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3 years ago

He figure PQRSTU represents the shape of the parking lot at a shopping mall. What is the area of the parking lot?

Mathematics

1 answer:

Murljashka [212]3 years ago

7 0

Answer:

I think is c

Step-by-step explanation:

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Graph the solution to the inequality 3(x+1)-3<21

kondor19780726 [428]

Answer is x<7 and the graph is in the attachment

5 0

4 years ago

Read 2 more answers

At a small liberal arts college, students can register for one to six courses. Let be the number of courses taken in the fall by

Maslowich

According to the probability distribution, there is a 0% probability that a randomly selected student earns more than 18 credits.

A student earns 3 credits per course.

Thus, the probability distribution for the number of credits earned is:

$P(X = 3) = 0.06$

$P(X = 6) = 0.06$

$P(X = 9) = 0.12$

$P(X = 12) = 0.2$

$P(X = 15) = 0.41$

$P(X = 18) = 0.15$

Since no person takes more than 6 courses, there is a 0% probability that a randomly selected student earns more than 18 credits.

A similar problem is given at brainly.com/question/25117113

5 0

3 years ago

Helppppppp! Right quick

Nesterboy [21]

Answer:

LN = 5x + 9

Step-by-step explanation:

LN = x + 9 + 4x

LN = 5x + 9

4 0

3 years ago

7/10-1/2= what is seven tenths minus one half

Setler [38]

You could make them have the same denominator, making it 7/10-5/10=2/10.
The answer is 2/10, or simplified, 1/5.

5 0

3 years ago

If sin theta = 2/5 and theta is in quadrant I, determine the following.

anzhelika [568]

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=\stackrel{hypotenuse}{5}\\ a=adjacent\\ b=\stackrel{opposite}{2}\\ \end{cases} \\\\\\ \pm\sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a\implies \stackrel{I~Quadrant}{+\sqrt{21}=a}$

recall that cosine is positive on the I Quadrant, so though we get a ± valid roots, only the positive one applies.

$\bf cos(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{hypotenuse}{5}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{\sqrt{21}}} \qquad \qquad cot(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{opposite}{2}} \\\\\\ csc(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{opposite}{2}} \qquad \qquad sec(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{\sqrt{21}}}$

now, for tangent and secant, let's rationalize the denominator.

$\bf tan(\theta)\implies \cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{\sqrt{21}}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{21} \\\\\\ sec(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{\sqrt{21}}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{5\sqrt{21}}{21}$

5 0

3 years ago