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noname [10]
3 years ago
12

What is 9 to the power of 4 + 6(7 x 3) - 3?

Mathematics
1 answer:
marshall27 [118]3 years ago
5 0

Answer: 127^9 so 8594754748609400000

Step-by-step explanation: i’m not sure but 4+6 (7x3)-3 = 127 127^9

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Which of the following points is in the solution set of y > -x2 + 5? (0, 5) (1, 3) (2, 4)
mylen [45]

Answer:

(2, 4)

Step-by-step explanation:

The only point that satisfies the inequality is (2, 4).

(0, 5) : -0^2 +5 = 5 . . . . . not > 5

(1, 3) : -1^2 +5 = 4 . . . . . . 3 is not > 4

(2, 4) : -2^2 +5 = 1 . . . . . . 4 is greater than 1, so this point is in the solution set.

5 0
3 years ago
Calculate the value of the expression 44-(5x4)
MrMuchimi

Answer:

24

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
*UNIT RATE AND PROPORTIONS*
Advocard [28]

Answer:

Unit rate is $7 per pound.

6 pounds will cost $42

You can buy 10 pounds for $70

Step-by-step explanation:

28/4 = $7 --> Unit rate is $7 per pound.

7 × 6 = $42 --> 6 pounds will cost $42

70/7= 10 --> You can buy 10 pounds for $70

3 0
2 years ago
124 ounces decreased by 67%
slavikrds [6]
40.92 ounces.
You can find it by multiplying 124 by 0.33
8 0
3 years ago
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What are the endpoint coordinates for the midsegment of △PQR that is parallel to PQ¯¯¯¯¯?
andriy [413]

Answer:

M(x₄ ,y₄) = (-3.5 , 0.5)  and

N (x₅ ,y₅) = ( -1 , -0.5 )

Step-by-step explanation:

Let the endpoint coordinates for the mid segment of △PQR that is parallel to PQ be

M(x₄ ,y₄) and N(x₅ ,y₅) such that MN || PQ

point P( x₁ , y₁) ≡ ( -3 ,3 )

point Q( x₂ , y₂) ≡ (2 , 1 )

point R( x₂ , y₂) ≡ (-4 , -2)  

To Find:

M(x₄ ,y₄) = ?  and

N (x₅ ,y₅) = ?

Solution:

We have Mid Point Formula as

Mid\ point(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})

As M is the mid point of PR and N is the mid point of RQ so we will have

Mid\ pointM(x_{4} ,y_{4})=(\frac{x_{1}+x_{3} }{2}, \frac{y_{1}+y_{3} }{2})

Mid\ pointN(x_{5} ,y_{5})=(\frac{x_{2}+x_{3} }{2}, \frac{y_{2}+y_{3} }{2})

Substituting the given value in above equation we get

Mid\ pointM(x_{4} ,y_{4})=(\frac{-3+-4 }{2}, \frac{3+-2} }{2})

∴ Mid\ pointM(x_{4} ,y_{4})=(\frac{-7} }{2}, \frac{1}{2})

∴ Mid\ pointM(x_{4} ,y_{4})=(-3.5, 0.5)

Similarly,

Mid\ pointN(x_{5} ,y_{5})=(\frac{2+-4 }{2}, \frac{1+-2 }{2})

∴ Mid\ pointN(x_{5} ,y_{5})=(\frac{-2 }{2}, \frac{-1}{2})

∴ Mid\ pointN(x_{5} ,y_{5})=(-1, -0.5)

∴ M(x₄ ,y₄) = (-3.5 , 0.5)  and

  N (x₅ ,y₅) = ( -1 , -0.5 )

3 0
3 years ago
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