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Hatshy [7]
2 years ago
15

A stockbroker has money in three accounts. The interest rates on the three accounts are 7%, 8%, 9%. If she has twice as much mon

ey invested at 8% as she does in 7% Three times as much at 9%
as she has at 7%, and the total interest for the year is $150 how much is invested at each rate?


The stockbroker invested $___at 7%,$___at 8%, and $___at 9%
Mathematics
1 answer:
Sveta_85 [38]2 years ago
5 0

Answer:

x=amount invested at 7%

2x=amount invested at 8%

3x=amount invested at 9%

interest=principal*rate*time (time=1 year)

$150=0.07x+0.08*2x+0.09*3x

$150=0.07x+0.16x+0.27x

$150=0.50x

$1500=5x

x=$300 invested at 7%

2x=$600 invested at 8%

3x=$900 invested at 9%

Step-by-step explanation:

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Answer please !! Due in 45 minutes m
weqwewe [10]

Answer:

(-3,3)

Step-by-step explanation:

Using elimination:

Multiply the first equation by 8 and the second by 5 which equals

40x+56y=48

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Now you add them together

40x+(-40x)=0 so that crosses out

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2 years ago
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kodGreya [7K]

Answer: 2.11 s

Step-by-step explanation:

Given

Equation of rockets height is given by

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So, we can write

\Rightarrow h(t)=0\ \text{or}\ -8t^2+16t+2=0\\\Rightarrow 8t^2-16t-2=0\\\\\Rightarrow t=\dfrac{16\pm \sqrt{(-16)^2-4(8)(-2)}}{2\times 8}\\\\\Rightarrow t=\dfrac{16\pm \sqrt{320}}{16}\\\\\text{Neglecting negative value of t}\\\\\Rightarrow t=\dfrac{16+17.888}{16}=2.11\ s

7 0
2 years ago
A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by ℎ = −16푡 ଶ+103푡+5 w
Liono4ka [1.6K]

Question:

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by h(t)=-16t^2+80t+5, where t is measured in seconds.  Write an equation to determine how long it will take for the ball to reach the ground.

Answer:

t = 5.0625

Step-by-step explanation:

Given

h(t)=-16t^2+80t+5

Required

Find t when the ball hits the ground

This implies that h(t) = 0

So, we have:

0=-16t^2+80t+5

Reorder

-16t^2+80t+5 = 0

Using quadratic formula, we have:

t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

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t = \frac{-80\±\sqrt{80^2 - 4*-16*5}}{2*-16}

t = \frac{-80\±\sqrt{6400 +320}}{-32}

t = \frac{-80\±\sqrt{6720}}{-32}

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t = -\frac{2}{32} or t = \frac{162}{32}

But time can not be negative.

So, we have:

t = \frac{162}{32}

t = 5.0625

<em>Hence, time to hit the ground is 5.0625 seconds</em>

3 0
2 years ago
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