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3 years ago

Which number represents a temperature below zero? –4° F 0° F 4° F 6° F

Mathematics

2 answers:

makkiz [27]3 years ago

8 0

The anwser is the first one you put , -4 is below zero

lidiya [134]3 years ago

6 0

-4 degrees Fahreheit

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The statement tan theta -12/5, csc theta -13/5, and the terminal point determined by theta is in quadrant 2."

Harlamova29_29 [7]

Answer:

Answer C:

Cannot be true because $csc(\theta)$ is greater than zero in quadrant 2.

Step-by-step explanation:

When the csc of an angle is negative, since the cosecant function is defined as:

$csc(\theta)=\frac{1}{sin(\theta)}$

that means that the sin of the angle must be negative, and such cannot happen in the second quadrant. The sine function is positive in the first and second quadrant.

Therefore, the correct answer is:

Cannot be true because $csc(\theta)$ is greater than zero in quadrant 2.

6 0

3 years ago

Please answer this simple question!!!

vladimir1956 [14]

Just substitute the value of x from equation 2 in equation 1,
6y = x - 12
6y = 22 - 12
y= 10/6
y = 5/3

In short, Your Answer would be Option Option B) (22, 5/3)

Hope this helps!

6 0

3 years ago

HELP: Will Award Brainliest! Simplify 17x - 12 = 114 + 3x

mrs_skeptik [129]

Answer:

x=9

Step-by-step explanation:

17x - 12 = 114 + 3x

First Subtract 3x from both sides

14x - 12 = 114

Add 12 to both sides

14x=126

Divide by GCF (14)

x=9

Hope this helps! Plz award Brainliest : )

5 0

3 years ago

Read 2 more answers

_____ compare one quantity to 100.

Alexxx [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

part 2. Find the value of the trig function indicated, use for that Pythagorean theorem to find the third side if you need it.

andreev551 [17]

Answer: $\bold{5)\ \cos \theta=\dfrac{\sqrt{11}}{6}\qquad 6)\ \tan \theta =\dfrac{8}{17}\qquad 7)\ \cos \theta = \dfrac{4}{3}\qquad 8)\ \cos \theta = \dfrac{\sqrt{10}}{10}}$

<u>Step-by-Step Explanation:</u>

Pythagorean Theorem is: a² + b² = c² , <em>where "c" is the hypotenuse</em>

$5)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{3\sqrt{11}}{18}\quad \rightarrow \large\boxed{\dfrac{\sqrt{11}}{6}}$

Note: (15)² + (3√11)² = hypotenuse² → hypotenuse = 18

$6)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{8}{17}\quad =\large\boxed{\dfrac{8}{17}}$

Note: 8² + 15² = hypotenuse² → hypotenuse = 17

$7)\ \tan \theta=\dfrac{\text{side opposite to}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{20}{15}\quad \rightarrow \large\boxed{\dfrac{4}{3}}$

Note: hypotenuse not needed for tan

$8)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{2}{2\sqrt{10}}\quad =\large\boxed{\dfrac{\sqrt{10}}{10}}$

Note: 2² + 6² = hypotenuse² → hypotenuse = 2√10

8 0

2 years ago