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3 years ago

What is the equation of a line that passes through the points (0, 5) and (4, 8)?

Mathematics

1 answer:

Andrew [12]3 years ago

6 0

Answer:

see explanation

Step-by-step explanation:

The equation of a line in slope - intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (0, 5) and (x₂, y₂ ) = (4, 8)

m = $\frac{8-5}{4-0}$ = $\frac{3}{4}$

Note the line crosses the y- axis at (0, 5) ⇒ c = 5

y = $\frac{3}{4}$ x + 5 ← equation in slope- intercept form

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3 years ago

Read 2 more answers

This is 22 points

AURORKA [14]

Let the initial number of girls be x, this represents 40% of the dancers.
Total number of dancers will therefore be:
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therefore the new percentage of girls can be expressed as follows:
(new number of girls)/(new number of dancers)×100
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3 years ago

Read 2 more answers

ramon earns $1735 each month and pays $53.10 for electricity. to the nearest tenth of a percent, what percent of ramon’s earning

Charra [1.4K]

Answer:

3.1%

Step-by-step explanation:

From the question, Ramon earns $1735 monthly

He pays $53.10 for electricity

% of Ramon’s salary spent on electricity each month will be

Amount spent on electricity divided by his monthly earning multiplied by 100%.

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=0.0306 x 100

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To the nearest tenth, it becomes 3.1%

So, Ramon spends 3.1% of his monthly earnings on electricity

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2 years ago

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3 years ago

Solve the equation: 2x^3-5x^2-6x+9=0 given that x=1 is a zero

olga_2 [115]

The given function

$2x^3+5x^2-6x+9=0$

Since x = 1 is one of its zeroes, then we will use it to find the other zeroes

$\begin{gathered} x=1 \\ x-1=1-1 \\ x-1=0 \end{gathered}$

The factor is x - 1

Use the long division to find the other factors

$\frac{2x^3-5x^2-6x+9}{x-1}$

Divide 2x^3 by x, then multiply the answer by (x - 1)

$\begin{gathered} \frac{2x^3}{x}=2x^2 \\ 2x^2(x-1)=2x^3-2x^2 \end{gathered}$

Subtract the product from the original equation

$\begin{gathered} 2x^3-5x^2-6x+9-(2x^3-2x^2)= \\ (2x^3-2x^3)+(-5x^2+2x^2)-6x+9= \\ 0-3x^2-6x+9 \\ \frac{2x^3-5x^2-6x+9}{x-1}=2x^2+\frac{-3x^2-6x+9}{x-1} \end{gathered}$

Now, divide -3x^2 by x, then multiply the answer by (x - 1)

$\begin{gathered} -\frac{3x^2}{x}=-3x \\ -3x(x-1)=-3x^2+3x \end{gathered}$

Subtract it from the denominator of the fraction

$\begin{gathered} (-3x^2-6x+9)-(-3x^2+3x)= \\ (-3x^2+3x^2)+(-6x-3x)+9= \\ 0-9x+9 \\ \frac{2x^2-5x^2-6x+9}{x-1}=2x^2-3x+\frac{-9x+9}{x-1} \end{gathered}$

Divide -9x by x and multiply the answer by (x - 1)

$\begin{gathered} \frac{-9x}{x}=-9 \\ -9(x-1)=-9x+9 \end{gathered}$

Subtract it from the numerator

$\begin{gathered} -9x+9-(-9x+9)= \\ (-9x+9x)+(9-9)= \\ 0+0 \\ \frac{2x^3-5x^2-6x+9}{x-1}=2x^2-3x-9 \end{gathered}$

Then we will factor this trinomial into 2 factors

$\begin{gathered} 2x^2=(2x)(x) \\ -9=(-3)(3) \\ (2x)(-3)+(x)(3)=-6x+3x=-3x\rightarrow middle\text{ term} \\ 2x^2-3x^2-9=(2x+3)(x-3) \end{gathered}$

Equate each factor by 0 to find the other zeroes of the equation

$\begin{gathered} 2x+3=0 \\ 2x+3-3=0-3 \\ 2x=-3 \\ \frac{2x}{2}=\frac{-3}{2} \\ x=-\frac{3}{2} \end{gathered}$ $\begin{gathered} x-3=0 \\ x-3+3=0+3 \\ x=3 \end{gathered}$

The zeroes of the equations are 1, 3, -3/2

The solutions of the equations are

$\begin{gathered} x=-\frac{3}{2} \\ x=1 \\ x=3 \end{gathered}$

7 0

10 months ago

What is the equation of a line that passes through the points (0, 5) and (4, 8)?​

What is the equation of a line that passes through the points (0, 5) and (4, 8)?