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2 years ago

If f(x) = x ^ 2 + 3 , then f(x+y)=?

Mathematics

1 answer:

kobusy [5.1K]2 years ago

5 0

Answer:

f(x)=x^2+3

f(x+y)=(x+y)^2+3

this is done by replacing x with (x+y)

f(x+y)=x^2+2xy+y^2+3

therefore : f(x+y)= x^2+y^2+2xy+3

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Ben is spending his summer driving across the country. He is going to spend the first day

KengaRu [80]

Answer:

The graph of the data presented is shown in the attached image to this solution.

It shows that the relationship between distance travelled in miles and the time take in hours is D = 55t

Step-by-step explanation:

The distance travelled with the corresponding time taken are presented in the table

Note: D - Distance travelled (miles)

t - Time (hours)

D | t

55 | 1

110 | 2

165 | 3

220 | 4

275 | 5

330 | 6

We are then told to plot these distance travelled on a graph with the time taken.

The distance travelled is plotted on the y-axis and the time in hours is plotted on the x-axis.

The image of the graph of this function will be attached to this answer

It is evident that the graph shows the relationship between the distance travelled and time taken to be D = 55t

Hope this Helps!!!

4 0

3 years ago

If 6 is subtracted from a number the result<br> is 39.

Paha777 [63]

Answer:

Step-by-step explanation:

To find the number you just add 6 to 39 which gets you 45.

5 0

2 years ago

Read 2 more answers

Evaluate each function: 1. p(a) = 3a-2; find p(-5) 2. h(t) = 4t+2; find h(-4) 3. f(a) = 3a-4; find f(1) 4. g(x) = 3x+ 1; find g(

frutty [35]

A-2x = 45 explained the theropod waysiver

3 0

2 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height

Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$

3 0

2 years ago

What is the solution to:<br><img src="https://tex.z-dn.net/?f=6%20%5Cgeqslant%20%20%5Cfrac%7B1%7D%7B2%7D%20x%20%2B%2021" id="Tex

Vladimir79 [104]

I really hope this helps. Btw x<-30

7 0

3 years ago