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Lostsunrise [7]
4 years ago
15

Given right triangle MNO, which represents the value of cos(M)?

Mathematics
2 answers:
laila [671]4 years ago
7 0

Answer:

Option B. MN/MO

Step-by-step explanation:

In a right angle triangle cosine of any acute angle = Base/ Hypotenuse

In the ∠ NMO Hypotenuse is MO and base of the triangle is MN.

Therefore cos (M) = Base / Hypotenuse = MN / MO

Therefore option B is the right answer.

zmey [24]4 years ago
3 0

Answer:  OPPTION B

Step-by-step explanation:

Given a right triangle, the cosine of an angle is:

cos\alpha=adjacent/hypotenuse

Therefore, given the triangle MNO, you can conclude that cos(M) is the one shown below:

adjacent=MN

hypotenuse=MO

Then:

cos(M)=MN/MO

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If p(x) = x2 – 1 and q (x) = 5 (x minus 1), which expression is equivalent to (p – q)(x)?
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5 0
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Read 2 more answers
Which is the solution set to the given inequality |x+3|&lt;12? x infinity(-15,9), x infinity {-15,9], x infinity (00,-15)U(9,00)
Viefleur [7K]

Answer:

Part 1) The solution set is (-15,∞) ∩ (-∞,9)=(-15,9)

Part 2) The ordered pair (1,3) is a solution of the system

Step-by-step explanation:

Part 1) we have

\left|x+3\right|

<u>First solution case Positive</u>

+(x+3)

x

x

The solution first case is the interval -------> (-∞,9)

<u>Second solution case Negative</u>

-(x+3)

-x-3

-x

-x ------> Multiply by -1 both sides

x>-15

The solution second case is the interval -------> (-15,∞)

The solution set is equal to

(-15,∞) ∩ (-∞,9)=(-15,9)  

Part 2) we have

y>-2 -------> inequality A

x+y\leq 4 -----> inequality B

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities

Verify each case

case a) (1,5)

Substitute the value of x and the value of y in the inequality and then compare

<u>Inequality A</u>

5>-2 ------> is true

<u>Inequality B</u>

1+5\leq 4

6\leq 4 -----> is not true

therefore

the ordered pair is not a solution

case b) (0,5)

Substitute the value of x and the value of y in the inequality and then compare

<u>Inequality A</u>

5>-2 ------> is true

<u>Inequality B</u>

0+5\leq 4

5\leq 4 -----> is not true

therefore

the ordered pair is not a solution

case c) (-2,-3)

Substitute the value of x and the value of y in the inequality and then compare

<u>Inequality A</u>

-3>-2 ------> is not true

therefore

the ordered pair is not a solution

case d) (1,3)

Substitute the value of x and the value of y in the inequality and then compare

<u>Inequality A</u>

3>-2 ------> is true

<u>Inequality B</u>

1+3\leq 4

4\leq 4 -----> is true

therefore

the ordered pair is  a solution

6 0
3 years ago
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