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2 years ago

When solving an equation, Francesca's first step is shown below. Which property

Mathematics

2 answers:

navik [9.2K]2 years ago

8 0

Would it be addition property of equality?

Yuliya22 [10]2 years ago

7 0

-2
the second equation is the result of subtracting 2

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How do you find the slope of (2, -6) & (-3, -4)

marta [7]

(-4)-(-6)/(-3)-2=

2/-5

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3 years ago

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List multiples of 6 between 20 and 50.

xenn [34]

Answer:

6,12,18,24,30,36,42,48,54,69

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3 years ago

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Which equation of the line that is parallel to the line 5x + 2y =12 and passes through point -2,4

Gnesinka [82]

Answer:

The equation of the line would be y = -5/2x - 1

Step-by-step explanation:

In order to find the equation of the line, we first need to find the slope of the original line. We can do that by solving for y.

5x + 2y = 12

2y = -5x + 12

y = -5/2x + 6

Now that we have a slope of -5/2, we know the new slope will be the same since parallel lines have the same slope. So we can use it along with the point in point-slope form to find the equation.

y - y1 = m(x - x1)

y - 4 = -5/2(x + 2)

y - 4 = -5/2x - 5

y = -5/2x - 1

3 0

3 years ago

What is the equation of the line that passes through (4, 2) and is parallel to 3x – 2y = -6?

Dmitrij [34]

Answer:

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

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 )

Rearrange 3x - 2y = - 6 into this form

Subtract 3x from both sides

- 2y = - 3x - 6 ( divide all terms by - 2 )

y = $\frac{3}{2}$ x + 3 ← in slope- intercept form

with slope m = $\frac{3}{2}$

• Parallel lines have equal slopes, hence

y = $\frac{3}{2}$ x + c ← partial equation of parallel line

To find c substitute (4, 2) into the partial equation

2 = 6 + c ⇒ c = 2 - 6 = - 4

y = $\frac{3}{2}$ x - 4 ← equation of parallel line

3 0

3 years ago

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

2 years ago