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stepladder [879]
2 years ago
13

4 quarts 2 cups − 1 quart 3 cups = ?

Mathematics
1 answer:
nevsk [136]2 years ago
6 0

Answer: 2 quarts 3 cups

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If the function h is defined by h(x)=<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" a
harkovskaia [24]

Given:

The function is:

h(x)=x^2-3x+5

To find:

The value of h(2x+1).

Solution:

We have,

h(x)=x^2-3x+5

Putting x=2x+1, we get

h(2x+1)=(2x+1)^2-3(2x+1)+5

h(2x+1)=(2x)^2+2(2x)(1)+(1)^2-3(2x)-3(1)+5

h(2x+1)=4x^2+4x+1-6x-3+5

On combining like terms, we get

h(2x+1)=4x^2+(4x-6x)+(1-3+5)

h(2x+1)=4x^2-2x+3

Therefore, the required function is h(2x+1)=4x^2-2x+3.

3 0
3 years ago
Anybody willing to help?
alexandr402 [8]

Answer:

16 is H.9

17. is A.9

Step-by-step explanation:

Hope it helped! ^_^

6 0
2 years ago
Find 310% of 150. a number blank
Ede4ka [16]

Answer:

465

Step-by-step explanation:

Of means multiply

310% * 150

Change to decimal form

3.10 * 150

465

7 0
3 years ago
Read 2 more answers
A number is selected from the set (1, 2, 3, 5, 15, 21, 29, 38, 500). If equal elemental probabilities are assigned, what is
Katen [24]

Answer:

\frac{7}{9} OR 77.78%

Step-by-step explanation:

So, we first need to find out how many numbers fit the "less than 29 or odd" group.

This would be: 1, 2, 3, 5, 15, 21, and 29, which is a total of 7 numbers.

Next, we figure out the total amount of numbers in the population.

This would be: 1, 2, 3, 5, 15, 21, 29, 38, and 500, which is a total of 9 numbers.

Finally, we put the first number (7) over the total (9) = \frac{7}{9}

\frac{7}{9} is now able to be written in a percentage, if you need.

The probability is \frac{7}{9} OR 77.78%

5 0
3 years ago
What is the probability of drawing a 3 or a 4 or a heart from a deck of cards?
Effectus [21]
\Omega=\{2\spadesuit;\ 3\spadesuit;...;A\spadesuit;2\heartsuit ;\ 3\heartsuit;...;A\heartsuit;\ 2\diamondsuit;\ 3\diamondsuit;...;A\diamondsuit;\ 2\clubsuit;\ 3\clubsuit;...;A\clubsuit\}\\\\|\Omega|=52\\\\A=\{3\spadesuit;\ 4\spadesuit;\ 3\diamondsuit;\ 4\diamondsuit;\ 3\clubsuit;\ 4\clubsuit;\ 2\heartsuit;\ 3\heartsuit;\ 4\heartsuit;...;A\heartsuit\}\\\\|A|=2+2+2+13=19\\\\P(A)=\dfrac{|A|}{|\Omega|}\to P(A)=\dfrac{19}{52}

Answer:\ \dfrac{19}{52}
5 0
3 years ago
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