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pshichka [43]
3 years ago
12

(1/4)^3+(3/4)^3+3(1/4)(3/4)(1/4+3/4)

Mathematics
1 answer:
bogdanovich [222]3 years ago
6 0
The answer should be 1
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Help me with my little brothers test i have a pdf document of the test and i need a answer key!!
mina [271]

Answer:

c

Step-by-step explanation:

it starts at (0,0) and follows the points best

5 0
3 years ago
Anybody have an answer?
lyudmila [28]

(3+2)^2+(5+4)^2=x^2\\ 5^2+9^2=x^2\\ 25+81=x^2\\ x^2=106\\ x\approx10.3

7 0
3 years ago
Type the correct answer in each box. use numerals instead of words. students in a reading class have three options for their boo
Alexus [3.1K]

The answers to the questions on how she would use the colored chips are given here:

  • brochure 4 blue chips
  • writing a new article 6 yellow chips
  • designing a computer presentation 10 red chips

<h3>How to solve for the values</h3>

For the first percentage

20 / 100 = x/20

= 20 * 20 = 100x

400 = 100x

divide through by 100

x = 400/100

x = 4

For the second percentage

30/100 = x/20

600 = 100x

x = 600/100

x = 6

For the last

50/100 = x/20

1000 = 100x

x = 10

The proof is that 6 + 4 + 10 still gives us 20

Read more on percentages here: brainly.com/question/24877689

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4 0
2 years ago
I don’t understand constant rates of change help!
REY [17]

Answer:

The (constant) rate of change with respect to the variable x of a linear function y = f(x) is the slope of its graph. If x and f have units in Definition 2, then the units of the rate of change are those of f divided by those of x.

Step-by-step explanation:

5 0
4 years ago
A car's position is given by s(t) = {3 – 5t? + 7t hundreds of meters with t in minutes.
Nadya [2.5K]

Answer:

A) (1 s, 2.3 s)

B) (-4 m/s², 3.8 m/s²)

Step-by-step explanation:

The car's position which is the distance is given by the equation;

s(t) = t³ - 5t² + 7t

A) Velocity is the first derivative of the distance. Thus;

v(t) = ds/dt = 3t² - 10t + 7

At v = 0, we have;

3t² - 10t + 7 = 0

Using quadratic formula, we have;

t = 1 and t = 2.3

Thus, time at velocity of 0 is t = (1 s, 2.3 s)

B) acceleration is the derivative of the velocity. Thus;

a(t) = dV/dt = 6t - 10

At velocity of 0, we got t = 1 and t = 2.3

Thus;

a(1) = 6(1) - 10 = -4 m/s²

a(2.3) = 6(2.3) - 10 = 3.8 m/s

Thus, a(t) at v = 0 gives; (-4 m/s², 3.8 m/s²)

8 0
3 years ago
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