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svp [43]
3 years ago
13

What is the equation of the line that passes through the point (6,3) and has a slope of -1/3

Mathematics
1 answer:
bezimeni [28]3 years ago
8 0

Answer:

y = -1/3x + 5

Step-by-step explanation:

y = -1/3x + b

3 = -1/3(6) + b

3 = -2 + b

b = 5

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What is the length of the unknown leg in the right triangle? A right triangle has a side with length 20 meters, hypotenuse with
galben [10]

The length of the unknown leg of the triangle is 15 m.

<u>Step-by-step explanation:</u>

Length of one leg = 20 m

Length of the hypotenuse= 25m

As it is a right angled triangle we can use pythogoras theorem.

Let the unknown length be y

(20) (20)  + y(y) = (25) (25)

400 + y(y) = 625

y(y) = 225

y = √225

y = 15

The length of the unknown leg is 15 m.

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Tell which value of the variable, if any, is a solution to the equation.
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Answer:

14

Step-by-step explanation:

14+9=25

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3 years ago
P(x)= 3x^3-5x^2-14x-4
nexus9112 [7]
   
\displaystyle\\&#10;P(x)=3x^3-5x^2-14x-4\\\\&#10;D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\&#10;\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\&#10;3x^3-5x^2-14x-4=0\\\\&#10;

\displaystyle\\&#10;\text{Verification}\\\\&#10;3x^3-5x^2-14x-4=\\\\&#10;=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\&#10;=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\&#10; =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\&#10;\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\&#10;\Longrightarrow~~~P(x)~\vdots~(3x+1)


\displaystyle\\&#10;3x^3-5x^2-14x-4=0\\&#10;~~~~~-5x^2 = x^2 - 6x^2\\&#10;~~~~~-14x =-2x-12x \\&#10;3x^3+x^2 - 6x^2-2x-12x-4=0\\&#10;x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\&#10;(3x+1)(x^2-2x -4)=0\\\\&#10;\text{Solve: } x^2-2x -4=0\\\\&#10;x_{12}= \frac{-b\pm  \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm  \sqrt{4+16}}{2}=\frac{2\pm  \sqrt{20}}{2}=\frac{2\pm  2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\&#10;x_1 =1+\sqrt{5}\\&#10;x_2 =1-\sqrt{5}\\&#10;\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}



7 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
pogonyaev

By using the general equation for a circle, we will see that the correct options are:

  • 1) This is a circle centered at (-3, - 5) with a radius of 4 units.
  • 2) The center is (-3, 5) and the radius is 2 units.
  • 3) The center is (3, 5) and the radius is 2 units.
  • 4) This is a circle centered at (3, - 5) with a radius of 4 units.

<h3>How to write a circle equation?</h3>

For a circle of radius R and center (a, b), the equation is given by:

(x - a)^2 + (y - b)^2 = R^2

With that in mind, if we look at the first equation:

(x + 3)^2 + (y + 5)^2 = 16 = 4^2

This is a circle centered at (-3, - 5) with a radius of 4 units.

For the second equation:

(x + 3)^2 + (y - 5)^2 = 4 = 2^2

The center is (-3, 5) and the radius is 2 units.

The third equation is:

(x - 3)^2 + (y - 5)^2 = 4 = 2^2

The center is (3, 5) and the radius is 2 units.

The fourth equation is:

(x - 3)^2 + (y + 5)^2 = 16 = 4^2

This is a circle centered at (3, - 5) with a radius of 4 units.

If you want to learn more about circles, you can read:

brainly.com/question/14283575

7 0
2 years ago
The graph of the function f(x) = tan x is given above for the interval x in[0,2 pi] NL Determine the one-sided limit. Then indic
damaskus [11]

Okay, here we have this:

Considering the provided function, we are going to calculate the requested one-side limits, so we obtain the following:

5 0
1 year ago
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