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3 years ago

How do i find the x intercet for y=6x−7.

Mathematics

1 answer:

kolezko [41]3 years ago

3 0

Answer:

Step-by-step explanation:

y=mx+b" is the formula of a straight line drawn on Cartesian coordinate system in which "y" is the vertical axis and "x" the horizontal axis.

In this formula :

y tells us how far up the line goes

x tells us how far along

m is the Slope or Gradient i.e. how steep the line is

b is the Y-intercept i.e. where the line crosses the Y axis

The X and Y intercepts and the Slope are called the line properties. We shall now graph the line y-6x+7 = 0 and calculate its properties.

Graph of a Straight Line

Calculate the Y-Intercept :

Notice that when x = 0 the value of y is -7/1 so this line "cuts" the y axis at y=-7.00000

y-intercept = -7/1 = -7.00000

Calculate the X-Intercept :

When y = 0 the value of x is 7/6 Our line therefore "cuts" the x axis at x= 1.16667

x-intercept = 7/6 = 1.16667

Calculate the Slope :

Slope is defined as the change in y divided by the change in x. We note that for x=0, the value of y is -7.000 and for x=2.000, the value of y is 5.000. So, for a change of 2.000 in x (The change in x is sometimes referred to as "RUN") we get a change of 5.000 - (-7.000) = 12.000 in y. (The change in y is sometimes referred to as "RISE" and the Slope is m = RISE / RUN)

Slope = 12.000/2.000 = 6.000

x-intercept = 7/6 = 1.16667

y-intercept = -7/1 = -7.00000

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3 years ago

What is the table of values for y=x+2

Marta_Voda [28]

Answer:

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Have a supercalifragilisticexpialidocious day!

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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,

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

Ignore, negative value

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

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