All categories

3 years ago

Giving brainliest for correct answer

Mathematics

2 answers:

yanalaym [24]3 years ago

7 0

Answer:

Option A 64/27 cm³

Step-by-step explanation:

Side of each cube=1/3 cm

Volume of a cube=side³=(1/3)³=1/27 cm³

Total no. of cubes in the prism=4*4*4=64 cubes

Volume of prism=Volume of 1 cube*Total no. of cubes

=(1/27)*64

=64/27 cm³

Ymorist [56]3 years ago

4 0

A) would be your right answer

Hope this helps

You might be interested in

Select the rational expression that is equivalent to the given expression below. 4/x-3

Aliun [14]

To find the correct option we need to calculate each of the given options.
Option (1):
$\frac{x+2}{x-3} / \frac{4}{x+2} = \frac{x+2}{x-3} * \frac{x+2}{4} = \frac{(x+2)^2}{4(x-3)}$

Option (2):
$\frac{x+2}{x-3} * \frac{4}{x+2} = \frac{4}{x-3}$

Option (3):
$\frac{x+2}{x-3} - \frac{4}{x+2} = \frac{(x+2)^2-4(x-3)}{(x-3)(x+2)}$

option (4)
$\frac{x+2}{x-3} + \frac{4}{x+2} = \frac{(x+2)^2+4(x-3)}{(x-3)(x+2)}$

So, the correct answer is option (2)

7 0

3 years ago

Read 2 more answers

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

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

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

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

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

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

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.