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jolli1 [7]
3 years ago
8

How do I solve number 1

Mathematics
1 answer:
cestrela7 [59]3 years ago
3 0
Because he ended up paying $24 more than the actual price of the drill, this is the amount of interest he paid.

In total, the repairman pays $324 in 6 equal payments. So, each payment was \dfrac{\$324}6=\$54.
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Bridgette had a whole pie. She ate 1/4 of the pie for a snack and ate another 25% of the pie for dessert. What fraction of the p
Kisachek [45]

Answer:

1/2

Step-by-step explanation:

Bridgette ate 1/4 of the pie, and 25% of the pie.

25% is converted to 1/4.

1/4 + 1/4 = 1/2.

1 - 1/2 = 1/2.

Bridge did NOT eat 1/2 of the pie.

8 0
3 years ago
Read 2 more answers
Find the value of n in the equation 6.2n – 3.7n = 85 + 45
qwelly [4]
Im not sure but i found the answer as n= 52
8 0
3 years ago
Solve this quadratic equation by completing the square.<br><br> x^2+10x=7
bogdanovich [222]
Ok, so the quadratic coefient is 1, so great

take 1/2 of the linear coefient and square it
10/2=5, (5)^2=25
add that to both sides
x^2+10x+25=7+25
factor perfect square trionomial
(x+5)^2=32
squaer root both sides
x+5=+/-4√2
minus 5
x=-5+/-4√2


x=-5+4√2 and -5-4√2
3 0
3 years ago
Given the function f(x)=4|x-5|+3 for what values of x is f(x)=15
Andreyy89

Answer:

  {2, 8}

Step-by-step explanation:

We want to find x for ...

  15 = 4|x -5| +3

  12 = 4|x -5| . . . . subtract 3

  3 = |x -5| . . . . . . divide by 4

  ±3 = x -5 . . . . . . show the meaning of absolute value

  5 ±3 = x = {2, 8} . . . . . add 5

The values of x for which f(x) = 15 are 2 and 8.

3 0
3 years ago
Help:<br> <img src="https://tex.z-dn.net/?f=1.%5C%20%26x%5E4-61x%5E2%2B900%3D0%20%5C%5C%202.%20%5C%20%26%20x%5E4-25x%5E2%2B144%3
yawa3891 [41]

\bf{1) \  x^4-61x^2+900=0 }

Applying the factorization method, for example:

      \bf{ 0=(x^2)^2-61(x^2)+900=(x^2-36)(x^2-25)}

               \bf{\\ &=(x+6)(x-6)(x+5)(x-5);    }

              \bf{ x_1=6 \quad x_2=-6 \quad \ \ \ x_3=5 \quad x_4=5  }

\bf{2) \  x^4-25x^2+144=0  }

Applying the factorization method:

          \bf{0&=(x^2)^2-25(x^2)+144=(x^2-16)(x^2-9) }

             \bf{ =(x-4)(x+4)(x-3)(x+3); }

            \bf{ x_1=4 \quad x_2=-4 \quad x_3=3 \quad x_4=-3  }

7 0
2 years ago
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