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3 years ago

What is 28 division 4

Mathematics

2 answers:

Morgarella [4.7K]3 years ago

8 0

Answer:

Step-by-step explanation:

frosja888 [35]3 years ago

8 0

Answer:

Step-by-step explanation:

28 / 4 = 7

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Simon can see two lights, light A and light B.

xz_007 [3.2K]

Answer:

2 Times

Step-by-step explanation:

Simon can see two lights A and B

Light A flashes every 15 seconds.
Light B flashes every 18 seconds.

First, we determine the next time both lights will flash together.

This is done by finding the Least Common Multiple of 15 and 18.

15=3X5

18= $2X3^2$

LCM of 15 and 18= $2X3^2X5=90$

This means that lights A and B will first flash together after 90 seconds.

Now, 240/90= $2.\overline{6}$ seconds

Therefore, the number of times more both light will flash together in the next 4 Minutes= 2 Times

3 0

3 years ago

Write at least 2 numerical expressions for each written phrase below. Then, solve.

zloy xaker [14]

Answer:

a. $\frac{3}{5}(7)\text{ or }\frac{3\times 7}{5}$

b. $\frac{1}{6}(4\times 8)\text{ or }\frac{4\times 8}{6}$

Step-by-step explanation:

In numerical expressions we write numbers with the mathematical symbols ( i.e. '+', '-' etc )

a. Given phrase,

Three fifths of seven,

$\frac{3}{5}(7)$

$=\frac{3\times 7}{5}$

$=\frac{21}{5}$

$=4\frac{1}{5}$

b. One sixth the product of four and eight,

$\frac{1}{6}(4\times 8)$

$=\frac{4\times 8}{6}$

$=\frac{32}{6}$

$=5\frac{1}{3}$

7 0

3 years ago

Math math help ASAP

tiny-mole [99]

Answer:

103.67 in³

Step-by-step explanation:

given Volume of cone is 1/3 * π * r² * h

here given r = 6/2 = 3 , height = 11 inch

using the formula = 1/3 * π * 3² * 11

=103.67 in³ or 33π in³

7 0

2 years ago

Read 2 more answers

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:

Given:

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

First derivative

$\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)$\\\end{minipage}}$

$\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}}$

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

Second derivative

$\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}$\\\end{minipage}}$

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

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

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

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

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

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

Proof

$(x^2+1)y_2+xy_1-m^2y$

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

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