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3 years ago

15

PLEASE HELPPP Which shows the graph of the solution y is less than or equal to -4x + 3

1 answer:

kicyunya [14]3 years ago

6 0

The less than or equal to means you are looking for a graph with a solid line first and foremost. Any graph with a dashed line is automatically eliminated.

Now because you look for a line with a y-intercept of 3 or crossed the y-axis at (0, 3).

Once you find a line fitting those two criteria, eliminate the others.

Now, for the slope. You are looking for a negative sloping line because of the negative attached to the 4x. Meaning from left to right, the y-values should be decreasing.

Lastly, because of the statement "y is less than", your shading should be on the left side of your line. This means every ordered pair in the shaded area and on your line will make your equation a true statement.

I list all of these because your picture does not include the rest of your options and the two featured are incorrect. Your graph should look like this:

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3 years ago

4 thousands less than 1.895

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Solve the equation to find where the treste ground levelEnter your answers from least to greatest and round to the nearest The m

DaniilM [7]

Answer:

The trestle meets ground level at 0.875 units and 9.125 units

Step-by-step explanation:

Poorly formatted question.

The given equation is:

$y = -x^2 + 10x - 8$

Required

The point where the trestle gets to the ground level

To do this, we set $y = 0$

So, we have:

$-x^2 + 10x - 8 = 0$

Multiply through by -1

$x^2 -10x + 8 = 0$

Solve using quadratic formula:

$x= \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where:

$a = 1; b =-10; c = 8$

So, we have:

$x= \frac{-(-10) \± \sqrt{(-10)^2 - 4*1*8}}{2*1}$

$x= \frac{-(-10) \± \sqrt{68}}{2*1}$

$x= \frac{-(-10) \± 8.25}{2}$

$x= \frac{10 \± 8.25}{2}$

Solve the fraction

$x= 5 \± 4.125$

Split

$x= 5 + 4.125\ or\ x= 5 - 4.125$

$x= 9.125\ or\ x= 0.875$

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3 years ago

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3 years ago

Solve using long division <br> Please

madreJ [45]

1. $Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}$

$Steps:$

$\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}$

$\mathrm{Divide}\:\frac{-2x^2-5}{x+3}:\quad \frac{-2x^2-5}{x+3}=-2x+\frac{6x-5}{x+3}$

$\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}$

$\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}$

$\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}$

2. $Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}$

$Steps:$

$\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}$

$\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}$

$\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

4 0

3 years ago