All categories

2 years ago

Answer these questions mathswatch

Mathematics

1 answer:

quester [9]2 years ago

7 0

Answer:

They have played a total of 28 games.

Step-by-step explanation:

For this question, we must start by adding up the total number of home goals scored, and away goals scored.

Home Goals:

9 + 6 + 8 + 7 = 30 goals

Mean (Average) of home goals: 2.5

$\frac{30total}{2.5 mean} = 12 games$

Away Goals:

2 + 7 + 4 + 5 = 28 goals

Mean (Average) of away goals: 1.8

$\frac{28total}{1.8mean} = approx. 15.6 games$ (rounded = 16 games)

Finally, add 12 + 16 and get your final answer, 28 games.

You might be interested in

Please help me i need help with this please give me the answer

vichka [17]

Part A:

$8.00+(r⋅$1.40)

Part B:

($26.00-($8.00)÷$1.40)=5

He can go on 5 rides.

8 0

3 years ago

PLEASE HELP WILL GIVE BRAINLIEST Abby purchased 4 2/5 gallons of paint to paint 4 walls of a room in her house. the cost of one

natima [27]

Answer:

Cost to paint 4 walls of a room = $24.20

Cost of painting 3 walls of a room =$18.15

Step-by-step explanation:

(a) Abby purchased 4 2/5 gallons of paint to paint 4 walls of a room in her house.

$4\frac{2}{5}gallons \ of \ paint = \frac{4*5+2}{5} =\frac{22}{5}$

1 gallon cost = 5.50

So 22/5 gallons cost = $\frac{22}{5} * 5.5= 24.20$

Cost to paint 4 walls of a room = $24.20

(b)cost of painting 4 walls of a room = $24.20

cost of painting 1 wall of a room = $\frac{24.20}{4} =6.05$

Cost of painting 3 walls of a room = 6.05 * 3= $18.15

7 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

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

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

Which of the following is equivalent to 5455 cm^3?

frosja888 [35]

Answer : B
==================================

1L = 1000cm^3

5.455L x 1000
= 5455 cm^3

5 0

2 years ago

Please help !!!!!!!!!

vodka [1.7K]

Answer:

A= 41° a=9.84... b=11.32...

Step-by-step explanation:

given B = 49° and C = 90°

and A + B + C = 180

so 49+90+A=180

A= 41°

we also know c (small) is 15 units

and maybe you know this SOH CAH TOA? no, i'm not calling anyone names :P this is just the mnemonic to memorize how sin , cos and tan fit sin(Ф)=Opp/Hyp Cos(Ф)=Adj/Hyp Tan(Ф)=Opp/Adj

Hyp is the hypontenuse of the triangle ( the long side). Opposite (Opp) means the side opposite from what ever angle you are starting at..adjacent (Adj) means what ever angle is next to where you are starting from.

use Cos(49) to find the side a

a is an adjacent side to the 49 angle. We know the Cos and we know the Hyp .. so solve for adj...

Cos(Ф)=Adj/ Hyp

Hyp*Cos(Ф)=Adj

15*Cos(49)= Adj

9.84...=Adj=a

now lets' use sin for the opposite side or b

sin(Ф)=Opp/Hyp

Hyp*Sin(Ф) = Opp

15*sin(49)= Opp

11.32...= Opp =b

7 0

3 years ago