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3 years ago

Is this relation a function?

Mathematics

1 answer:

kobusy [5.1K]3 years ago

5 0

Answer:

yes though i am not completely sure

Step-by-step explanation:

You might be interested in

Write a 4 -digit number that is divisible by both 5 and 9

vodomira [7]

4500 is both divisible by 5 and 9
4500/9=500
4500/5=900

6 0

3 years ago

SinA=√1-x÷√1+X<br> Find the value of cosA

Irina-Kira [14]

The value of cos A is √(1 + x²)/ (1 - x²) /√1 + x

<h3>Trigonometric ratios</h3>

It is important to note that

sin A = opposite/ hypotenuse

cos A = adjacent/ hypotenuse

Then,

opposite = $\sqrt{1} - x$

Hypotenuse = $\sqrt{1} + x$

Let's find the adjacent side using the Pythagorean theroem

$(\sqrt{1} + x)^2 = (\sqrt{1 -x } )^2 + x^2$

$1 + x^2 = 1 - x^2 + x^2$

$x = \sqrt{\frac{1 + x^2}{1 -x^2} }$

cos A = x/hypotenuse

$cos A = \frac{\sqrt{\frac{1+x^2}{1 -x^2} } }{\sqrt{1} +x}$

cos A = √(1 + x²)/ (1 - x²) /√1 + x

Thus, the value of cos A is √(1 + x²)/ (1 - x²) /√1 + x

Learn more about trigonometric identities here:

brainly.com/question/7331447

#SPJ1

6 0

2 years ago

C. Find the value of x below:<br> 15.r + 5<br> -2 + 16.

nydimaria [60]

Answer:

Step-by-step explanation:

(15x + 5) and (-2 + 16x) represents the measures of corresponding angles.

Measures of corresponding angles are equal.

Therefore,

- 2 + 16x = 15x + 5

16x - 15x = 5 + 2

x = 7

5 0

3 years ago

Simplify the expression: 2/3 divided by -4 minus (1/6 - 8/6)

guapka [62]

Answer:

Step-by-step explanation:

$\frac{2}{3} \div - 4 - ( \frac{1}{6} - \frac{8}{6} )$

$- \frac{2}{3} \times \frac{1}{4} - ( - \frac{7}{6} )$

$- \frac{1}{3} \times \frac{1}{2} + \frac{7}{6}$

$- \frac{1}{6} + \frac{7}{6}$

$\boxed{\green{= 1}}$

5 0

2 years ago

6.4934 rounded to 6.493

iris [78.8K]

Answer:

To round you take the last diget and round it either up or down based on the number

if the number is 0-5 you round down

if the number is 6-9 you round up

so in this case you would round down to 6.4.930

Step-by-step explanation:

4 0

3 years ago

Is this relation a​ function?

Is this relation a function?