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3 years ago

Please help me ASAP!!

Mathematics

2 answers:

Sergio [31]3 years ago

7 0

Answer:

x = 16

Step-by-step explanation:

The sum of measures of these three angles of any triangle is equal to 180 °

54 + 90 + (2x + 4) = 180

144 + 2x + 4 = 180

148 + 2x = 180

2x = 180 - 148

2x = 32

$x= \frac{32}{2} \\\\x=16$

Olin [163]3 years ago

7 0

Focus on triangle KLM. Since JKLM is a rectangle, angle KLM is right. Since the sum of the interior angles of every triangle is 180, we deduce that

$\hat{K}+\hat{L}+\hat{M}=180$

Now plug the values for each angle and you have

$54+90+(2x+4)=180$

Move 54 and 90 to the right side to get

$2x+4=36 \iff 2x=32 \iff x=16$

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

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2 years ago

What is the answer for -|-5x(-7)|

Airida [17]

Answer will be -35

Since in an absolute equation, the inside will all be positive. So get rid of the negatives. 5x7. Then, outside of the absolute sign, there is a negative sign. Since that is outside, you cannot turn it into positive. Now, the answer is -35.

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2 years ago

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The formula for the resistance, R, of a conductor with voltage, V, and current , I , is R= V/I. Solve for V.

borishaifa [10]

Answer:

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Step-by-step explanation:

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2 years ago

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