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3 years ago

WILL MARK BRAINLIEST!!!!Which of the following relations is not a function?

Mathematics

1 answer:

Inessa [10]3 years ago

3 0

Answer:

The third (C) option is the answer.

Because 4 is repeating in x coordinate,

that is called Domain.

It's allowed to be repeated in y coordinates,

that is called Range.

Good luck!

Intelligent Muslim,

From Uzbekistan.

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Is 6 + p polynominal

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I belive that 6 + p is polynominal so yes it is

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Solve the equation 5/7d=2/3

Oduvanchick [21]

Equation: 5/7d=2/3

Work:

5d/7 = 2/3

5d = 2/3 x 7

5d = 14/3

d = 14/3/5

d = 14 / 3 x 5

d = 14/15

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3 years ago

If a 10 foot piece of electrical tape has 0.037 feet cut from it,then is the new length of tape ?

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3 years ago

The missing number in the series 1,4,27,__,3125 is:

sveta [45]

Answer:

D ) 256

Step-by-step explanation:

The given sequence is 1,4,27, ---, 3125

This is sequence of given problem

$1^{2},2^{2} ,3^{3} ,4^{4} ,5^{5}$

so here missing is '4' term is

The fourth term is 256.

5 0

3 years ago

Find the EXACT value of sin(A−B) if cos A = 3/5 where A is in Quadrant IV and cos B = 12/13 where B is in Quadrant IV. Assume al

MissTica

$\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)$

well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.

$\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5$

$\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}$

4 0

2 years ago