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3 years ago

What is the slope of a line perpendicular to 5y + 3x = 15

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

7 0

Answer:

Using the slope-intercept form, the slope is −35 . The equation of a perpendicular line to y=−3x5+3 y = - 3 x 5 + 3 must have a slope that is the negative reciprocal of the original slope.

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7x+2y=-28<br><br><br>Help write slope-intercept form pls

kap26 [50]

Answer:

y = - $\frac{7}{2}$ x - 14

Step-by-step explanation:

Slope intercept form: y = mx + b

Isolate the y:

7x + 2y = -28

2y = -7x - 28

y = - $\frac{7}{2}$ x - 14

3 0

3 years ago

For the functions f and g, find (f o g)(8). f(x) = 17x^2 - 10x, g(x) = 9x - 9

Andreyy89

Hey mate. Here is your answer.

Set up the composite function and evaluate.

g (17x^2 - 10x) = 153x^2 - 90x - 9

Hope this helps.

5 0

3 years ago

Is a hectare roughly equal to 2 tennis courts or to 2 soccer fields

lys-0071 [83]

It is most likely equal to two soccer fields

6 0

3 years ago

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PLEEEEEEEAAAAAAASSSSSSSSSSEEEEEEEEEEE help me!!!!!!!!!!

defon

Answer:

$\boxed{\sf \ \ \ \dfrac{18}{(x-7)(x+12)(x+30)} \ \ \ }$

Step-by-step explanation:

hello,

first of all we will study the quadratic expressions

we can write that, (the different answers provide good clues :-) )

$x^2+5x-84=(x-7)(x+12)$

and

$x^2+23x-210=(x-7)(x+30)$

so first of all as we cannot divide by 0 we need to take x different from 7, -12 and -30 and then we can write

$\dfrac{1}{x^2+5x-48}-\dfrac{1}{x^2+23x-210}=\dfrac{1}{(x-7)(x+12)}-\dfrac{1}{(x-7)(x+30)}\\\\=\dfrac{(x+30) -(x+12)}{(x-7)(x+12)(x+30)}=\dfrac{x+30-x-12}{(x-7)(x+12)(x+30)}\\\\=\dfrac{18}{(x-7)(x+12)(x+30)}$

hope this helps

7 0

3 years ago

Use the give points and line. determine the slope of the line.

wariber [46]

Slope=1 because 1/1=1

5 0

3 years ago

Read 2 more answers