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3 years ago

Denzel wants to work out the density of a block of wood the block of wood is in the shape of the cuboid

Mathematics

1 answer:

marishachu [46]3 years ago

5 0

Answer:

$Density = 1.094g/cm^3$

Step-by-step explanation:

Given

$Length = 17.7cm$

$Width = 8.3cm$

$Height = 13.0cm$

$Mass = 2090g$

See attachment

Required

Determine the density of the wood

First, we calculate the volume of the wood.

$Volume = Length * Width * Height$

Substitute values for Length, Width and Height

$Volume = 17.7cm * 8.3cm * 13.0cm$

$Volume = 1909.83cm^3$

The density is:

$Density = \frac{Mass}{Volume}$

$Density = \frac{2090g}{1909.83cm^3}$

$Density = 1.094g/cm^3$ --- approximated

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Graph the image of the given triangle under a dilation with a scale factor of 1/4 and center of dilation (0, 0)

schepotkina [342]

Answer:

The image in the attached figure

Step-by-step explanation:

we know that

The dimensions of the image of the given triangle are equal to the original dimensions of the pre-image multiplied by the scale factor

In this problem the scale factor is 1/4

The height of the pre-image is 4 units

The base of the pre-image is 8 units

Find out the dimensions of the image

The height of the image is $(4)\frac{1}{4}=1\ unit$

The base of the image is $(8)\frac{1}{4}=2\ units$

The image in the attached figure

6 0

4 years ago

Find the probability.

KATRIN_1 [288]

Using it's concept, it is found that there is a 0.0366 = 3.66% probability that your coach and your friend get orange and you get a fruit-punch.

<h3>What is a probability?</h3>

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem, there are 15 bottles.

5 are orange, hence the is a 5/15 = 1/3 probability that the coach gets orange, hence P(A) = 1/3.

After the coach, there will be 14 bottles remaining, of which 4 are orange, hence the probability that the friend gets orange is of P(B) = 4/14 = 2/7.

For you, there will be 13 bottles remaining, of which 5 will be of fruit-punch, hence the probability you get fruit-punch is of P(C) = 5/13.

The probability of the three outcomes occurring is given by:

$p = \frac{1}{3} \times \frac{2}{7} \frac{5}{13} = \frac{10}{273} = 0.0366$

0.0366 = 3.66% probability that your coach and your friend get orange and you get a fruit-punch.

More can be learned about probabilities at brainly.com/question/14398287

#SPJ1

6 0

2 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:

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

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

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

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

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

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

$= 0$

8 0

2 years ago

South School has a circular pattern on the playground that has a radius of 6 feet. Find a. the circumference of the pattern and

katrin [286]

Answer:

Circumference: 37.68

Area: 113.04

Step-by-step explanation:

the formula to find the circumference is 2 x 3.14 x r

r is the radius so you plug in 6 for that

2 x 3.14 x 6 = 37.68

the formula to find the area of a circle is 3.14 x r ²

r is still the radius so you plug in 6 for that

3.14 x 6² = 113.04

6 0

3 years ago

What is the simplest form of the radical expression 8?

Kisachek [45]

The answer oils be 2 square root 2 because 4 times 2 is 8 and because the square root of 4 is 2 you would move that 2 to the outside and keep the other two on the inside of the square root.

5 0

3 years ago