Which equation represents the graph of the linear function?

1 answer:

7 0

I think you forgot to attach the graph along with the question. I am answering the question based on my knowledge and research. "y = −2x − 1" is the one equation among the following choices given in the question that represents the graph of the linear function. The correct option among all the options that are given in the question is option "B".

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Help please!! Solve for X.

Mekhanik [1.2K]

9514 1404 393

Answer:

3) x = 9

4) x = 3

Step-by-step explanation:

3) The two short segments are indicated as having a sum equal to the long segment.

(x +2) +(-5 +x) = 15

2x = 18 . . . . . . . . . . . . add 3

x = 9 . . . . . . . . . divide by 2

(This makes the segments be 9+2 = 11 and -5+9 = 4, which total 15.)

4) Same deal.

3x +3 = 4x

3 = x . . . . . . . . subtract 3x

(This makes the segments be 3(3) = 9 and 4(3) = 12, where 9+3=12.)

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]

Question:

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

Answer:

It takes 2 seconds for object to hit the ground

Solution:

The given equation is:

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

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,

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