What is 15% of 60? please help me

Assoli18 [71]

Answer:

What is it you need help with for this question? Please provide all information in your question so we can know exactly how to help you. Thank you!

8 0

2 years ago

Questlon 4 of 10

Kisachek [45]

Answer:

c

Step-by-step explanation:

To find the midpoint the formulas

x=x1+x2/2 and y=y1+y2/2 are used

in this case x1 is 5,x2 is 5,y1 is 4 and y2 is -3

therefore

x=5+5/2.

=10/2

=5

y=4+(-3)/2

=1/2

3 0

3 years ago

What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S

Gekata [30.6K]

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is $y=\frac{1}{3} x+2$

Solution:

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3

And as the line passing through (-3,1) and is perpendicular to 3x + y = -8, product of slopes of two line will be -1 as lies are perpendicular.

Let required slope = x

$\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}$

So we need to find the equation of a line whose slope is $\frac{1}{3}$ and passing through (-3,1)

Equation of line passing through $(x_1 , y_1)$ and having lope of m is given by

$\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)$

$\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}$

Substituting the values we get,

$\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}$

Hence the required equation of line is found using slope intercept form

4 0

2 years ago

ABC was dilated to create A' B' C'. What is the center of dilation? And what is the scale factor of the dilation?

Virty [35]

Given:

In the given figure the vertices of triangle ABC are A(-2,2), B(4,2), C(-2,-2).

The vertices of triangle A'B'C' are A'(-4,4), B'(8,4), C'(-4,-4).

To find:

The center of dilation and the scale factor.

Solution:

In the given graph, draw the lines AA', BB' and CC'. The intersection point of these lines is the center of dilation.

From the below graph, it is clear that the lines AA', BB' and CC' intersect each other at (0,0).

Therefore, the center of dilation is (0,0).

If a figure is dilated by factor k and the center of dilation is origin, then

$(x,y)\to (kx,ky)$

$A(-2,2)\to A'(k(-2),k(2))$

$A(-2,2)\to A'(-2k,2k)$

It is given that A'(-4,4). So,

$(-2k,2k)=(-4,4)$

On comparing both sides, we get

$-2k=-4$

$2k=4$

$k=\dfrac{4}{2}$

$k=2$

Therefore, the scale factor is 2.

4 0

3 years ago

The scale on a trail map is 0.5 cm : 1 km. The straight distance between 2 huts on the trail is 16.9 cm. What is the actual dist

sesenic [268]

We have been given that the scale on a trail map is 0.5 cm : 1 km. The straight distance between 2 huts on the trail is 16.9 cm. We are asked to find the actual distance between 2 huts.

We will use proportions to solve our given problem.

$\frac{\text{Actual length}}{\text{Scale length}}=\frac{\text{1 km}}{\text{0.5 cm}}$