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3 years ago

Show all ways to make 52 as tens and ones

1 answer:

4 0

There are a very large amount of possible solutions. A few examples.
10 + 10 + 10 + 10 + 10 + 1 + 1 = 52.
10 + 10 + 10 + 10 + (12 ones here) = 52.
10 + 10 + 10 + (22 ones here) = 52.
10 + 10 + (32 ones here) = 52
10 + (42 ones here) = 52
(52 ones here) = 52

I'm fairly sure those are all possible ways.

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Factor completely. 16x^4 - 81.<br><br> Please show all work.

Jet001 [13]

Answer:

$\large\boxed{16x^4-81=(2x-3)(2x+3)(4x^2+9)}$

Step-by-step explanation:

$Use\\\\ (a^n)^m=a^{nm}\\\\a^2-b^2=(a-b)(a+b)\\------------------------\\\\16x^4-81=4^2(x^2)^2-9^2=(4x^2)^2-9^2=(4x^2-9)(4x^2+9)\\\\=(2^2x^2-3^2)(4x^2+9)=\left[(2x)^2-3^2\right](4x^2+9)\\\\=(2x-3)(2x+3)(4x^2+9)$

5 0

3 years ago

Read 2 more answers

How do I get the result for 1/8 out of 40?

velikii [3]

Answer:

Step-by-step explanation:

You divide your number by the denominator and then times it by the numerator

3 0

2 years ago

Read 2 more answers

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$