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Natalija [7]
2 years ago
11

The median of the values:29,24,30,23,28,18,

Mathematics
2 answers:
joja [24]2 years ago
6 0

Answer:

26

Step-by-step explanation:

18,23,24,28,29,30

(6 in total + 1)/2=3.5.

- the number which is 3.5 is between 24 and 28.

- middle number between these two is 26.

DiKsa [7]2 years ago
4 0
The answer to your question is 26
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B-4/6 how do I solve this? It's hard
serious [3.7K]
                                     
if you are looking for.... 2(B-4)= (6)b
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                                     2b - 6b= 8
                                     -4b=8
                                      so b= -2

the question is b-4 divided by 6 equals b/2

8 0
3 years ago
Anne has an aquarium in the shape of a rectangular prism that measures 10 in. By 20 in. By 10 in. How many cubic inches of water
kaheart [24]

Answer:

Volume = 2000\ in^3

Step-by-step explanation:

Given

Dimension: 10in by 20in by 10in

Required

Determine the capacity of the aquarium

This question implies that, we calculate the volume of the prism and this is calculated by multiplying the dimensions of the aquarium. i.e.

Volume = Length * Width * Height

Substitute values for the dimensions

Volume = 10 * 20 * 10

Volume = 2000\ in^3

Hence, the volume of the aquarium is 2000 cubic inches

6 0
3 years ago
X
slava [35]

Answer:

Equation 1 x=-16

Equation 2 m=-3

Step-by-step explanation:

Equation 1

3x-x=-24-6

2x=-32

x=-32/2

x=-16

Equation 2

-2m=16-10

-2m=6

m=6/-2

m=-3

6 0
3 years ago
If <img src="https://tex.z-dn.net/?f=%5Cmathrm%20%7By%20%3D%20%28x%20%2B%20%5Csqrt%7B1%2Bx%5E%7B2%7D%7D%29%5E%7Bm%7D%7D" id="Tex
Harman [31]

Answer:

See below for proof.

Step-by-step explanation:

<u>Given</u>:

y=\left(x+\sqrt{1+x^2}\right)^m

<u>First derivative</u>

\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If  $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}

<u />

<u />\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If  $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}

<u />

\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1}  \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}

<u>Second derivative</u>

<u />

\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If  $y=uv$  then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}

\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}

\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}

\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m

\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m

\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}

              = \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)

<u>Proof</u>

  (x^2+1)y_2+xy_1-m^2y

= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m

= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m

= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]

= \left(x+\sqrt{1+x^2}\right)^m\left[0]

= 0

8 0
1 year ago
18. You are standing beside the Colorado River, 400 feet
Veseljchak [2.6K]
<h3>Answer: B. 781.6 feet approximately</h3>

=======================================================

Work Shown:

The horizontal portion is 400+166 = 566 feet. Label this as 'a', so a = 566. The vertical side is unknown, so b = x. The hypotenuse is c = 965

Use the pythagorean theorem

a^2+b^2 = c^2

566^2+x^2 = 965^2

x^2 = 965^2 - 566^2

x = sqrt( 965^2 - 566^2 )

x = 781.58108984289 which is approximate

x = 781.6 feet when rounding to one decimal place

6 0
3 years ago
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