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mrs_skeptik [129]
3 years ago
12

Helpppppp meeeeee plzzzzzzz!!!!!

Mathematics
1 answer:
loris [4]3 years ago
3 0
50 is the answer if 30 is 60% of the club that means 10 is 20% and you just skip count until it reaches 100%
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How many faces does the solid made from the net below have?<br>​
nikdorinn [45]

Answer:We need to see the net.

Step-by-step explanation:

5 0
3 years ago
How is this solved using trig identities (sum/difference)?
GenaCL600 [577]
FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

cos \beta = -\frac{1}{2}  \sqrt{3}

tan \beta= \frac{1}{3}  \sqrt{3}

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2}  \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )
sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )
sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )
cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )
cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )

Find tan (α - β)
tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha  tan \beta }
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3}   }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}

Simplify the denominator
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3}   }{1+(\frac{5\sqrt{3}}{24})}
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }

Simplify the numerator
tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }
tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }

Simplify the fraction
tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})
tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}

7 0
3 years ago
View the lane below and calculate the slope<br> by using the slope formula
velikii [3]

Answer:

slope = 4

Step-by-step explanation:

Calculate the slope m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (0, 0) and (x₂, y₂ ) = (3, 12) ← 2 points on the line

m = \frac{12-0}{3-0} = \frac{12}{3} = 4

6 0
2 years ago
Read 2 more answers
Help me please !<br> 4,5 and 6
andreev551 [17]

9514 1404 393

Answer:

  4a. ∠V≅∠Y

  4b. TU ≅ WX

  5. No; no applicable postulate

  6. see below

Step-by-step explanation:

<h3>4.</h3>

a. When you use the ASA postulate, you are claiming you have shown two angles and the side between them to be congruent. Here, you're given side TV and angle T are congruent to their counterparts, sides WY and angle W. The angle at the other end of segment TV is angle V. Its counterpart is the other end of segment WY from angle W. In order to use ASA, we must show ...

  ∠V≅∠Y

__

b. When you use the SAS postulate, you are claiming you have shown two sides and the angle between them are congruent. The angle T is between sides TV and TU. The angle congruent to that, ∠W, is between sides WY and WX. Then the missing congruence that must be shown is ...

  TU ≅ WX

__

<h3>5.</h3>

The marked congruences are for two sides and a non-included angle. There is no SSA postulate for proving congruence. (In fact, there are two different possible triangles that have the given dimensions. This can be seen in the fact that the given angle is opposite the shortest of the given sides.)

  "No, we cannot prove they are congruent because none of the five postulates or theorems can be used."

__

<h3>6.</h3>

The first statement/reason is always the list of "given" statements.

1. ∠A≅∠D, AC≅DC . . . . given

2. . . . . vertical angles are congruent

3. . . . . ASA postulate

4. . . . . CPCTC

8 0
3 years ago
Which is equal to (2x − 3y + 5z) − (−4x − y + 8z)?
Fittoniya [83]
(2x - 3y + 5z) - (-4x - y + 8z)
= 2x - 3y + 5z + 4x + y - 8z
= 6x - 2y - 3z

In short, Your Answer would be Option D

Hope this helps!
8 0
3 years ago
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