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mamaluj [8]
3 years ago
14

Write the contrapositive of the conditional statement. If it is January, then there are 31 days this month.

Mathematics
2 answers:
guajiro [1.7K]3 years ago
8 0

Answer:

If there are not 31 days this month, then it is not January.

Step-by-step explanation:

Contrapositive of the conditional statement means switching the hypothesis and conclusion of a conditional statement and negating both.

Given statement is :

If it is January, then there are 31 days this month.

Its contrapositive will be :

If there are not 31 days this month, then it is not January.

kifflom [539]3 years ago
5 0
If there aren't 31 days this month, then it isn't January. 

Hope this helps!
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What is 3 3/4 - 1 1/4
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2 1/2

Step-by-step explanation:

The answer is 2 1/2 simply because 3/4 minus 1/4 equals 2/4 which is equivalent equivalent to 1/2. The you would subtract 3 by 1 which is 2 so the answer is 2 1/2

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The number of nails in a 5 lb box is normally distributed with a mean of 325 nails and a standard deviation of 15 nails. suppose
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Step-by-step explanation:

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3 0
2 years ago
Is (-a + b)² = a² - 2ab + b² right or wrong, and why?
antiseptic1488 [7]

Answer:

\huge\boxed{\sf Right.}

Step-by-step explanation:

(-a+b)^2

Let's apply the formula (x+y)² = x² + 2xy + y²

Here, x = -a and y = b

So,

= (-a)² + 2(-a)(b) + (b)²

= a² - 2ab + b²

Hence, it has been proved that (-a + b)² = a² - 2ab + b².

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
6 0
3 years ago
Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,
sweet [91]

Answer:

Hence, the particular solution of the differential equation is y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x.

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

f'' = x - \frac{3}{2}

f' = \int {\left(x-\frac{3}{2}\right) } \, dx

f' = \int {x} \, dx -\frac{3}{2}\int \, dx

f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C, where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative (f'(4) = 1):

1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C

C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)

C = -1

The first derivative is y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1, and the particular solution is found by integrating one more time and using the initial condition (f(0) = 0):

y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1  \right)} \, dx

y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx

y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C

C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0

C = 0

Hence, the particular solution of the differential equation is y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x.

5 0
3 years ago
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