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Veseljchak [2.6K]
3 years ago
14

Please help me! 20 point question :)

Mathematics
1 answer:
bezimeni [28]3 years ago
5 0
(x^4/5)^3/7
= x^(4/5 * 3/7)
= x^12/35

Answer is B. second option
x^12/35
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Five times the difference of a number, x, and 3 plus 7 is equal to -45.
miskamm [114]

Answer:

5(x-3)+7=-45

Step-by-step explanation:

The only possibly way for 5 to multiply 2 numbers is if those numbers are in parenthesis

7 0
3 years ago
What is the slope for the graph y - 4x =-2 ?
dem82 [27]

Answer:

4.000

Step-by-step explanation:

6 0
3 years ago
Solve by the substitution method.
GalinKa [24]

Answer:

The unique solution to the system is (x,y){(-9,-3)}

Step-by-step explanation:

x - y = -6     equation 1

5x + 6y = -63     equation 2

We will find the value of x from equation 1.

x= y-6

Now put the value of x in equation 2.

5(y-6)+6y = -63

5y-30+6y = -63

Combine the like terms:

5y+6y= -63+30

11y = -33

Divide both sides by 11.

11y/11 = -33/11

y = -3

Now put the value y = -3 in x=y-6

x = y-6

x= -3-6

x= -9

Therefore The unique solution to the system is (x,y){(-9,-3)} ....

7 0
3 years ago
Read 2 more answers
Help plss !! Will give branliest to who ever has the best answer
ozzi

Answer:

a c and e hope it helps :D

Step-by-step explanation:

5 0
2 years ago
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As part of quality-control program, 3 light bulbs from each bath of 100 are tested. In how many ways can this test batch be chos
hichkok12 [17]

Answer:

<h3>By 161700 ways this test batch can be chosen.</h3>

Step-by-step explanation:

We are given that total number of bulbs are = 100.

Number of bulbs are tested = 3.

Please note, when order it not important, we apply combination.

Choosing 3 bulbs out of 100 don't need any specific order.

Therefore, applying combination formula for choosing 3 bulbs out of 100 bulbs.

^nCr = \frac{n!}{(n-r)!r!} read as r out of n.

Plugging n=100 and r=3 in above formula, we get

^100C3 = \frac{100!}{(100-3)!3!}

Expanding 100! upto 97!, we get

=\frac{100\times 99\times 98\times 97!}{97!3!}

Crossing out common 97! from top and bottom, we get

=\frac{100\times 99\times 98}{3!}

Expanding 3!, we get

=\frac{100\times 99\times 98}{3\times 2\times 1}

= 100 × 33  × 49

= 161700 ways.

<h3>Therefore,  by 161700 ways this test batch can be chosen.</h3>
3 0
4 years ago
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