All categories

2 years ago

Someone give me the answers to this

Mathematics

2 answers:

strojnjashka [21]2 years ago

6 0

Range: the set of all y-values of a graph or equation.

Y-intercept: the y-value where the graph intersects the y-axis.

Domain: the set of all x-values of a graph or equation.

X-intercept: the x-values where the graph intersects the x-axis.

Hope this helps! :)

kolezko [41]2 years ago

5 0

Answer:

The set of all y-values of a graph or equation.

You might be interested in

One side of a square is 3x + 2. Which expression represents the perimeter?

MissTica

Your answer would be 12x + 8 because there are 4 sides to a square and they are all 3x + 2, so you would need to multiply 3x + 2 by 4 which you can do by expanding brackets which gives you 4(3x + 2) and thus when you multiply it out you get 12x + 8. I hope this helps!

4 0

2 years ago

Read 2 more answers

HELP ASAP WILL GIVE BRAINLYIST AND 30 POINTS

Ilia_Sergeevich [38]

Answer:

Angles A and E

Step-by-step explanation:

If you see the angles size you would see that A and E are in opposite sides meaning it would be the same so Angles A and E are the same

6 0

2 years ago

Please Help! 20 Points!

iVinArrow [24]

$\text {x - coordinate = } \bigg( 5\dfrac{1}{2} + 3\dfrac{3}{4} \bigg) \div 2$

$\text {x - coordinate = } \bigg( 5\dfrac{2}{4} + 3\dfrac{3}{4} \bigg) \div 2$

$\text {x - coordinate = } 8\dfrac{5}{4} \div 2$

$\text {x - coordinate = }\dfrac{37}{4} \times \dfrac{1}{2}$

$\text {x - coordinate = }\dfrac{37}{8}$

$\text {x - coordinate = } 4\dfrac{5}{8}$

$\text {y - coordinate = } \bigg( -4\dfrac{1}{4} - 1\dfrac{1}{4} \bigg) \div 2$

$\text {y - coordinate = } -5\dfrac{2}{4} \div 2$

$\text {y - coordinate = } -\dfrac{22}{4} \times \dfrac{1}{2}$

$\text {y - coordinate = } -\dfrac{11}{4}}$

$\text {y - coordinate = } -2\dfrac{3}{4}}$

$\text {The coordinate is }\bigg( 4\dfrac{5}{8}, -2\dfrac{3}{4}} \bigg)$

6 0

2 years ago

Need help ASAP now plz now z now now now

Viktor [21]

Root root 20^2 - 4^2
X = 2

6 0

2 years ago

Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

1 year ago