All categories

3 years ago

Which equations represent this fact family ? Check all that apply

Mathematics

1 answer:

yanalaym [24]3 years ago

6 0

Answer:

9-11=20

and

20-11=9

Step-by-step explanation:

20+11=31, not 9- the first one is not the answer

20+9-29, not 11- the third is incorrect

11-20=-9, not 9- fifth is not it

You might be interested in

Plz answer.............

spayn [35]

Answer:

( - x, - y )

Step-by-step explanation:

The starting is always Quadrant I.

270 degrees clockwise from Quadrant I is Quadrant III.

In Quadrant III, the points will be in the form ( - x, - y ).

7 0

3 years ago

Given the piece wise function show below, select all that are true

NeTakaya

Answer:

Option (C) and (D)

Step-by-step explanation:

Given piecewise function is,

f(x) = 2x, x < 1

5, x = 1

$x^{2}$ , x > 1

Option (A),

x = 5 means x > 1

So the function will be,

f(x) = $x^{2}$

f(5) = (5)²

= 25

Therefore, f(5) = 1 is not correct.

Option (B),

x = -2 means x < 1

f(x) = 2x will be applicable.

f(-2) = 2(-2) = -4

Therefore, f(-2) = 4 is not correct.

Option (C)

For x = 1,

f(1) = 5

Therefore, f(1) = 5 is the correct option.

Option (D)

x = 2 means x > 1 and the function defined will be,

f(x) = x²

f(2) = 2²

= 4

Therefore, f(2) = 4 will be the correct option.

Options (C) and (D) will be the answer.

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=f%20-%20354%20%3D%201221" id="TexFormula1" title="f - 354 = 1221" alt="f - 354 = 1221" align="

alukav5142 [94]

$f-354=1221$ <em>add 354 to both sides</em>

$f+354-354=1221+354\\\\\boxed{f=1,575}\to\boxed{D.}$

6 0

3 years ago

Eight years ago the daughter age was thrice the son age

coldgirl [10]

Answer:

what's the question here ?

8 0

3 years ago

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

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