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fenix001 [56]
3 years ago
10

3+5 plz help will gibe brain

Mathematics
2 answers:
ololo11 [35]3 years ago
7 0
8 because i said so and i’m smart
LUCKY_DIMON [66]3 years ago
4 0

Answer:

8

Step-by-step explanation:

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3a+a+2a-5a? <br><br><br> Can someone give me the answer please? <br><br> Ty
Shkiper50 [21]

Answer:

a

Step-by-step explanation:

=3a + a + 2a - 5a

=4a - 3a

= a

hope it helps u tate!

5 0
3 years ago
Read 2 more answers
An acute angle, θ, is in a right triangle such that cos θ = 12/13. What is the Value of csc θ? (100 Points!)
Hatshy [7]

Answer:

csc θ = 13/12

Step-by-step explanation:

cos θ = 12/13

Formula: cos θ = 1/cscθ

Its just the reciprocal.

solve:

1/csc θ = 12/13

csc θ = 13/12

4 0
2 years ago
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Can anyone help ?!<br> Please
Vaselesa [24]

Answer:

MARK AS BRAINLIST

Step-by-step explanation:

x=v*(y-3)

x=vy/4w

x=(3+w)/v

x=(y-6w)/v

6 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
Evaluate 0 · |-6|. -6 0 6
natita [175]

Answer:

0

Step-by-step explanation:

0 · |-6|

Absolute value means take the non negative number

0* 6

0

7 0
3 years ago
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