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3 years ago

(0.04)^3/2. ...............

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

4 0

Answer:

Step-by-step explanation:

hello :

(0.04)^3/2=√(0.04)^3 =√(2²/10^2)^3 = √((2/10)^2)^3

(0.04)^3/2=√((2/10)^3)^2=8/1000 = 0.008

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Jim paddles from one shore of a lake three miles wide at 4 mph, and John paddles from the opposite shore at 5 mph. How long will

AleksandrR [38]

D= distance
t= time
s= speed

FORMULA
Distance= speed * time

STEP 1:
Find total distance
Jim's D + John's D= 3 miles

STEP 2:
Distance= (Jim's speed * time) + (John's speed * time)
D= 4t + 5t
3 miles= 9t
divide both sides by 9
3/9= t
reduce by 3
1/3 hour= t

STEP 3:
1 hour= 60 minutes
=1/3 * 60
=(60/3)
= 20 minutes

ANSWER: They will travel 20 minutes before they meet.

Hope this helps! If it does, please mark me brainliest! :)

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3 years ago

Which series of transformations would transform triangle A into similar triangle A' in Quadrant IV? A. Rotate triangle A 90 cloc

VikaD [51]

Answer:

option D

Step-by-step explanation:

It is required to transform triangle A into similar triangle A' in Quadrant IV.

The given triangle in Quadrant II.

We will check the options:

A. Rotate triangle A 90 clockwise about the origin then translate five units up.

Wrong because triangle A' will be at Quadrant I.

B. Dilate triangle A by a scale factor of 4. Translate triangle A to the left.

Wrong because triangle A' will be at Quadrant II.

C. Reflect triangle A across the y-axis and then translate the triangle 3 units up and 2 units to the right.

Wrong because triangle A' will be at Quadrant I.

D. Reflect triangle A across the x-axis and then reflect the triangle across the y-axis.

True Because reflection triangle A across the x-axis, the first image will be at Quadrant III, then reflection the first image across the y-axis, we will get triangle A' will be at Quadrant IV.

So, the answer is option D

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3 years ago

(Area of triangle).

pochemuha

Your answer is B for this one

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3 years ago

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Find the surface area to the nearest square foot. how to do this?

klio [65]

$\bold{ANSWER:}$
1026.78

$\bold{SOLUTION:}$

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2 years ago

Plzzz I need help with this question I tried to solve it many times but I can't

blsea [12.9K]

Answer and Step-by-step explanation: Area of a right triangle, (as any other triangle), is calculated as: $A1=\frac{(base)(height)}{2}$

Area of a rectangle is calculated as: $A2=(side)(side)$

Area of a right trapezoid is: $A3=\frac{(a+b)h}{2}$ , where:

a is short base

b is long base

h is height

1) Expressing areas in terms of x:

Area of triangle S1:

$S1=\frac{(2x-3)(4x-6)}{2}$

$S1=4x^{2}-12x+9$

Area of rectangle S2:

$S2 = (4x-6)(3x-2)$

$S2=12x^{2}-26x+12$

Area of trapezoid S3:

$S3=\frac{(2x+3+4x+1)(2x-3)}{2}$

$S3=\frac{(6x+4)(2x-3)}{2}$

$S3=6x^{2}-5x-6$

2) a) $S=4x^{2}-12x+9+12x^{2}-26x+12-(6x^{2}-5x-6)$

$S=4x^{2}-12x+9+12x^{2}-26x+12-6x^{2}+5x+6$

$S=10x^{2}-33x+37$

Which is the same as S = (2x-3)(5x-9)

b) For the areas to be the same:

$\frac{(3x-2+3x-2+2x-3)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$\frac{(8x-7)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$32x^{2}-48x-28x+42=12x^{2}+8x-18x-12$

$20x^{2}-66x+54=0$

Using Bhaskara to solve the second degree equation:

$\frac{66+\sqrt{(-66)^{2}-(4.20.54)} }{2(20)}$

$x_{1}=\frac{66+6}{40}$ = 1.8

$x_{2}=\frac{66-6}{40}$ = 1.5

For the areas of AFGC and ADEB to be equal, x has to be 1.5 or 1.8.

c) Expand a polynomial (or equation) is to multiply all the terms, remiving the parenthesis. Reduce a polynomial (or equation) is to combine terms alike,e.g.:

$S=(2x-3)(5x-9)$