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3 years ago

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1+x+x²please explain

1 answer:

oee [108]3 years ago

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The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha

Xelga [282]

Answer:

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

Step-by-step explanation:

Let suppose that function $f$ is continuous and integrable in the given intervals, by integral definition of average we have that:

$\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6$ (1)

$\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20$ (2)

By Fundamental Theorems of Calculus we expand both expressions:

$\frac{F(3)-F(-2)}{3-(-2)} = -6$

$F(3) - F(-2) = -30$ (1b)

$\frac{F(5)-F(3)}{5-3} = 20$

$F(5) - F(3) = 40$ (2b)

We obtain the average value of $f$ over the interval $[-2, 5]$ by algebraic handling:

$F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)$

$F(5) - F(-2) = 10$

$\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}$

$\bar f = \frac{10}{7}$

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

4 0

3 years ago

How much more interest will Maria receive if she invests $1,000 for one year at x percent annual interest, compounded semiannual

notka56 [123]

Answer:

$0.025x² . . . where x is a number of percentage points

Step-by-step explanation:

The multiplier for semi-annual compounding will be ...

(1 + x/2)² = 1 + x + x²/4

The multiplier for annual compounding will be ...

1 + x

The multiplier for semiannual compounding is greater by ...

(1 + x + x²/4) - (1 + x) = x²/4

Maria's interest will be greater by $1000×(x²/4) = $250x², where x is a decimal fraction.

If x is a percent value, as in x = 6 when x percent = 6%, then the difference amount is ...

$250·(x/100)² = $0.025x² . . . where x is a number of percentage points

_____

<u>Example</u>:

For x percent = 6%, the difference in interest earned on $1000 for one year is $0.025×6² = $0.90.

3 0

3 years ago

10 PTS!! (5 + 2/8) - (3 7/8)

Artyom0805 [142]

The answer should be 1 3/8

7 0

3 years ago

Read 2 more answers

If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that

Ivan

Answer:

$\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}$

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

<u>Given</u><u> </u><u>:</u><u>-</u>

• $\sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}$

<u>To</u><u> </u><u>Prove</u><u> </u><u>:</u><u>-</u><u> </u>

• $\sf\implies cos2A =\dfrac{\sqrt3}{2}$

<u>Proof </u><u>:</u><u>-</u><u> </u>

We know that ,

$\sf\implies cos2A = 1 - 2sin^2A$

Therefore , here substituting the value of sinA , we have ,

$\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2$

Simplify the whole square ,

$\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}$

Add the numbers in numerator ,

$\sf\implies cos2A = 1-2\times \dfrac{4-2\sqrt3}{8}$

Multiply it by 2 ,

$\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}$

Take out 2 common from the numerator ,

$\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}$

Simplify ,

$\sf\implies cos2A = 1 -\dfrac{ 2-\sqrt3}{2}$

Subtract the numbers ,

$\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}$

Simplify,

$\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }$

Hence Proved !

8 0

3 years ago

How is the graph of y=3(x+1)^2 related to its parent function y=x^2

creativ13 [48]

The parent function is y = x^2

There is a compression of 3:

y = 3x^2

There is a shift by 1 unit to the left:

y = 3(x + 1)^2

3 0

3 years ago

Read 2 more answers