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3 years ago

A motor cyclist completes a journey at an average speed of 65 mph in 3½ hours. Calculate the distance travelled.

Mathematics

2 answers:

sweet-ann [11.9K]3 years ago

8 0

Answer:

227.5

Step-by-step explanation:

multiply 65 mph to 3 1/2 [3.5 (estimated)] to calculate the distanbce travelled

frosja888 [35]3 years ago

6 0

Answer:

227.5

Step-by-step explanation:

65 mhp

after 3.5 h

Just multiply

65 x 3.5 = 227.5

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2. What is the volume of the regular pyramid to the nearest whole number?<br> 14 cm<br> 12 cm

Yuki888 [10]

291 cm^3

The volume of a pyramid is (area of base x height) / 3
In this case the base is an equilateral triangle with side 12cm.
Using the semi perimeter formula gives
Base = sr( 18 x 6 x 6 x 6) = 62.3538...
Volume = 62.3538... x 14 / 3 = 290.9845...
= 291 to nearest whole number

You could find the base area using Pythagoras and 1/2bh or the Sine formula

5 0

3 years ago

If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value o

almond37 [142]

If the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

<h3>What is area of the region bounded by the curve ?</h3>

An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.

Area bounded by the curve $=\int\limits^a_b {x} \, dx$

We have,

$y^2 =4ax$

⇒ $y=\sqrt{4ax}$

$x= 4a$ ,

Area of the region $=\frac{256}{3}$ Sq units

Now comparing both given equation to get the intersection between points;

$y^2=16a^2$

$y=4a$

So,

Area bounded by the curve $= \[ \int_{0}^{4a} y \,dx \]$

$\frac{256}{3} =\[ \int_{0}^{4a} \sqrt{4ax} \,dx \]$

$\frac{256}{3}= \[\sqrt{4a} \int_{0}^{4a} \sqrt{x} \,dx \]$

$\frac{256}{3}= \[2\sqrt{a} \int_{0}^{4a} x^{\frac{1}{2} } \,dx \]$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} *\frac{2}{3} \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}$

On applying the limits we get;

$\frac{256}{3}= \frac{4}{3} \sqrt{a} \left[\begin{array}{ccc}(4a)^{\frac{3}{2} \end{array}\right]$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} *\sqrt{(4a)^{3} }$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} * 8 *a^{2} \sqrt{a}$

$\frac{256}{3}= \frac{4}{3} * 8 *a^{3}$

⇒ $a^{3} =8$

$a=2$

Hence, we can say that if the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

To know more about Area bounded by the curve click here

brainly.com/question/13252576

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8 0

2 years ago

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How can i prove this? Please someone help i get 100points

MA_775_DIABLO [31]

Typing.................

3 0

3 years ago

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