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3 years ago

Pls help :(

Mathematics

1 answer:

user100 [1]3 years ago

4 0

Answer:

A candle holder holds a candle

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A book club charges an initial joining fee of $18.00. The cost per book is $10.25. The graph shows the cost of belonging to the

Alex73 [517]

The question doesn't show any graph, but I have included it to help you with the problem.

Answer:

The new graph will be shifted up by 3 units

Step-by-step explanation:

Graph of a Real Function

The total cost of belonging to the book club depends on one fixed cost, the initial joining fee, and a variable cost, depending on the number of books purchased. Call C the total cost and x the number of purchased books, then

$C=18+10.25x$

By giving x some values, we get the corresponding values of C and then we can plot the graph as a blue line, in the image provided below.

If the joining fee goes up by $3, the new function is

$C'=18+10.25x+3$

$C'=21+10.25x$

The graph is also shown in the image, but with a red line. We can see the graph is shifted up by 3 units in the y-axis, but the slope remains the same, as the cost per book hasn't changed

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3 years ago

I'm having trouble with unit rates. Please help.

dezoksy [38]

The answer to your question is "C. 4 feet/ 1 second."

When the rate has a quantity of 1, it is a unit rate. So if the rate is 4 feet to 1 second (or four feet per second), it would be considered a unit rate.

5 0

3 years ago

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How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

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3 years ago

Find the surface area of a cylinder with a radius of 5 cm and a height of 30 cm. Use 3.14

Juli2301 [7.4K]

Whole formula 2*pi*radius^2 + 2*pi*radius*height:

157+942= 1099

SA= 1099

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3 years ago

The measure of an angle formed by two secants intersecting inside the circle equals...

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The measure of an angle formed by two secants intersecting inside the circle equals ½ the difference of the intercepted arcs. The correct option among all the options that are given in the question is the second option or option "B". I hope that this is the answer that has come to your desired help.

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4 years ago

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