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baherus [9]
3 years ago
8

On three 150-point geography tests, you earned grades of 88%, 94%, and 90%. The final test is worth 250 points. What percent do y

ou need on the final to earn 93% of the total points on all tests?
Mathematics
2 answers:
Semmy [17]3 years ago
7 0
You must earn 243 points to have earned 93% of all possible points.

We first multiply each percentage on the previous tests by 150:

0.88*150 = 132
0.94*150 = 141
0.9*150 = 135

The total number of points possible is given by adding up the possible points for the three previous tests and the 250 for the last test:
150+150+150+250 = 700

93% of the 700 points would be 0.93(700) = 651 points.

Now we have 132+141+135+x (last test) = 651
408 + x = 651

Subtract 408 from both sides:
408+x-408 = 651-408
x = 243
user100 [1]3 years ago
3 0

Answer:

You must earn 243 points to have earned 93% of all possible points.

We first multiply each percentage on the previous tests by 150:

0.88*150 = 132

0.94*150 = 141

0.9*150 = 135

The total number of points possible is given by adding up the possible points for the three previous tests and the 250 for the last test:

150+150+150+250 = 700

93% of the 700 points would be 0.93(700) = 651 points.

Now we have 132+141+135+x (last test) = 651

408 + x = 651

Subtract 408 from both sides:

408+x-408 = 651-408

x = 243

Step-by-step explanation:

You must earn 243 points to have earned 93% of all possible points.

We first multiply each percentage on the previous tests by 150:

0.88*150 = 132

0.94*150 = 141

0.9*150 = 135

The total number of points possible is given by adding up the possible points for the three previous tests and the 250 for the last test:

150+150+150+250 = 700

93% of the 700 points would be 0.93(700) = 651 points.

Now we have 132+141+135+x (last test) = 651

408 + x = 651

Subtract 408 from both sides:

408+x-408 = 651-408

x = 243

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The length of the side of square A is 50% of the length of the side of square b
galben [10]

Answer:

25%

Step-by-step explanation:

let length of 1st square be x and second be y

then

A/q

x = y/2

area of first square = x^2 = (y/2)^2 = (y^2)/4

and

area of second square = y^ 2

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area of first square = 1/4 * area of second square

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8 0
3 years ago
How can I factor 5x^2 + 26x - 24 = 0 using the completing the square method.
kifflom [539]
<h2><u>QUADRATIC EQUATION</u></h2>

<h2>\mathbb{ANSWER:}</h2>
  • \bold{x = 0.8 \: } \:  \sf \:  {\color{grey}or} \:  \:  \:  \bold{ x=  \frac{4}{5} } \\

  • \bold{x =  - 6}

— — — — — — — — — —

<h3>Step-by-step explanation:</h3>

<u>How can </u><u>we</u><u> factor 5x^2 + 26x - 24 = 0 using the completing the square </u><u>method?</u>

Let's solve your equation step-by-step.

\bold{Given \:  Equation: \color{brown} 5x²+26x-24=0}

First, add 24 to both sides.

  • \bold{5x²+26x-24 - \purple{ 24} = 0 +  \purple{24}}

  • \bold{ \implies \: 5x²+26x = 24 }

Since the coefficient of 5x² is 5, divide both sides by 5.

  • \bold{ \frac{5 {x}^{2} + 26x }{5} =  \frac{24}{5}  } \\

  • \bold{ \implies \:  {x}^{2} +  \frac{26}{5} x =  \frac{24}{5}  } \\

The coefficient of 26/5x is 26/5. So, let b=26/5.

Then we need to add (b/2)²=169/25 to both sides to complete the square.

Add 169/25 to both sides.

  • \bold{ {x}^{2} +  \frac{26}{5} x +  \frac{  \purple{169}}{ \purple{25}}=  \frac{24}{5}    +  \frac{ \purple{169}}{ \purple{25}} } \\

  • \bold{ \implies \:\bold{ {x}^{2} +  \frac{26}{5} x +  \frac{  169}{ {25}}=  \frac{289}{25}  } } \\

Factor the left side.

  • \bold{(x +  \frac{13}{5} ) {}^{2} =  \frac{289}{25}  } \\

Take square root.

  • \bold{x +  \frac{13}{5} = ±  \:  \sqrt{ \frac{289}{25}  }} \\

Then, add (-13)/5 to both sides.

  • \bold{x +  \frac{13}{5} +  \frac{\purple{ - 13} }{\purple{ 5}} = \frac{{ - 13} }{{ 5}} ± \:      \sqrt{ \frac{289}{25}}} \\

  • \bold{  \: x =  \frac{ - 13}{5} ± \sqrt{ \frac{289}{25} } } \\

  • \implies \: \underline{ \boxed{ \bold{   \frac{4}{5}  } \sf  \:  \: or \:  \:  \bold{ x =  - 6}}} \\

_______________❖_______________

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Answer:  -p^2-(q+r)^2

Work Shown:

I'm assuming your teacher wants you to factor as much as possible.

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