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3 years ago

Can someone help me with this please

Mathematics

1 answer:

alexgriva [62]3 years ago

6 0

The answer to the question is c

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Find the area of the figure. 12 ft 18 ft 18 ft 36 ft

Nataly_w [17]

Answer:

139968

Step-by-step explanation:

Multiply the all of the numbers together and you have your answer (:

7 0

3 years ago

8(− 2 3 x + 6) = −9( 3 4 x − 8)

Travka [436]

Answer:

x=12/61

Step-by-step explanation:

subtract 48 from both sides

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add 306 x to both sides

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4 0

3 years ago

Which of the following inequalities represents all values of x for which the quotient below is defined? sqrt 21(x-1) divided by

Triss [41]

Answer:

the answer to your question is D

4 0

3 years ago

If the hypotenuse of a right triangle is 2 <img src="https://tex.z-dn.net/?f=%5Csqrt%7B3%7D" id="TexFormula1" title="\sqrt{3}" a

Svetradugi [14.3K]

Answer:

Step-by-step explanation:

a = the other leg:

$a^{2} =(2\sqrt{3} )^{2} -(\sqrt{3}) ^{2} =(4)(3)-3=12-3$

$a^{2} =9$

$a=\sqrt{9}$

$a=3$

Hope this helps

5 0

2 years ago

Find values of  that satisfy the equation for 0º    360º and 0  θ  2 . Give answers in degrees and radians.

suter [353]

Answer:

$\theta = 30^\circ, 330^\circ$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

Step-by-step explanation:

Given [Missing from the question]

Equation:

$cos\theta = \frac{\sqrt 3}{2}$

Interval:

$0 \le \theta \le 360$

$0 \le \theta \le 2\pi$

Required

Determine the values of $\theta$

The given expression:

$cos\theta = \frac{\sqrt 3}{2}$

... shows that the value of $\theta$ is positive

The cosine of an angle has positive values in the first and the fourth quadrants.

So, we have:

$cos\theta = \frac{\sqrt 3}{2}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{\sqrt 3}{2})$

$\theta = 30$ --- In the first quadrant

In the fourth quadrant, the value is:

$\theta = 360 -30$

$\theta = 330$

So, the values of $\theta$ in degrees are:

$\theta = 30^\circ, 330^\circ$

Convert to radians (Multiply both angles by $\pi/180$ )

So, we have:

$\theta = \frac{30 * \pi}{180}, \frac{330 * \pi}{180}$

$\theta = \frac{\pi}{6}, \frac{33 * \pi}{18}$

$\theta = \frac{\pi}{6}, \frac{11 * \pi}{6}$

$\theta = \frac{\pi}{6}, \frac{11\pi}{6}$

8 0

3 years ago

Can someone help me with this please ​

Can someone help me with this please