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insens350 [35]
3 years ago
14

Can someone please explain to me how to solve this problem? I have the answer already all I need is to know how to solve the pro

blem.

Mathematics
2 answers:
omeli [17]3 years ago
5 0

Answer:

Step-by-step explanation:

you have two supplementary angles,

both added up=180

7y+5y=12y

12y=180y=15

angle DBC=7y=15*7=105

goldenfox [79]3 years ago
3 0

Answer:

Step-by-step explanation:

You know that angle DBA = 180 degrees.

Therefore you can write an equation:

7y+5y=180

12y=180

y=15

Now, you know that y = 15, you can multiply 7 * 15 = 105 degrees

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Answer:

y=40

Step-by-step explanation:

A direct variation is a relationship between two numbers where the output equals the input times a constant.

The equation for a direct variation can be written as

y=kx

We can figure out the y value of the second coordinate by finding out what the constant (k) is.  We can do that by plugging in the first coordinates into the equation.

y = kx

10 = 6k

10/6 = k

We can solve for y by plugging this constant, and the second coordinate into the equation.

y=kx

y=10/6(24)

y=40

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4 years ago
Given: ∠1 and ∠2 are complementary angles.<br> Prove: ∠3 and ∠4 are complementary angles.
Sergeeva-Olga [200]

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Step-by-step explanation:

If the measures of angles 1 and 2 are congruent, then the measures of 3 and 4 should also be congruent because they are vertical angles.

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3 years ago
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4 years ago
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Look at the picture.

r=\sqrt{x^2+y^2}

\sin\theta=\dfrac{y}{r}\\\\\cos\theta=\dfrac{x}{r}\\\\\tan\theta=\dfrac{y}{x}\\\\\cot\theta=\dfrac{x}{y}\\\\\sec\theta=\dfrac{r}{x}\\\\\csc\thera=\dfrac{r}{y}

We have the point \left(\dfrac{8}{17},\ \dfrac{15}{17}\right)\to x=\dfrac{8}{17},\ y=\dfrac{15}{17}.

Calculate r:

r=\sqrt{\left(\dfrac{8}{17}\right)^2+\left(\dfrac{15}{17}\right)^2}=\sqrt{\dfrac{64}{289}+\dfrac{225}{289}}=\sqrt{\dfrac{289}{289}}=\sqrt1=1

\sin\theta=\dfrac{\frac{15}{17}}{1}=\dfrac{15}{17}\\\\\cos\theta=\dfrac{\frac{8}{17}}{1}=\dfrac{8}{17}\\\\\tan\theta=\dfrac{\frac{15}{17}}{\frac{8}{17}}=\dfrac{15}{17}:\dfrac{8}{17}=\dfrac{15}{17}\cdot\dfrac{17}{8}=\dfrac{15}{8}\\\\\cot\theta=\dfrac{\frac{8}{17}}{\frac{15}{17}}=\dfrac{8}{17}:\dfrac{15}{17}=\dfrac{8}{17}\cdot\dfrac{17}{15}=\dfrac{8}{15}\\\\\sec\theta=\dfrac{1}{\frac{8}{17}}=1:\dfrac{8}{17}=\dfrac{17}{8}\\\\\csc\theta=\dfrac{1}{\frac{15}{17}}=1:\dfrac{15}{17}=\dfrac{17}{15}

5 0
4 years ago
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