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3 years ago

Help ASAP!!!!! ;))) tyyy

Mathematics

1 answer:

hodyreva [135]3 years ago

3 0

3. dude these are easy. I think you got it.

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The functions f(x) = −(x − 1)2 5 and g(x) = (x 2)2 − 3 have been rewritten using the completing-the-square method. apply your kn

ale4655 [162]

The vertex of the function f(x) exists (1, 5), the vertex of the function g(x) exists (-2, -3), and the vertex of the function f(x) exists maximum and the vertex of the function g(x) exists minimum.

<h3>How to determine the vertex for each function is a minimum or a maximum? </h3>

Given:

$\mathrm{f}(\mathrm{x})=-(\mathrm{x}-1)^{2}+5$ and

$\mathrm{g}(\mathrm{x})=(\mathrm{x}-2)^{2}-3$

The generalized equation of a parabola in the vertex form exists

$$y=a(x-h)^{2}+k$

Vertex of the function f(x) exists (1, 5).

Vertex of the function g(x) exists (-2, -3).

Now, if (a > 0) then the vertex of the function exists minimum, and if (a < 0) then the vertex of the function exists maximum.

The vertex of the function f(x) exists at a maximum and the vertex of the function g(x) exists at a minimum.

To learn more about the vertex of the function refer to:

brainly.com/question/11325676

#SPJ4

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2 years ago

A student estimated a mass to be 250 grams but upon carefully measuring it, he found the actual mass to be 240 grams. What is th

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3 years ago

Your are 5 feet tall. What is your height in meters?i

Svetach [21]

Answer: 5 ft = 1.5240 m

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3 years ago

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2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

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3 years ago

Stacy owes $40 on her credit card. After she made a payment, her new balance was $10. What was the amount of her

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Answer:

Step-by-step explanation:

x + 40 = 10

x = -30

She paid $30 on her credit card

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