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3 years ago

Which graph represents a function?

Mathematics

1 answer:

Vilka [71]3 years ago

8 0

Answer:

Bottom right

Step-by-step explanation:

For a graph to represent a function, each input must only have one output.

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Answer:

y=0.6 x=3.6

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3 years ago

The price of a movie ticket in a given year can be modeled by the regression equation y= 6.94(1.02^x) where y is the ticket pric

Harlamova29_29 [7]

Answer:

The answer is B) $10.31

Step-by-step explanation:

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3 years ago

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60 POINTS ! BRAINLIEST IF YOU ANSWER NOW

Pavlova-9 [17]

Answer:

y=5x

Step-by-step explanation:

The table is given as:

x | y

2 | 10

4 | 20

6 | 30

8 | 40

A proportional relation satisfy the relation: $y=kx$

We can substitute any x value with the corresponding y-value and solve for k.

When x=2,y=10

This implies that:

$10=2k$

Divide both sides by 2 to get:

$k=5$

Hence the equation is $y=5x$

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4 years ago

I neeed heeelppppppppp

zheka24 [161]

Answer:

P(x) = $-2x^{2}$ + 304x - 8510

x = 37 smallest B.E.P

Step-by-step explanation:

P = R - E

P = $-2x^{2}$ + 317 x

E = 13x + 8510

P = $-2x^{2}$ + 317 x - 13x - 8510

P(x) = $-2x^{2}$ + 304x - 8510

B.E.P.(X) where R=E

$-2x^{2}$ + 304 x - 8510 = 0

2(x-37)(x-115)

4 0

3 years ago

Please help me!!!!!!!!!!!!!

monitta

Answer: see proof below

<u>Step-by-step explanation:</u>

Use the following Half-Angle Identities: tan (A/2) = (sinA)/(1 + cosA)

cot (A/2) = (sinA)/(1 - cosA)

Use the Pythagorean Identity: cos²A + sin²B = 1

Use Unit Circle to evaluate: cos 45° = sin 45° = $\frac{\sqrt2}{2}$

<u>Proof LHS → RHS</u>

Given: $cot\ (22\frac{1}{2})^o-tan\ (22\frac{1}{2})^o$

Rewrite Fraction: $cot\ (\frac{45}{2})^o-tan\ (\frac{45}{2})^o$

Half-Angle Identity: $\dfrac{sin(45)^o}{1-cos(45)^o}-\dfrac{sin(45)^o}{1+cos(45)^o}$

Substitute: $\dfrac{\frac{\sqrt2}{2}}{1-\frac{\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{1+\frac{\sqrt2}{2}}$

Simplify: $\dfrac{\frac{\sqrt2}{2}}{\frac{2-\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{\frac{2+\sqrt2}{2}}$

$=\dfrac{\sqrt2}{2-\sqrt2}-\dfrac{\sqrt2}{2+\sqrt2}$

$=\dfrac{\sqrt2}{2-\sqrt2}\bigg(\dfrac{2+\sqrt2}{2+\sqrt2}\bigg)-\dfrac{\sqrt2}{2+\sqrt2}\bigg(\dfrac{2-\sqrt2}{2-\sqrt2}\bigg)$

$=\dfrac{2\sqrt2+2}{4-2}-\dfrac{2\sqrt2-2}{4-2}$

$=\dfrac{4}{2}$

= 2

LHS = RHS: 2 = 2 $\checkmark$

7 0

4 years ago