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3 years ago

13

Graph triangle xyz with vertices x (0,-4) y (4,-1) and z (6,-3) on the coordinate grid

1 answer:

marishachu [46]3 years ago

5 0

How are we supposed to graph this, it's easy the first number in the parenthesis is where it is on the x-axis, the second number in there is where it is at on the y-axis, that should help you so you can graph it yourself.

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urgente ordenar de mas grande a mas pequeño estos numeros: 85,94 85,490 , 85,93, 85,035 ,85

Ilia_Sergeevich [38]

Largest to Smallest:-

First lets write all the numbers down. 45, 94, 85, 490, 85, 93, 85, 035, 85.

035 is 35.

So lets order them now.

The biggest number is 490 and smallest is 35 (035).

490 - 94 - 93 - 85 - 85 - 85 - 85 - 035 (35)

4 0

3 years ago

Read 2 more answers

Here are 5 cards with a picture of a rose and 3 cards with a picture of a daisy. Alia keeps all the cards down on a table with t

Sergio039 [100]

4 over 7 because, she takes away a rose card from the total 8 cards she has, but also takes away a card from the total amount of cards.

Then again, I might be wrong :P

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3 years ago

Photography World has a portrait special. A three-pose portrait package has a sitting fee of $35, and a six-pose package has a s

ki77a [65]

We know that a three-pose portrait package has a sitting fee of $35 and <span> a six-pose package has a sitting fee of $60. The total amount they collect is $690.

</span> $\text{Therefore, the equation can be}$ $35x + 60y = 690$

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3 years ago

What is the answer plzzzzzz help a kid out

Komok [63]

Answer:

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3 years ago

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:

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

3 0

3 years ago