All categories

2 years ago

Question 2 of 10

Mathematics

1 answer:

KIM [24]2 years ago

3 0

hi my name is afsshenn I am stealing your points hahaha

You might be interested in

Anty was riding his bike to school at a speed of 12 mph. When he was of the way there, he got a flat tire. His mother drove him

NemiM [27]

Complete question:

Anty was riding his bike to school at a speed of 12 mph. When he was half of the way there, he got a flat tire. His mother drove him the rest of the way at a speed of 48 mph. What was his average speed?

Answer:

His average speed is 19.2 mph

Step-by-step explanation:

Given;

speed of Anty to school half of the way, v₁ = 12 mph

speed of Anty when his mother drove him half of the way, v₂ = 48 mph

The average speed of Anty is calculated as;

$V_{average} = \frac{Total \ distance }{Total \ time }$

The value of the average speed is closer to 12 mph than 48 mph because Anty spent more time moving at 12 mph than at 48 mph.

Let the equal distance travel at each speed = 96 miles

Time taken at 12 mph = $\frac{96 \ mile}{12 \ m/h} = 8 \ hours$

Time taken at 12 mph = $\frac{96 \ mile}{48 \ m/h} = 2 \ hours$

Average speed = $\frac{Total \ distance}{Total \ time } = \frac{96 \ mile \ + \ 96 \ mile}{2\ hr \ + \ 8\ hr} = \frac{192 \ miles}{10 \ hrs} = 19.2 \ mph$

Also, you can assume any other equal distance traveled at each speed, the average speed will still be 19.2 mph.

Check: let the equal distance traveled at each speed = 480 miles

Time taken at 12 mph = 480/12 = 40 hours

Time taken at 48 mph = 480/ 48 = 10 hours

Average speed = (480 + 480) / (40 + 10)

= 19.2 mph

8 0

2 years ago

In photo, please help I am timed!!

Furkat [3]

Answer:

Step-by-step explanation:

1/4 x 2 = 2/4

2/4 simplified = 1/2

3 0

3 years ago

Read 2 more answers

Write 6:58 pm in a 24-hour time format

Gre4nikov [31]

Answer:

Step-by-step explanation:

Military Time 0658 is: 06:58 AM using 12-hour clock notation, 06:58 using 24-hour clock notation.

5 0

3 years ago

can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc

mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

$\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx$
$\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx$
$\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)$

and so on.

6 0

3 years ago

Help me please !!! Determine the measure

Crank

Answer:

Step-by-step explanation:

3 0

2 years ago