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3 years ago

A salt march harvest mouse ran a 360 cm straight course race in 9 seconds. How fast did it learn

Mathematics

2 answers:

Makovka662 [10]3 years ago

6 0

Answer: $40c m/s$

Solution:

Mouse ran 360 cm straight course race in 9 seconds.

We need to find the rate or speed of mouse

We know the speed formula is distance covered by given time.

$Speed=\frac{Distance}{Time}$

Where,

Distance = 360 cm

Time = 9 seconds

$\therefore$ speed $=\frac{360}{9}$

Speed=40 cm/s

Mouse ran with speed 40 cm/s

artcher [175]3 years ago

3 0

Distance covered by by the harvest mouse= 360 cm

Or distance covered by by the harvest mouse= 360 /100= 3.60 m

Time taken to cover the above distance= 9 s

We know,

Speed= distance covered/time taken to cover that distance

Speed= 3.60/9= 0.4 m/s

Therefore, the mouse ran 0.4 m/s fast.

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Terry burnt a total of 654 calories working out. If Terry does the same workout routine a total of 7 times over the course of th

Flauer [41]

Answer: 4,578

Step-by-step explanation:

654 calories x 7

7 0

3 years ago

Look at the problems 2.3×3.68 and 23×368. How are they similar? Which problem has a greater product?

Margarita [4]

Here is your answer:

1. These equations are similar because they "both share the same numbers but one is a decimal and the other is a whole number."

2. You have to solve each equation in order to determine which number is greater:

$2.3\times 3.68$
$2.3 \times 3.68 = 8.464$
$= 8.464$

And:

$23 \times 368 =$
$23 \times 368 = 8464$
$= 8464$

Finally:

$8.464 < 8464$

Therefore "8,464 is greater than 8.464."

Hope this helps!

3 0

3 years ago

Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^

Gemiola [76]

Answer:

$\rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x}$

Step-by-step explanation:

we would like to figure out the differential coefficient of $e^{2x}(1+\ln(x))$

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

$\displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )$

to do so distribute:

$\displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x}$

take derivative in both sides which yields:

$\displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )$

by sum derivation rule we acquire:

$\rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x}$

Part-A: differentiating $e^{2x}$

$\displaystyle \frac{d}{dx} {e}^{2x}$

the rule of composite function derivation is given by:

$\rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)$

so let g(x) [2x] be u and transform it:

$\displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x$

differentiate:

$\displaystyle {e}^{u} \cdot 2$

substitute back:

$\displaystyle \boxed{2{e}^{2x} }$

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

$\displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)$

let

$f(x) \implies \ln(x)$
$g(x) \implies {e}^{2x}$

substitute

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x}$

differentiate:

$\rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }$

Final part:

substitute what we got:

$\rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }$

and we're done!

6 0

3 years ago

Please I would be grateful if someone answer this question

Alina [70]

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5 0

2 years ago

The sum of a number is<br> 4 less then 2 times a<br> number is 35 what is the number?

Artist 52 [7]

<h2>Answer:</h2>

<h2>Step-by-step explanation:</h2>

35 divided by 2 = 17.5
17.5 + 4 = 21.5
<u>Just do the inverse</u>

5 0

2 years ago