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love history [14]

3 years ago

8

Use the given line and the point not on the line to answer the question. What is the point on the line perpendicular to the give

n line, passing through the given point that is also on the y-axis? (−3.6, 0) (−2, 0) (0, −3.6) (0, −2))

2 answers:

svp [43]3 years ago

5 0

C. (0,-3.6) is the answer.

DaniilM [7]3 years ago

5 0

Answer:

0, -3.6

Step-by-step explanation:

c on edg

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What is 5,183.6 X 0.0016 =

DENIUS [597]

Answer:

8.3 (nearest tenth)

Step-by-step explanation:

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3 years ago

The point P on the circle x2 + y2 = r2 that is also on the terminal side of an angle θ in standard position is given. Find the i

Brut [27]

Okay so you do not really need the circle equation. if you make a triangle with it's x length as 3 and it's y length as 4, you will be able to find the third length. Do Pythagorean theorem to find the hypotenuse.
the hypotenuse will be five after you calculate it.

sin is opposite over hypotenuse or Y over R

so... the sin is 4/5

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6 0

3 years ago

Which of the following values have 2 significant figures? Check all that apply.

katen-ka-za [31]

Answer:

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Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

How many 1/16 pound servings of chocolate are in a 1/2 pound chocolate bar

Keith_Richards [23]

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6 0

3 years ago

What is the answer please.... need help

mina [271]

Answer:

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

$f\left(x\right)=x^3-6x^2+3x+10$

As the highest power of the x-variable is 3 with the leading coefficients of 1.

So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

$f\left(x\right)=x^3-6x^2+3x+10$

$0=x^3-6x^2+3x+10$ ∵ $f(x)=0$

as

$Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

so

$\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

if $ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x=-1,\:x=2,\:x=5$

Therefore, the zeros of the function are:

$x=-1,\:x=2,\:x=5$

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.

8 0

2 years ago