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3 years ago

Which of the following is a correct tangent ratio for the figure?

Mathematics

1 answer:

kow [346]3 years ago

5 0

Answer:

C) tan(39°) = 11/15

Step-by-step explanation:

SohCahToa

tangent = opposite / adjacent

The given angle is 39°. The angle directly opposite of 39° is 11 and the angle adjacent to 39° is 15.

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A bus travels on an east-west highway connecting two cities A and B that are 100 miles apart. There are 2 services stations alon

melamori03 [73]

Answer:

51/4

Step-by-step explanation:

To begin with you have to understand what is the distribution of the random variable. If X represents the point where the bus breaks down. That is correct.

X~ Uniform(0,100)

Then the probability mass function is given as follows.

$f(x) = P(X=x) = 1/100 \,\,\,\, \text{if} \,\,\,\, 0 \leq x \leq 100\\f(x) = P(X=x) = 0 \,\,\,\, \text{otherwise}$

Now, imagine that the D represents the distance from the break down point to the nearest station. Think about this, the first service station is 20 meters away from city A, and the second station is located 70 meters away from city A then the mid point between 20 and 70 is (70+20)/2 = 45 then we can represent D as follows

$D(x) =\left\{ \begin{array}{ll} x & \mbox{if } 0\leq x \leq 20 \\ x-20 & \mbox{if } 20\leq x < 45\\ 70-x & \mbox{if } 45 \leq x \leq 70\\ x-70 & \mbox{if } 70 \leq x \leq 100\\ \end{array}\right.$

Now, as we said before X represents the random variable where the bus breaks down, then we form a new random variable $Y = D(X)$ , $Y$ is a random variable as well, remember that there is a theorem that says that

$E[Y] = E[D(X)] = \int\limits_{-\infty}^{\infty} D(x) f(x) \,\, dx$

Where $f(x)$ is the probability mass function of X. Using the information of our problem

$E[Y] = \int\limits_{-\infty}^{\infty} D(x)f(x) dx \\= \frac{1}{100} \bigg[ \int\limits_{0}^{20} x dx +\int\limits_{20}^{45} (x-20) dx +\int\limits_{45}^{70} (70-x) dx +\int\limits_{70}^{100} (x-70) dx \bigg]\\= \frac{51}{4} = 12.75$

3 0

3 years ago

10x - 80 = 16x - 12 <br> what is x?

AnnyKZ [126]

I just did it on paper but hope that helps:)

7 0

3 years ago

I need help with this question please answer asap

saveliy_v [14]

Answer:

9 months

Step-by-step explanation:

8 sculptures in 6 months is 1 sculpture in 0.75 months.

8/8=1

6/8=0.75

1×12=12

0.75×12=9

8 0

3 years ago

Please help me with the worksheet

aivan3 [116]

Answer:

9) a = ¾, <u>vertex</u>: (-4, 2), <u>Equation</u>: y = ¾|x + 4| + 2

10) a = ¼, <u>vertex</u>: (0, -3), <u>Equation</u>: y = ¼|x - 0| - 3

11) a = -4, <u>vertex</u>: (3, 1), <u>Equation</u>: y = -4|x - 3| + 1

12) a = 1, <u>vertex</u>: (-2, -2), <u>Equation</u>: y = |x + 2| - 2

Step-by-step explanation:

I could <u><em>only</em></u> work on questions 9, 10, 11, 12 in accordance with Brainly's rules. Nevertheless, the techniques demonstrated in this post applies to all of the given problems in your worksheet.

<h2><u>Definitions:</u></h2>

The given set of graphs are examples of absolute value functions. The <u>general form</u> of absolute value functions is: y = a|x – h| + k, where:

|a| = determines the vertical stretch or compression factor (wideness or narrowness of the graph).

(h, k) = vertex of the function

x = h represents the axis of symmetry.

<h2><u>Solutions:</u></h2><h3>Question 9) ⇒ Vertex: (-4, 2)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (0, 5) and substitute into the general form (equation):

y = a|x – h| + k

5 = a| 0 - (-4)| + 2

5 = a| 0 + 4 | + 2

5 = a|4| + 2

5 = 4a + 2

Subtract 2 from both sides:

5 - 2 = 4a + 2 - 2

3 = 4a

Divide both sides by 4 to solve for <em>a</em>:

$\LARGE\mathsf{\frac{3}{4}\:=\:\frac{4a}{4}}$

a = ¾

Therefore, given the value of a = ¾, and the vertex, (-4, 2), then the equation of the absolute value function is:

<u>Equation</u>: y = ¾|x + 4| + 2

<h3>Question 10) ⇒ Vertex: (0, -3)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (4, -2) and substitute into the general form (equation):

y = a|x – h| + k

-2 = a|4 - 0| -3

-2 = a|4| - 3

-2 = 4a - 3

Add 3 to both sides:

-2 + 3 = 4a - 3 + 3

1 = 4a

Divide both sides by 4 to solve for <em>a</em>:

$\LARGE\mathsf{\frac{1}{4}\:=\:\frac{4a}{4}}$

a = ¼

Therefore, given the value of a = ¼, and the vertex, (0, -3), then the equation of the absolute value function is:

<u>Equation</u>: y = ¼|x - 0| - 3

<h3>Question 11) ⇒ Vertex: (3, 1)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (4, -3) and substitute into the general form (equation):

y = a|x – h| + k

-3 = a|4 - 3| + 1

-3 = a|1| + 1

-3 = a + 1

Subtract 1 from both sides to isolate <em>a</em>:

-3 - 1 = a + 1 - 1

a = -4

Therefore, given the value of a = -4, and the vertex, (3, 1), then the equation of the absolute value function is:

<u>Equation</u>: y = -4|x - 3| + 1

<h3>Question 12) ⇒ Vertex: (-2, -2)</h3>

<u>Solve for a:</u>

In order to solve for the value of <em>a</em>, choose another point on the graph, (-4, 0) and substitute into the general form (equation):

y = a|x – h| + k

0 = a|-4 - (-2)| - 2

0 = a|-4 + 2| - 2

0 = a|-2| - 2

0 = 2a - 2

Add 2 to both sides:

0 + 2 = 2a - 2 + 2

2 = 2a

Divide both sides by 2 to solve for <em>a</em>:

$\LARGE\mathsf{\frac{2}{2}\:=\:\frac{2a}{2}}$

a = 1

Therefore, given the value of a = -1, and the vertex, (-2, -2), then the equation of the absolute value function is:

<u>Equation</u>: y = |x + 2| - 2

5 0

2 years ago

Solve<br> 5t < -15 <br><br> T< 3<br> T< -3<br> T> 3<br> T > -3

OLga [1]

Answer:

Step-by-step explanation:

5t < -15 (divide by 5 on each side)

t < -3

7 0

3 years ago

Read 2 more answers