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3 years ago

The mean, median and mode of the five numbers are same. if four numbers are 7,7,5 and 7, what must be the fifth number?

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

7 0

Answer:

The fifth number is <u>9</u>

Step-by-step explanation:

7 is the mode because it appears the most.

7 has to be the median, so the fifth number could be anything.

7 has to be the mean, so you would do:

7 x 5 = 35

35 - 7 - 7 - 7 - 5 = <u>9</u>

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Subtracing 3x2+4x-5 from 7x2-3x from this polynomial the difference is

kolezko [41]

7x^2-3x - (3x^2 +4x -5) multiply negative one across the entire second term

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4 0

3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

15. Rewrite the given functions in

Lisa [10]

15. The vertex form of f(x) = x² + 8x - 10 is f(x) = (x + 4)² - 26

17. The standard form of g(x) = (x + 3)² - 4 is g(x) = x² + 6x + 5

Step-by-step explanation:

Any quadratic function has:

1. A vertex form form f(x) = a(x - h)² + k, where (h , k) are the

coordinates of its vertex point

2. A standard form g(x) = ax² + bx + c, where a , b , c are constant and

a is the coefficient of x², b is the coefficient of x and c is the

y-intercept

3. the x-coordinate of the vertex point h = $\frac{-b}{2a}$ and

k = f(h)

15.

∵ f(x) = x² + 8x - 10

∴ a = 1 and b = 8

∵ h = $\frac{-b}{2a}$

∴ h = $\frac{-8}{2(1)}=-4$

∴ h = -4

∵ k = f(h)

∴ k = f(-4)

- Substitute x by -4 in f(x)

∴ k = (-4)² + 8(-4) - 10

∴ k = 16 - 32 - 10

∴ k = -26

∴ f(x) = (x - -4)² + (-26)

∴ f(x) = (x + 4)² - 26

The vertex form of f(x) = x² + 8x - 10 is f(x) = (x + 4)² - 26

17.

∵ g(x) = (x + 3)² - 4

- Lets solve the power 2 of the bracket:

(1st + 2nd)² = (1st)(1st) + 2(1st)(2nd) + (2nd)(2nd)

∵ (x + 3)² = (x)(x) + 2(x)(3) + (3)(3)

∴ (x + 3)² = x² + 6x + 9

∴ (x + 3)² - 4 = (x² + 6x + 9) - 4

∵ 9 - 4 = 5

∴ (x + 3)² - 4 = x² + 6x + 5

The standard form of g(x) = (x + 3)² - 4 is g(x) = x² + 6x + 5

Learn more:

You can learn more about the vertex form of quadratic function in brainly.com/question/9390381

#LearnwithBrainly

4 0

4 years ago