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2 years ago

Ind the equation of the tangent line to 9x^2+16y^2=52 through (2, -1)

1 answer:

6 0

Here are the steps.

In order to get the answer, first we need to get the x and y value of the center of the circle by using the equation of a circle (x^2 + y^2 = r^2)

<span>The center is (-9, -4)</span>

Once you get the center point, find the equation of the line (in a form of a radius) from the center point to point (2, -1). Get the slope.

The slope from center to point (2, -1) is 11/3

Finally, get the negative reciprocal of the slope from the previous step and form the equation using slope-intercept form ( y=mx + b). Then, that's the equation of the tangent line.

The slope of the tangent line is -3/11
The equation is y = -3/11x - 5/11

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Find the difference -11 - -5, 25 - (-3) ,-7 - (-2), 4 - 15, 6 - (-9)

Leviafan [203]

Answer:

-11-(-5)= -6

25-(-3)= 28

-7-(-2)= -5

4-15= -11

6-(-9)= 15

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3 years ago

Over what interval will the immediate value theorem apply

koban [17]

Answer:

Any [a,b] that does NOT include the x-value 3 in it.

Either an [a,b] entirely to the left of 3, or

an [a,b] entirely to the right of 3

Step-by-step explanation:

The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.

Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.

6 0

2 years ago

Answer please i will give brainliest

il63 [147K]

Step-by-step explanation:

simply add both equations

x + y = 2

-3x - y = 5

----------------

-2x + 0 = 7

x = -7/2

x + y = 2

-7/2 + y = 2

y = 2 + 7/2 = 4/2 + 7/2 = 11/2

1/2 x + 2y = -13

x - 4y = 8

first multiply the first equating by 2

x + 4y = -26

x - 4y = 8

--------------------

2x + 0 = -18

x = -18/2 = -9

x - 4y = 8

-9 - 4y = 8

-4y = 17

y = -17/4

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2 years ago

Equation for cost of <br> beans and rice

Natasha_Volkova [10]

Answer:

is there a passage or sum?

Step-by-step explanation:

3 0

2 years ago

Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL

Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

J(2;5); K(1;1)

=> JK = $\sqrt{(1-2)^{2} + (1-5)^{2} } = \sqrt{(-1)^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

J(2;5); L(5;2)

=> JL = $\sqrt{(5-2)^{2} + (2-5)^{2} } = \sqrt{3^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

K(1;1); L(5;2)

=> KL = $\sqrt{(5-1)^{2} + (2-1)^{2} } = \sqrt{4^{2}+1^{2} } = \sqrt{1+16}=\sqrt{17}$

In the triangle QNP, the sides can be calculate as following:

Q(-4;4); N(-3;0)

=> QN = $\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

Q (-4;4); P(-7;1)

=> QP = $\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

N(-3;0); P(-7;1)

=> NP = $\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}$

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0

2 years ago

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