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3 years ago

Plssss help (ALGEBRA 1 ) will mark brainlest

Mathematics

2 answers:

Dimas [21]3 years ago

7 0

Answer:

I'm confused by the layout of the first two, but I can help you with the last one!

The last one is C!

Step-by-step explanation:

Keith_Richards [23]3 years ago

3 0

Answer:

B) (x+2)

D) x-1

B) x^2+2x+1

Step-by-step explanation:

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Ralph sold brownies for $0.44 a piece in order to earn money to buy baseball cards. If the cards cost $0.60 per pack, and if

Dima020 [189]

He had to sell 16 brownies

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3 years ago

Cómo resolver sopa de números encontrando resultados de divisiones por ejemplo 436/2

Rzqust [24]

Answer:

Lo resuelves como de costumbre :)

Step-by-step explanation:

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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:

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

3 0

3 years ago

Pls help I need a good grade

Nikitich [7]

Answer:

If it is number 5. the answer is 13

Step-by-step explanation:

take the largest number and subtract the smallest number.

7 0

2 years ago

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The lengths of two sides of a right triangle are 5 inches and 8 inches. What is the difference between the two possible

3241004551 [841]

Answer:

Option 2 - 3.2 inches.

Step-by-step explanation:

Given : The lengths of two sides of a right triangle are 5 inches and 8 inches.

To find : What is the difference between the two possible lengths of the third side of the triangle?

Solution :

According to question, it is a right angle triangle

Applying Pythagoras theorem,

$H^2=P^2+B^2$

Where, H is the hypotenuse the longer side of the triangle

P is the perpendicular

B is the base

Assume that H=8 inches and B = 5 inches

Substitute the value in the formula,

$8^2=P^2+5^2$

$64=P^2+25$

$P^2=64-25$

$P^2=39$

$P=\sqrt{39}$

$P=6.24$

Assume that P=8 inches and B = 5 inches

Substitute the value in the formula,

$H^2=8^2+5^2$

$H^2=64+25$

$H^2=89$

$H=\sqrt{89}$

$H=9.43$

Therefore, The possible length of the third side of the triangle is

$L=H-P$

$L=9.43-6.24$

$L=3.19$

Therefore, The difference between the two possible lengths of the third side of the triangle is 3.2 inches.

So, Option 2 is correct.

3 0

3 years ago

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