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EastWind [94]
3 years ago
7

Angela’s average for six math tests is 87. on her first four tests she had scores of 93, 87, 82, and 86. on her last tests she s

cored 4 points lower than she did on her fifth test what scores did Angela receive on her firth and sixth tests?
Mathematics
1 answer:
Annette [7]3 years ago
7 0

Answer:

the scores on her last test is x (x > 0)

because on her last tests she scored 4 points lower than she did on her fifth test

=> the scores in the 5th test is x + 4

because Angela’s average for six math tests is 87, we have:

\frac{93 + 87 + 82 + 86 + x + x + 4}{6} = 87 \\  \\  <  =  >  \frac{352 + 2x}{6}   = 87 \\  \\  <  =  > 352 + 2x = 522 \\  \\  <  =  > 2x = 170 \\  \\  <  =  > x = 85

=> on her last test, she had 85

=> on her 5th test, she had 85 + 4 = 89

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Answer:

-0.53571428571

Step-by-step explanation:

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3 years ago
The expression 120-15x represents how many invitations Luanne has to address after x days. The expression 120 + 15(7-x) represen
Aloiza [94]

Answer:

not possible.

Step-by-step explanation:

number of invitations luanne  has to address after x days = 120-15x

number of invitations  Darius has to address after x days =  120 + 15(7-x)

so, if the number of invitations should be the same for both of them,

we have to equate their number of invitations

120-15x =  120 + 15(7-x)

subtracting 120 from both the sides,

-15x = 15(7-x) = 105 -15x

adding we 15x on both sides, we wont find any solution for x.

so, this isnt possible or the question must be wrong.

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3 years ago
I need a answer please
8_murik_8 [283]

Answer: 6/24 ????????????

Step-by-step explanation:

8 0
2 years ago
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A circle with center (-3,5) passes through point (-9,-3). Find the circumference of the circle
Aliun [14]

Answer:20π

Step-by-step explanation:

equation of the circle

(x+3)² + (y-5)² = r²

now, (-9, -3) lies on it

So, (-9+3)² + (-3-5)² =r²

36+ 64 = r² =100

hence, r =10

So, circumference = 2πr = 20π

4 0
3 years ago
Assume that k and V0 are constant and are, respectively, 0.02, and 7. The parameter CA varies from an initial value CA0 = 100 to
julia-pushkina [17]

Answer:

V = 929.7

Step-by-step explanation:

Given the equation:

\frac{dCA}{dV} = \frac{-kCA}{V0}

The integral of the above equation is:

\int\limits {\frac{dCA}{dV} } = \int\limits{\frac{-kCA}{V0} }

Re-organizing the integrals:

\int\limits  {\frac{dCA}{CA} } = \int\limits {\frac{-kdV}{V0} }

Integrating:

ln(CA) - ln(CA0) = \frac{-kV}{V0}

Inputting the initial conditions of CA and the values of k and V0:

ln(7) - ln(100) = \frac{-0.02V}{7}

1.946 - 4.605 = -0.00286V

-2.659 = -0.00286V

=> V = \frac{2.659}{0.00286}

V = 929.720

Approximating to one decimal place,

V = 929.7

7 0
2 years ago
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