the mentioned sports.

This leaves only 29-10=19 players of either baseball, basketball, or both sports. If one student is randomly selected, then the propability that they play basketball or baseball is:

\displaystyle P=\frac{19}{29}

P = 0.66

Note: if we are to calculate the probability to choose one student who plays only one of the sports, then we proceed as follows:

We also know 7 students play basketball and 14 play baseball. Since 14+7 =21, the difference of 21-19=2 students corresponds to those who play both sports.

Thus, there 19-2=17 students who play only one of the sports. The probability is:

\displaystyle P=\frac{17}{29}

P = 0.59

3 0
3 years ago
there are 30 students in a math class of the 30 students 30% have an a in math 40% have a b in math how many students have a c i
stepladder [879]

the answer is 9

30 percent of 30 is 9

40 percent of 30 is 12

12+9 =21

30-21 =9

3 0
3 years ago
Read 2 more answers
Five more than a number is three
Pavel [41]

I'm not sure if you use "x" as your variable, so:

# + 5 = 3 or x + 5 = 3

7 0
3 years ago
Read 2 more answers
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