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Advocard [28]
3 years ago
15

1+x+x²please explain​

Mathematics
1 answer:
oee [108]3 years ago
4 0
What is the question?
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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
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