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2 years ago

Which expression is equivalent to 1/2 (13f+5)?

Mathematics

1 answer:

jeka942 years ago

4 0

The answer is B. 6.5f+2.5 because 13/2 is 6.5 and 5/2 is 2.5

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What is the slope- intercept form of 11x-2y=-14

ivann1987 [24]

The answer is y= 11/2x+7

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3 years ago

A riverboat travels downriver (with the current) at a speed of 15 km/h. The current is 3km/h. To someone on the shore, how fast

Rom4ik [11]

Answer:

18km/h

Step-by-step explanation:

if the riverboat is already going 15 km/h and the current is 3km/h then you should add and you get 18 km/h

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3 years ago

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10Hello, can you help me with this? How much is 5 raised to 5?

vesna_86 [32]

Answer:

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3 years ago

For the first 30 km, the bicyclist rode with a speed of v km/hour. For the remaining 17 km he rode with a speed which was 2 km/h

damaskus [11]

Answer:

t= 2.52 hours

Step-by-step explanation:

It is given that for first 30 km, the speed of bicyclist is v km/hour

time taken to cover first 30 km is given by

$t_{1} =\frac{30}{v}$ ( $time=\frac{distance}{speed}$ )

for next 17 km the speed of bicyclist is 2 km/hour greater than his original speed

so the speed to cover next 17 km = v+2

time taken to cover next 17 km is given by

$t_{2} =\frac{17}{v+2}$

now total time t spent by the bicyclist to cover entire trip is given by

$t=t_{1} +t_{2}$

$t=\frac{30}{v} +\frac{17}{v+2}$

now if v=18 , we have

$t=\frac{30}{18} +\frac{17}{20}$

now to add fractions we make the denominator same

Hence we will find the LCM of 18 and 20

LCM of 18 and 20 = 180

now we need to make both the denominator equal to 180

$t=\frac{30(10)}{18(10)} +\frac{17(9)}{20(9)}$ $t=\frac{300}{180} +\frac{153}{180} =\frac{300+153}{180}$

$t=\frac{453}{180} =2.52$ hours (approx)

7 0

3 years ago

Fundamental theorem of calculus<br> <img src="https://tex.z-dn.net/?f=g%28s%29%3D%5Cint%5Climits%5Es_6%20%7B%28t-t%5E4%29%5E6%7D

mr_godi [17]

Answer:

$\displaystyle g'(s) = (s-s^4)^6$

Step-by-step explanation:

The Fundamental Theorem of Calculus states that:
$\displaystyle \frac{d}{dx}\left[ \int_a^x f(t)\, dt \right] = f(x)$

Where <em>a</em> is some constant.

We can let:
$\displaystyle g(t) = (t-t^4)^6$

By substitution:

$\displaystyle g(s) = \int_6^s g(t)\, dt$

Taking the derivative of both sides results in:
$\displaystyle g'(s) = \frac{d}{ds}\left[ \int_6^s g(t)\, dt\right]$

Hence, by the Fundamental Theorem:

$\displaystyle \begin{aligned} g'(s) & = g(s) \\ \\ & = (s-s^4)^6\end{aligned}$

3 0

2 years ago