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3 years ago

Which table shows ordered pairs that satisfy the function y=x^2+1

Mathematics

1 answer:

ohaa [14]3 years ago

3 0

Where is the table? You could potentiallysolve this by deconstructing the function.

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javier has 3 math books that are the same size. he makes a stack by placing the books on top of each other. The total stack is 1

Alexeev081 [22]

Answer:
The height of one book is $\frac{10}{3}$ in

Explanation:
We know that the 3 books are of the same size. This means that the 3 books have the same height.
Assume that each book has a height x in

Now, we know that the total height is 10 in.
Therefore:
x + x + x = 10
3x = 10
x = 10/3 in

Based on the above, the height of one book is 10/3 in

Hope this helps :)

4 0

3 years ago

Fill in the blank (there was no word bank)

Elena L [17]

angle ADC = 50° + 85°

= 135°

<h3>PLEASE MARK ME BRAINLIEST</h3>

3 0

3 years ago

Explain why arctan(1) = 45° or π/4.

xeze [42]

Answer:

Explanation below

Step-by-step explanation:

Inverse Trigonometric Functions

They are defined as the inverse of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant.

The corresponding inverse functions are called arcsin, arccos, arctan, arccot, arcsec, arccsc.

These inverse functions in trigonometry are used to get the angle with any of the trigonometry ratios.

For example,

$\displayystyle \sin 30^\circ=\frac{1}{2}$

Then:

$\displaystyle arcsin \frac{1}{2}=30^\circ$

Since:

$\tan 45^\circ=1$

Or, equivalently:

$\displaystyle \tan \frac{\pi}{4}=1$

Then:

$\displaystyle arctan(1)=45^\circ$

Or:

$\displaystyle arctan(1)=\frac{\pi}{4}$

6 0

3 years ago

At the beginning of a lesson, a piece of chalk is 4.875 inches long. At the end of the lesson, it is 3.125 inches long. Write tw

natima [27]

Its probably 4 & 875/1000, 3 & 125/1000....hope this helps

7 0

4 years ago

Special right triangles

garik1379 [7]

Answer: The answers are given below.

Step-by-step explanation: The calculations are as follows.

(1) We have in the given right-angled triangle,

$\tan 30^\circ=\dfrac{10}{x}\\\\\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{10}{x}\\\\\Rightarrow x=10\sqrt3,$

and

$y^2=(10)^2+x^2=100+(10\sqrt3)^2=100+300=400\\\\\Rightarrow y=20.$

∴ x = 10√3 and y = 20.

(2) We have in the given right-angled triangle,

$\cos 60^\circ=\dfrac{2}{x}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{2}{x}\\\\\Rightarrow x=4,$

and

$y^2=x^2-2^2=16-4=12\\\\\Rightarrow y=2\sqrt3.$

∴ x = 4 and y = 2√3.

(3) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{7}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{7}{y}\\\\\Rightarrow y=\dfrac{14\sqrt3}{3},$

and

$\sin 30^\circ=\dfrac{7}{x}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{7}{x}\\\\\Rightarrow x=14.$

∴ x = 14 and y = 14√3.

(4) We have in the given right-angled triangle,

$\sin 30^\circ=\dfrac{x}{6}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{6}\\\\\Rightarrow x=3,$

and

$\cos 30^\circ=\dfrac{y}{6}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{y}{6}\\\\\Rightarrow y=3\sqrt3.$

∴ x = 3 and y = 3√3.

(5) We have in the given right-angled triangle,

$\sin 30^\circ=\dfrac{y}{10}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{y}{10}\\\\\Rightarrow y=5,$

and

$\cos 30^\circ=\dfrac{x}{6}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{x}{10}\\\\\Rightarrow y=5\sqrt3.$

∴ x = 5 and y = 5√3.

(6) We have in the given right-angled triangle,

$\sin 30^\circ=\dfrac{x}{8}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{8}\\\\\Rightarrow x=4,$

and

$\cos 30^\circ=\dfrac{y}{8}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{y}{8}\\\\\Rightarrow y=4\sqrt3.$

∴ x = 4 and y = 4√3.

(7) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{7\sqrt3}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{7\sqrt3}{y}\\\\\Rightarrow y=14,$

and

$\sin 30^\circ=\dfrac{x}{y}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{14}\\\\\Rightarrow x=7.$

∴ x = 7 and y = 14.

(8) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{6\sqrt3}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{6\sqrt3}{y}\\\\\Rightarrow y=12$

and

$\sin 30^\circ=\dfrac{x}{y}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{12}\\\\\Rightarrow x=6$

∴ x = 6 and y = 12.

(9) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{\sqrt3}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{\sqrt3}{y}\\\\\Rightarrow y=2$

and

$\sin 30^\circ=\dfrac{x}{y}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{2}\\\\\Rightarrow x=4.$

∴ x = 4 and y = 2.

Thus, all are completed.

3 0

3 years ago