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nirvana33 [79]
3 years ago
9

Find the slope of the line through the points(5,-1) and (-2,-3)

Mathematics
1 answer:
Talja [164]3 years ago
7 0

Answer:

\frac{2}{7}

Step-by-step explanation:

Plug in points to equation m =  \frac{ y_2 - y _ 1}{x_ 2 - x_ 1}

m =  \frac{ -3 - -1}{-2 - 5}

Subtract to get \frac{2}{7}

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Find the exact values of the remaining five trigonometric functions of o satisfying the given conditions.
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Recall the Pythagorean identity:

sin²(<em>θ</em>) + cos²(<em>θ</em>) = 1

Then

sin(<em>θ</em>) = -√(1 - cos²(<em>θ</em>)) = -4/5

and so

tan(<em>θ</em>) = (-4/5) / (3/5) = -4/3

The remaining trig ratios are just reciprocals of the ones found already:

sec(<em>θ</em>) = 1/cos(<em>θ</em>) = 5/3

csc(<em>θ</em>) = 1/sin(<em>θ</em>) = -5/4

cot(<em>θ</em>) = 1/tan(<em>θ</em>) = -3/4

4 0
3 years ago
Given that ABCD is a rhombus. Prove ∠1≅∠2.
natka813 [3]

Answer:

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4) SSS

5) CPCTC

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3 years ago
A number is k units to the left of 0 on the number line . Describe the location of its opposite
shusha [124]
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LekaFEV [45]
The zero product property tells us that if the product of two or more factors is zero, then each one of these factors CAN be zero.

For more context let's look at the first equation in the problem that we can apply this to: (x-3)(x+4)=0

Through zero property we know that the factor (x-3) can be equal to zero as well as (x+4). This is because, even if only one of them is zero, the product will immediately be zero.

The zero product property is best applied to factorable quadratic equations in this case.

Another factorable equation would be 2x^{2}+6x=0 since we can factor out 2x and end up with 2x(x+3)=0. Now we'll end up with two factors, 2x and (x+3), which we can apply the zero product property to.

The rest of the options are not factorable thus the zero product property won't apply to them.
3 0
4 years ago
Read 2 more answers
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Nataly_w [17]

Answer: C

Step-by-step explanation:

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Now transform the expression using a^m/n = n root a raised to a power of m.

And that's how you get your answer.

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