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stepladder [879]

3 years ago

8

Need help with 2 and 3 and express I’m radical form

1 answer:

kondaur [170]3 years ago

4 0

This is just simple Pythagorean’s theorem. a^2 + b^2 = c^2

Number 2:
5^2 + x^2 = 10^2
25 + x^2 = 100
x^2 = 75
x = sqrt 75 = 5*sqrt(3)

Number 3:
4^2 + x^2 = 7^2
16 + x^2 = 49
x^2 = 33
x = sqrt 33

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Write an equation in standard form of the line that contains the point (-1,2) amd is perpendicular to y = 3x - 1

Strike441 [17]

Since it is perpendicular to y=3x-1, the slope must be a negative reciprocal.
there the slope of the perpendicular line is $-\frac{1}{3} x$

So now we have,
$y=- \frac{1}{3} x+b$

To find b, we use the point that the line passes through and substitute it to the y and x values.

$2=- \frac{1}{3} (-1)+b$

which would give us b=5/3

Therefore, the equation of the line is
$y=- \frac{1}{3} x+ \frac{5}{3}$

7 0

2 years ago

Vanessa is solving the equation -3 + 4x=9

mina [271]

Answer:

The -3

Step-by-step explanation:

Hope this helps :)

8 0

3 years ago

Read 2 more answers

PLEASE ANSWER QUICKLY

jenyasd209 [6]

Answer:

It should be D because I have done that a couple days ago

6 0

2 years ago

Insert one term between -13 and -119 to create an arithmetic sequence of three terms.

daser333 [38]

Answer:

-66

Step-by-step explanation:

- 13, ____, -119

first, find the common difference:

-13 - (-119) = 106

next, divide 106 by 2: $\frac{106}{2}$ = 53

so, the common difference is 53.

now, subtract 53 from -13:

-13 - 53 = -66

5 0

2 years ago

Read 2 more answers

What is the result of a dilation of scale factor 3 centered at the origin of the line 2y + 3x=10?? PLEASE HELP PLEASEEEEEEEEE

maks197457 [2]

Given:

The equation of a line is:

$2y+3x=10$

The line is dilated by factor 3.

To find:

The result of dilation.

Solution:

The equation of a line is:

$2y+3x=10$

For $x=0$ ,

$2y+3(0)=10$

$2y+0=10$

$y=\dfrac{10}{2}$

$y=5$

For $x=2$ ,

$2y+3(2)=10$

$2y+6=10$

$2y=10-6$

$2y=4$

Divide both sides by 2.

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

$y=2$

The given line passes through the two points A(0,5) and B(2,2).

If the line dilated by factor 3 with origin as center of dilation, then

$(x,y)\to (3x,3y)$

Using this rule, we get

$A(0,5)\to A'(3(0),3(5))$

$A(0,5)\to A'(0,15)$

Similarly,

$B(2,2)\to B'(3(2),3(2))$

$B(2,2)\to B'(6,6)$

The dilated line passes through the points A'(0,15) and B'(6,6). So, the equation of dilated line is:

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

$y-15=\dfrac{6-15}{6-0}(x-0)$

$y-15=\dfrac{-9}{6}(x)$

$y-15=\dfrac{-3}{2}x$

Multiply both sides by 2.

$2(y-15)=-3x$

$2y-30=-3x$

$2y+3x=30$

Therefore, the equation of the line after the dilation is $2y+3x=30$ .

3 0

2 years ago