All categories

3 years ago

What is the domain of the function y= (x+4? - O -4 O OSX 00 O 45x<0

Mathematics

1 answer:

asambeis [7]3 years ago

6 0

Answer:

What is the domain of the function y= Vx+4? O -4 O OSX 500 O 45x<

You might be interested in

Complete the square please.

Aleks04 [339]

Answer:

(x-3)^2 + (y+2)^2 = 9^2

Step-by-step explanation:

x^2 -6x+y^2+4y-68 = 0

Complete the square

x^2 -6x+y^2+4y-68+68 = 0+68

x^2 -6x+y^2+4y = 68

Find the term to add for x

-6 /2 = -3 -3^2 = 9

Find the term to add for y

4/2 =2 2^2 = 4

Add 9 and 4

x^2 -6x+9+y^2+4y+4 = 68+9+4

(x-3)^2 + (y+2)^2 = 81

(x-3)^2 + (y+2)^2 = 9^2

The standard form is

(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius

3 0

3 years ago

SECOND TIME ASKING PLS HELP: A multiple choice question has 3 options. What is the percentage that you will chose the WRONG opti

DiKsa [7]

Answer:

2/3 or 66.67%

Step-by-step explanation:

2 wrong options, 3 total options. 2 over 3.

<em>Big Brain</em>

Hope this helps!! :)

5 0

3 years ago

Read 2 more answers

Which shape has a line of symmetry that can be drawn through the figure?

zimovet [89]

Answer: Option D.

Step-by-step explanation:

A figure has a line of symmetry if we can draw a line through the figure, in such way that the line cuts the figure in exactly two equal halves.

Then any figure that can be cutin exactly two halves, is a correct answer to this question.

Then:

For a circle, the line of symmetry is the diameter of the circle.

For the square, the line of symmetry can be obtained by cutting the square with a line that is perpendicular to one of the sides, and cuts that side exactly on the midpoint.

For an equilateral triangle, the line of symmetry is the line that cuts any base in the midpoint and also passes through the opposite vertex.

Then all the figures are correct options, then the correct option is D

7 0

4 years ago

Find tan2θ if θ terminates in Quadrant IV and cosθ = 3/5.

natali 33 [55]

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$\cos( \alpha ) = \frac{3}{5} \\$

${sin}^{2}( \alpha ) = 1 - {cos}^{2} ( \alpha )$

${sin}^{2}( \alpha ) = 1 - ({ \frac{3}{5} })^{2} \\$

${sin}^{2}( \alpha ) = 1 - \frac{9}{25} \\$

${sin}^{2}( \alpha ) = \frac{25}{25} - \frac{9}{25} \\$

${sin}^{2}( \alpha ) = \frac{16}{25} \\$

$sin( \alpha )= ± \sqrt{ \frac{16}{25} } \\$

In Quadrant IV , sin is negative .

Thus ;

$sin( \alpha ) = - \sqrt{ \frac{16}{25} } \\$

$\sin( \alpha ) = - \frac{4}{5} \\$

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$\tan(2 \alpha ) = \frac{ \sin(2 \alpha ) }{ \cos(2 \alpha ) } \\$

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$\sin(2 \alpha ) = 2. \sin( \alpha ). \cos( \alpha )$

$\sin(2 \alpha ) = 2 \times ( - \frac{4}{5} ) \times ( \frac{3}{5} ) \\$

$\sin(2 \alpha ) = - \frac{24}{25} \\$

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$\cos(2 \alpha ) = {cos}^{2}( \alpha ) - {sin}^{2}( \alpha )$

$\cos(2 \alpha ) = ({ \frac{3}{5} })^{2} - ({ - \frac{4}{5} })^{2} \\$

$\cos(2 \alpha ) = \frac{9}{25} - \frac{16}{25} \\$

$\cos(2 \alpha ) = - \frac{7}{25} \\$

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

$\tan(2 \alpha ) = \frac{ \sin(2 \alpha ) }{ \cos(2 \alpha ) } \\$

$\tan(2 \alpha ) = \frac{ - \frac{24}{25} }{ - \frac{7}{25} } \\$

$\tan(2 \alpha ) = - \frac{24}{25} \div - \frac{7}{25} \\$

$\tan(2 \alpha ) = - \frac{24}{25} \times - \frac{25}{7} \\$

$\tan(2 \alpha ) = \frac{24}{7} \\$

Thus the correct answer is (( C )) .

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

4 0

3 years ago

Read 2 more answers

In DEF, DE = 15, and m angle F=32 Find EF to the nearest tenth.

iren [92.7K]

the answer would be 24.0.

3 0

3 years ago