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2 years ago

Which set of ordered pairs could be generated by an exponential function?

Mathematics

2 answers:

KIM [24]2 years ago

8 0

Hello,

Answer C with y=2^x

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a_sh-v [17]2 years ago

8 0

By visually assessing we can eliminate the other choices until we have the ff:

(–1, –1), (0, 0), (1, 1), (2, 8)(–1, 1), (0, 0), (1, 1), (2, 4)
Well, it's definitely not the second set of points. Exponential functions are monotonic (either always increasing, or decreasing). In the second set, the function has decreased from (-1,1) to (0,0), and then increased from (0,0) to (1,1).

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Natasha_Volkova [10]

6/5=1,

11/7=2,

21/4=5,

32/7=5,

24/9=3

25/6=4

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5 0

2 years ago

What does 3x^3 equal when x=2

victus00 [196]

Answer:

Step-by-step explanation:

Given

3x³ ← substitute x = 2 into the expression

= 3 × 2³ = 3 × 8 = 24

8 0

2 years ago

Read 2 more answers

Question 3 of 10 What is the slope of the line shown below? C O A. 5 5 (3,7) B. 2 O D.-2

scoundrel [369]

Slope = Y2-Y1/X2-X1 = 7-(-3)/3-(-2)= 10/5 = 2

8 0

2 years ago

Read 2 more answers

A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra

Harman [31]

Answer:

A.

The student made an error in step 3 because a is positive in Quadrant IV; therefore,

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

Step-by-step explanation:

Given

$P\ (a,b)$

$r = \± \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}$

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

$r = \sqrt{(a)^2 + (b)^2}$

Since a belongs to the x axis and b belongs to the y axis;

$cos\theta$ is calculated as thus

$cos\theta = \frac{a}{r}$

Substitute $r = \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}$

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}}$

Rationalize the denominator

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

So, from the list of given options;

The student's mistake is that a is positive in quadrant iv and his error is in step 3

3 0

2 years ago

Which table represents a linear function?

OleMash [197]

Answer:

the last option

Step-by-step explanation:

can you show the rest of the last option? if there isn't anything cutoff, then the last option would be correct. also, you can use desmos .com next time to help you graph

5 0

2 years ago