All categories

2 years ago

Which ordered pair is a solution of the equation?

Mathematics

1 answer:

uysha [10]2 years ago

5 0

Only b

apply -1 to x and it will give you y=-5

You might be interested in

Find f(x)+g(x), given f(x)=x^2-8 and g(x)=-2-x^2

Gnoma [55]

$\left(x^2-8\right)+\left(-2-x^2\right)=-10$

$\left(x^2-8\right)+\left(-2-x^2\right)$

$=x^2-8-2-x^2$

$=x^2-x^2-8-2$

$=-8-2$

$=-10$

8 0

2 years ago

Jean transformed a point by using the rule (x,y)(x-6,y+8) the image point is (-4,1) which point is the pre image

Xelga [282]

To find the pre image you need to back track on the image. To get to the image you used (x-6,y+8). Now you need to use the exact opposite to get back to the pre image. For this you would change the signs to look like (x+6,y-8). Now we just apply this to (-4,1).

(-4+6,1-8)

(2,-7) should be the pre image point.

8 0

3 years ago

Read 2 more answers

Multiple choice!! There's only one correct answer.

GalinKa [24]

The solution is where the red line crosses the blue line, which on the X axis is between the 1 and the 2, so about 1.5.

Convert the fractions to decimal and see which one is close to 1.5:

13/8 = 1.625

25/16 = 1.5625

7/4 = 1.75

27/16 = 1.6875

The closest one to 1.5 is 25/16

5 0

2 years ago

Please Help! This is a trigonometry question.

liraira [26]

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

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

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

________

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

________

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

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

<span>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: <em>trigonometry trig function cosecant csc double angle identity geometry</em>

</span>

8 0

3 years ago

A line is defined by the equation y=2/3x-6. The line passes through a point whose y-coordinate is 0. What is the x- coordinate o

NISA [10]

Answer: Its 9

Thats the answer

5 0

2 years ago