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Juli2301 [7.4K]
3 years ago
6

What does the line -5+8y=8 look like?

Mathematics
2 answers:
algol133 years ago
8 0

Answer:

Is a horizontal line

Step-by-step explanation:

we have

-5+8y=8

Solve for y

Adds 5 both sides

-5+8y+5=8+5

8y=13

Divide by 8 both sides

y=13/8

y=1.625

This is a horizontal line parallel to the x-axis

The slope is equal to zero

see the attached figure to better understand the problem


miskamm [114]3 years ago
5 0
It's across below two a little so a straight line across hope this helps
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Which states the following?
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Step-by-step explanation:

If a point lies  in between  two end points of a line then the sum of distance between the points in between is always equal to the whole line.

If A is in between two points B and C then the sum of line segments BA and AC will always equal to the length of BC.

Or BA+AC=BC in all cases whatever be the position of point A between the points Band C.

Always is the right option.

3 0
3 years ago
Which sets does square 7 belong to.​
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Step-by-step explanation:

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3 years ago
GRAVITATION The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial vel
zhuklara [117]
<h3>Option B</h3><h3>At 2 second and 1.75 second, the object be at a height of 56 feet</h3>

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

Given that,

<em><u>The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation:</u></em>

h(t) = -16t^2 + 60t

<em><u>At what times will the object be at a height of 56 feet</u></em>

<em><u>Substitute h = 56</u></em>

56 = -16t^2 + 60t\\\\16t^2 - 60t + 56 = 0\\\\Divide\ the\ equation\ by\ 4\\\\4t^2 - 15t + 14=0

Solve the above equation 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=4,\:b=-15,\:c=14\\\\x =\frac{-\left(-15\right)\pm \sqrt{\left(-15\right)^2-4\cdot \:4\cdot \:14}}{2\cdot \:4}\\\\Simplify\\\\x = \frac{15 \pm \sqrt{1}}{8}\\\\x = \frac{15 \pm 1}{8}\\\\We\ have\ two\ solutions\\\\x = \frac{15+1}{8} \text{ and } x = \frac{15-1}{8}\\\\x = 2 \text{ and } 1.75

Thus, at 2 second and 1.75 second, the object be at a height of 56 feet

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