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3 years ago

5

2. The height of one square pyramid is 12 m. A similar pyramid has a height of 6 m. The volume of the larger pyramid is 400 m3.

Determine each of the following, showing all your work and reasoning.

1 answer:

kvv77 [185]3 years ago

5 0

a) 6 m / 12 m = 1/2 is the right answer

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Perform the computation and write the result in scientific notation:

gtnhenbr [62]

Answer:

6.06 x $10^{9}$

Step-by-step explanation:

(5.25 x $10^{5}$ ) / (8.7 x $10^{-5}$ )

= (5.25 x $10^{5}$ ) x( $10^{5}$ )/ (8.7 )

= (5.25 x $10^{5 + 5}$ )/ (8.7 )

= 0.606 x $10^{10}$

= 6.06 x $10^{9}$

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3 years ago

4x/3+3=x-2 <br>value of x ?

Gekata [30.6K]

Answer:

x = - 15

Step-by-step explanation:

Given

$\frac{4x}{3}$ + 3 = x - 2

Multiply through by 3 to clear the fraction

4x + 9 = 3x - 6 ( subtract 3x from both sides )

x + 9 = - 6 ( subtract 9 from both sides )

x = - 15

6 0

3 years ago

Read 2 more answers

Expand the binomial<br> <img src="https://tex.z-dn.net/?f=%28x-%5Cfrac%7B2%7D%7B5%7D%29%5E%7B2%7D" id="TexFormula1" title="(x-\f

kkurt [141]

Answer:

$(x- \frac{2}{5})^{2} = x^2 -\frac{4}{5}x + \frac{4}{25}$

Step-by-step explanation:

You have two methods to expand this binomial.

Method 1

If you have the expression:

$(x- \frac{2}{5})^{2}$

You can write the expression it in the following way:

$(x-\frac{2}{5})^{2}=(x-\frac{2}{5})(x-\frac{2}{5})$

Then, apply the distributive property:

$(x-\frac{2}{5})(x-\frac{2}{5}) = x^2 -\frac{2}{5}x -\frac{2}{5}x+ (\frac{2}{5})\frac{2}{5}$

Simplify the expression:

$(x-\frac{2}{5})^2= x^2 -\frac{4}{5}x+ (\frac{4}{25})$

---------------------------------------------------------------------------------------

Method 2

For any expression of the form:

$(a-b)^2$

Its expanded form will be:

$(a-b)^2= a^2 -2ab + b^2$

If

$a = x$

$b =\frac{2}{5}$

$(x- \frac{2}{5})^{2} = x^2 - 2x\frac{2}{5} + (\frac{2}{5})^2$

$(x- \frac{2}{5})^{2} = x^2 -\frac{4}{5}x + \frac{4}{25}$

3 0

3 years ago

A man travels 20 km by car from Town P to Town Q at an average speed of x km/h. He finds that the time of the journey would be s

yuradex [85]

Answer:

x = 20.

Step-by-step explanation:

First, you should remember the relation:

Distance = Speed*Time.

First, we know that a man travels a distance of 20km at a speed of x km/h, in a time T.

We can write this as:

20km = (x km/h)*T

We know that the time is shortened by 12 minutes if the speed is increased by 5km/h

Rewriting these 12 minutes in hours (remember that 60min = 1 hour)

12 min = (12/60) hours = 0.2 hours

Then from this, he can travel the same distance of 20km in a time T minus 0.2 hours if the speed is increased by 5 km/h

We can write this as:

20km = (x + 5 km/h)*(T - 0.2 h)

Then we have a system of two equations, and we want to find the value of x:

20km = (x km/h)*T

20km = (x + 5 km/h)*(T - 0.2 h)

First, we should isolate the variable T in one of the equations, if we isolate it in the first one, we will get:

20km/(x km/h) = T

Replacing that in the other equation we get:

20km = (x + 5 km/h)*(T - 0.2 h)

20km = (x + 5 km/h)*( 20km/(x km/h) - 0.2 h)

Now we can solve this for x.

Removing the units (that we know that are correct) so the math is easier to read, we get:

20 = (x + 5)*(20/x - 0.2)

We only want to solve this for x.

20 = x*20/x - x*0.2 + 5*20/x - 5*0.2

20 = 20 - 0.2*x + 100/x - 1

subtracting 20 in both sides we get:

20 - 20 = 20 - 0.2*x + 100/x - 1 - 20

0 = -0.2*x + 100/x - 1

If we multiply both sides by x we get:

0 = -0.2*x^2 + 100 - x

-0.2*x^2 - x + 100 = 0

This is just a quadratic equation, we can solve it using the Bhaskara's equation, the solutions are:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4*(-0.2)*100} }{2*-0.2} = \frac{1 \pm 9 }{-0.4}$

Then the two solutions are:

x = (1 + 9)/-0.4 = -25

x = (1 - 9)/-0.4 = 20

As x is used to represent a speed, the negative solution does not make sense, so we should use the positive one.

x = 20

then the average speed initially is 20 km/h

3 0

3 years ago

What is 3.31 + 1 kilometer

MArishka [77]

Answer:

4.31 kilometer

..........................................

5 0

3 years ago