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3 years ago

6

A number is less than 20 units away from 7 on the number line. What numbers are within these parameters? Choose the inequality t

hat represents the situation.

2 answers:

lara [203]3 years ago

3 0

1.) |x-7|<20
2.) -13<x<27
3.) D

miv72 [106K]3 years ago

3 0

Answer:

it is B:l x-7 l<20 Than D:-13<x<27 and lastly Graph D these are correct on EDU

hope this helps

Step-by-step explanation:

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1 The metal used to make a coin is a mixture of steel and copper. The ratio mass of steel

Masteriza [31]

$\huge{ \underline{ \underline{ \gray{ \bf{Solution:}}}}}$

Let,

Mass of the steel be 108x
Mass of the copper be 7x
Then, Total mass = 108x + 7x = 115x

Given,

Total mass of the coin = 230 mg

That means,

⇛ Total mass of the coin = 230 mg

⇛ 115x = 230 mg

⇛ x = 230 mg/115

⇛ x = 2 mg

Then,

Mass of the steel = 108(2) = 216 mg
Mass of the copper = 7(2) = 14 mg

☁️ ANSWER - 14 mg (Mass of copper)

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7 0

3 years ago

Find (f o g)(0) *serious answers only*

jeyben [28]

I think the answer should be 0 but I’m not so sure!

5 0

2 years ago

?÷(−7)=(−4) what is the missing number

Hunter-Best [27]

Answer:

The answer is 28.

Step-by-step explanation:

To find the solution, you must multiply the other two numbers together to find the original number that is divided.

-7 * -4 = 28 Answer

28/-7 = -4

Hope this helps!

3 0

3 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

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

And the magnitude of the derivative of unit tangent vector is

$||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2$

The curvature is

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1$

The tangential component of acceleration is given by the formula

$a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}$

We know that $\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ and $||\boldsymbol{r}'(t)}||=2$

$\frac{d}{dt}\left(2\right)\: = 0$ so

$a_{\boldsymbol{T}}=0$

The normal component of acceleration is given by the formula

$a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2$

We know that $\kappa=1$ and $\frac{ds}{dt}=2$ so

$a_{\boldsymbol{N}}=1 (2)^2=4$

3 0

3 years ago

What is 0.0000981 in scientific notation?

tatiyna

9.81 or 98.1 pretty sure it's the first one

3 0

3 years ago