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3 years ago

How to write (X+3)(x+5) in a equivalent expression

Mathematics

2 answers:

Blizzard [7]3 years ago

5 0

Answer:

x^2 +8x+15

Step-by-step explanation:

(X+3)(x+5)

FOIL

first x^2

outer 5x

inner 3x

last 15

Add together

x^2 +5x+3x+15

x^2 +8x+15

Alex Ar [27]3 years ago

5 0

Answer:

Step-by-step explanation:

(x+3)(x+5)=x²+5x+3x+15

x²+8x+15 is equivalent to: (x+3)(x+5)

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4 years ago

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Part A. Write an equation of a line that is perpendicular to the line y = 2/3x + 5 and passes through the point (4, 0).

fenix001 [56]

Answer:

$y = -\frac{3}{2}x+6$

Step-by-step explanation:

Given:

The given equation of the line $y = \frac{2}{3} x + 5$ that passes through the point (4, 0).

Part A.

The equation of the line.

$y=mx+c$ -----------(1)

Where:

m = Slope of the line

c = y-intercept

The given equation of the line.

$y = \frac{2}{3} x + 5$

Comparing the given equation with equation 1.

The slope of the line is $m=\frac{2}{3}$ and y-intercept $c = 5$

We know that the slope of the perpendicular line is $(-\frac{1}{m})$ .

So the slope of the perpendicular line is $(m=-\frac{1}{\frac{2}{3}}=-\frac{3}{2})$ .

Using point slope formula we write the equation of the perpendicular line that passes through the point (4, 0).

$y-y_{1} = m(x-x_{1}$

Now we substitute the slope of the perpendicular line $m=-\frac{3}{2}$ and $y_{1} = 0, x_{1}=4$ from point (4, 0) in above equation.

$y-0 = -\frac{3}{2}(x-4)$

$y = -\frac{3}{2}x-(-\frac{3}{2}\times 4)$

$y = -\frac{3}{2}x-(-3\times 2)$

$y = -\frac{3}{2}x-(-6)$

$y = -\frac{3}{2}x+6$

Therefore the perpendicular line equation is $y = -\frac{3}{2}x+6$ .

Part B.

1. The slope of the perpendicular line is $m=-\frac{3}{2}$ .

2. The y-intercept of the perpendicular line is 6.

3. The equation of the line $y = -\frac{3}{2}x+6$ is perpendicular to the equation of line $y = \frac{2}{3} x + 5$ .

6 0

3 years ago

Write the equation in siope-intercept form for the line that passes through the given pointand is perpendicular to the given equ

yulyashka [42]

For a line in the form:

$\begin{gathered} ax+by=c \\ The\text{ slope m is:} \\ m=-\frac{a}{b} \end{gathered}$

In this case, for the line 2x + 10y = 20 with a=2 and b=10 the slope is:

$\begin{gathered} m_1=-\frac{a}{b}=-\frac{2}{10} \\ m_1=-\frac{1}{5} \end{gathered}$

Now, two lines are perpendiculars if the slopes satisfy the following equation:

$m_2=-\frac{1}{m_1}$

So, for the line we want the slope is:

$\begin{gathered} m_1=-\frac{1}{5} \\ m_2=-\frac{1}{m_1}=-\frac{1}{(-\frac{1}{5})}=5 \end{gathered}$

Finally, the line pass througth the point (2, 3) with slope m=5, so the equation is:

$\begin{gathered} P_1=(2,3),m=5 \\ y=mx+b \\ \text{The P1 must satisfy the equation:} \\ 3=5\cdot2+b \\ b=3-10 \\ b=-7 \end{gathered}$

The equation of the line is y = 5x - 7

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1 year ago