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2 years ago

11

Christine's grade in the three quarters are 88,91,90. if she targets to have an average grade not less then 92 in the four quart

er .what is the least grade she should have for fourth quarter

1 answer:

dezoksy [38]2 years ago

8 0

Total marks C needs to get = 92 x 4 = 368

Least grade (fourth quarter) = 368 - 88 - 91 - 90 = 99

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1 year ago

Alex777 [14]

Answer:

<h2>A)t=6.7</h2>

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

quadratic equation
quadratic equation word problems
solving quadratic

<h3>given:</h3>

h(t) = -4t² + 12t + 100

<h3>to solve:</h3>

t

<h3>tips and formulas:</h3>

<u>the</u><u> </u><u>Ball</u><u> </u><u>will</u><u> </u><u>hit </u><u>the</u><u> ground</u><u> </u><u>when</u><u> </u><u>the</u><u> height</u><u> is</u><u> </u><u>0</u>
<u>solving</u><u> </u><u>quadratics </u><u>using</u><u> </u><u>quadratic</u><u> formula</u>
<u>PEMDAS</u>

<h3>let's solve: </h3>

$step - 1 : \: define$

$- 4 {t}^{2} + 12t + 100 = 0$

$step - 2 : \\ divide \: both \: sides \: by \: - 4$

${t}^{2} - 3t - 25 = 0$

$step - 3 : \\ solve \: the \: quadratic$

$formula : \\ x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a}$

$define \: a ,b \: and \: c \\ which \: are \: 1, - 3 \: and \: - 25 \: respectively$

$t = \frac{ - ( - 3)± \sqrt{ {( - 3)}^{2} - 4.1. - 25 } }{2.1}$

$t = \frac{ 3± \sqrt{ 9 + 100} }{2}$

$t = \frac{3 + \sqrt{ 109 } }{2}$

$t = \frac{3 - \sqrt{ 109 } }{2}$

$t = 6.7 \\ t = - 3.7$

$as \: we \: know \: time \: cannot \: be \: negative$

$\huge \therefore \: t = 6.7$

5 0

2 years ago