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Olin [163]
3 years ago
10

Match the values based on parallelogram ABCD, shown in the figure. Tiles length of value of value of Pairs 56 arrowBoth 4 arrowB

oth 44 arrowBoth 2 arrowBoth
Mathematics
1 answer:
borishaifa [10]3 years ago
7 0
Given a parallelogram ABCD with the measure of segment BC = (6 - x) units, the measure of segment AD = (x + 2) units.
Angle A measures (100 - y) degrees and angle C measures (12 + y) degrees.

Recall that for a parallelogram, <span>the opposite sides are of equal length and the opposite angles are of equal measure.

From the given parallelogram, segment BC is opposite to segment AD and angle A is opposite to angle C.

Thus, 6 - x = x + 2
2x = 4
x = 2

Therefore, the value of x is 2 and the length of segment BC is 6 - 2 = 4 units.

Similarly, 100 - y = 12 + y
2y = 88
y = 44

Therefore, value of y is 44 and the measure of angle DAB is 100 - 44 = 56 degrees.
</span>
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X^2-X-6=(X-3)(X+2)
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Which just leaves (x-3)
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5 0
3 years ago
4TH TIME ASKING THIS!!! Please help me! Someone pleaseeee. I need the correct answers. I don’t want to fail
Alexeev081 [22]

Answer:

The functions are inverses; f(g(x)) = x ⇒ answer D

h^{-1}(x)=\sqrt{\frac{x+1}{3}} ⇒ answer D

Step-by-step explanation:

* <em>Lets explain how to find the inverse of a function</em>

- Let f(x) = y

- Exchange x and y

- Solve to find the new y

- The new y = f^{-1}(x)

* <em>Lets use these steps to solve the problems</em>

∵ f(x)=\sqrt{x-3}

∵ f(x) = y

∴ y=\sqrt{x-3}

- Exchange x and y

∴ x=\sqrt{y-3}

- Square the two sides

∴ x² = y - 3

- Add 3 to both sides

∴ x² + 3 = y

- Change y by f^{-1}(x)

∴ f^{-1}(x)=x^{2}+3

∵ g(x) = x² + 3

∴ f^{-1}(x)=g(x)

∴ <u><em>The functions are inverses to each other</em></u>

* <em>Now lets find f(g(x))</em>

- To find f(g(x)) substitute x in f(x) by g(x)

∵ f(x)=\sqrt{x-3}

∵ g(x) = x² + 3

∴ f(g(x))=\sqrt{(x^{2}+3)-3}=\sqrt{x^{2}+3-3}=\sqrt{x^{2}}=x

∴ <u><em>f(g(x)) = x</em></u>

∴ The functions are inverses; f(g(x)) = x

* <em>Lets find the inverse of h(x)</em>

∵ h(x) = 3x² - 1 where x ≥ 0

- Let h(x) = y

∴ y = 3x² - 1

- Exchange x and y

∴ x = 3y² - 1

- Add 1 to both sides

∴ x + 1 = 3y²

- Divide both sides by 3

∴ \frac{x + 1}{3}=y^{2}

- Take √ for both sides

∴ ± \sqrt{\frac{x+1}{3}}=y

∵ x ≥ 0

∴ We will chose the positive value of the square root

∴ \sqrt{\frac{x+1}{3}}=y

- replace y by h^{-1}(x)

∴ h^{-1}(x)=\sqrt{\frac{x+1}{3}}

4 0
3 years ago
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IgorLugansk [536]

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4 0
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6 0
3 years ago
Read 2 more answers
AB is tangent to circle D. Find the value of x.
Svet_ta [14]

Answer:

c     15

Step-by-step explanation:

Tangent AB is tangent to circle D at point A. The angle made by a tangent to a circle and a radius of the circle at the point of tangency is a right angle. That means that angle is a right angle, and triangle ABD is a right triangle.

We can use the Pythagoren theorem.

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x^2 + 400 = x^2 + 10x + 10x + 100

400 = 20x + 100

20x = 300

x = 15

6 0
3 years ago
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