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lianna [129]
3 years ago
6

Solve for a. 2(a+4)+6a=48

Mathematics
2 answers:
andreev551 [17]3 years ago
3 0
2a+8+6a=48
8a=40
a=5Hope this answer helps!
Dafna1 [17]3 years ago
3 0
<span>2(a + 4) + 6a = 48

2a + 8 + 6a = 48

8a = 48 - 8

8a = 40

a = </span>\frac{40}{8}

a = 5
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Step-by-step explanation:

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Finally, the leading coefficient is the same as the standard form, a.

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What are the real zeros of y=(x+3)^3+10
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<span> y - x3 - 9x2 - 27x - 37 = 0 is this what you were looking for?</span>
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3 years ago
Tarik rides a stationary bike and Janet rides her bike on a trail. Their data is shown
KiRa [710]

Answer:

Part a. The graph does not model a proportional relationship.

Part b. The values in table model a proportional relation.

3.5 minutes per mile.

Step-by-step explanation:

Part a.

The graph shown in the question representing Janet's data is not a straight line although it passes through the origin.

That is why the rate of change of distance with time is not constant.

Therefore, the graph does not model a proportional relationship.

Part b.

If we plot the data in the table using distance in miles along the y-axis and time in minutes along the x-axis, then we will get a straight line passing through the origin.

So, the values in the table model a proportional relation.

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6 0
3 years ago
Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the
Rama09 [41]

<em><u>Question:</u></em>

Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

1s   2s   3s   4s

<em><u>Answer:</u></em>

It takes 2 seconds for object to hit the ground

<em><u>Solution:</u></em>

<em><u>The given equation is:</u></em>

h(t) = -16t^2 + v_0t+h_0

Initial velocity = 27 feet/sec

h_0 = 10\ feet

Therefore,

h(t) = -16t^2 +27t+10

At the point the object hits the ground, h(t) = 0

-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0

Solve by quadratic formula,

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125

Ignore, negative value

Thus, it takes 2 seconds for object to hit the ground

7 0
3 years ago
Solve for y. <br> Ax + By = C; Solve for y.
IgorC [24]

Step-by-step explanation:

ur answer is there in image

8 0
3 years ago
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