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3 years ago

Which of the following sets represents the solution of the equation below?

Mathematics

1 answer:

Novay_Z [31]3 years ago

8 0

Answer:

I don't see anything

Step-by-step explanation:

ya srry

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Could anyone help me with this problem i'm stuck h(17)= ?

nikdorinn [45]

Answer:

$\sqrt{17t}$

Step-by-step explanation:

Consider the right side of the table

It informs us of the values for h(t) for different values of t

The top one informs us what h(t) is when t = 17, that is

h(t) = $\sqrt{17t}$ , when t = 17, thus

h(17) = $\sqrt{17t}$

4 0

3 years ago

I NEED HELP. What is the answer?

Lina20 [59]

Answer:

Step-by-step explanation:

Here’s what B has:

Min: 24

Q1: 27

Median: 31

Q3: 33

Max: 34

Although Choice A and B were really close with their five number summaries, out of the two of them, only B had a maximum value of 34, like the one in the box plot shown above.

7 0

2 years ago

Please Help! This is a trigonometry question.

liraira [26]

$\large\begin{array}{l} \textsf{From the picture, we get}\\\\ \mathsf{tan\,\theta=\dfrac{2}{3}}\\\\ \mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{2}{3}}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\qquad\mathsf{(i)} \end{array}$

$\large\begin{array}{l} \textsf{Square both sides of \mathsf{(i)} above:}\\\\ \mathsf{(3\,sin\,\theta)^2=(2\,cos\,\theta)^2}\\\\ \mathsf{9\,sin^2\,\theta=4\,cos^2\,\theta}\qquad\quad\textsf{(but }\mathsf{cos^2\theta=1-sin^2\,\theta}\textsf{)}\\\\ \mathsf{9\,sin^2\,\theta=4\cdot (1-sin^2\,\theta)}\\\\ \mathsf{9\,sin^2\,\theta=4-4\,sin^2\,\theta}\\\\ \mathsf{9\,sin^2\,\theta+4\,sin^2\,\theta=4} \end{array}$

$\large\begin{array}{l} \mathsf{13\,sin^2\,\theta=4}\\\\ \mathsf{sin^2\,\theta=\dfrac{4}{13}}\\\\ \mathsf{sin\,\theta=\sqrt{\dfrac{4}{13}}}\\\\ \textsf{(we must take the positive square root, because }\theta \textsf{ is an}\\\textsf{acute angle, so its sine is positive)}\\\\ \mathsf{sin\,\theta=\dfrac{2}{\sqrt{13}}} \end{array}$

________

$\large\begin{array}{l} \textsf{From (i), we find the value of }\mathsf{cos\,\theta:}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{2}\,sin\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{\sqrt{13}}}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\sqrt{13}}}\\\\ \end{array}$

________

$\large\begin{array}{l} \textsf{Since sine and cosecant functions are reciprocal, we have}\\\\ \mathsf{sin\,2\theta\cdot csc\,2\theta=1}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{sin\,2\theta}\qquad\quad\textsf{(but }}\mathsf{sin\,2\theta=2\,sin\,\theta\,cos\,\theta}\textsf{)}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\,sin\,\theta\,cos\,\theta}}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\cdot \frac{2}{\sqrt{13}}\cdot \frac{3}{\sqrt{13}}}} \end{array}$

$\large\begin{array}{l} \mathsf{csc\,2\theta=\dfrac{~~~~1~~~~}{\frac{2\cdot 2\cdot 3}{(\sqrt{13})^2}}}\\\\ \mathsf{csc\,2\theta=\dfrac{~~1~~}{\frac{12}{13}}}\\\\ \boxed{\begin{array}{c}\mathsf{csc\,2\theta=\dfrac{13}{12}} \end{array}}\qquad\checkmark \end{array}$

If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2150237

$\large\textsf{I hope it helps.}$

Tags: trigonometry trig function cosecant csc double angle identity geometry

8 0

3 years ago

16x 2 + 8x - 15 Mathematics?

Gala2k [10]

If you're factoring, your answer is

(4x - 3)(4x + 5)

You'll work it out this way:

16x^2 + 8x - 15

Since there's no GCF here, you multiply the first and third terms, giving you -240. Your next step is to find two numbers that have a sum of 8 and a product of -240. These numbers are 20 and -12. Plug those in and you've got:

16x^2 + 20x - 12x - 15

From here you divide it into two binomials. These are (16x^2 + 20x) and (-12x - 15).
When you take out the greatest common factors (4x and -3), they become:

4x(4x + 5) - 3(4x + 5)

Then you group what's outside of the parentheses together (4x - 3)
And bring what's inside of the parentheses down (4x + 5).
This brings your answer to

(4x - 3)(4x + 5)

5 0

3 years ago

Can someone help with this question? thanks

Licemer1 [7]

Answer:

Regular: C & E

Irregular: A & B

Not a polygon: F & D

Step-by-step explanation:

A polygon is a close sided shape with no curves. Since F and D have curves they are not polygons. Regular polygons have side lengths which are all equal to each other. C and E are regular polygons. Irregular polygons are polygons whose sides have different values from each other. A and B are irregular.

7 0

3 years ago