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OLEGan [10]
3 years ago
10

Which of the following illustrates the truth value of the given conditional statement?

Mathematics
1 answer:
bazaltina [42]3 years ago
6 0
Given p: 10 < 7, clearly p is false and ~p is true. Also, given q: 10 < 5, clearly q is false and ~q is true.

Thus, the truth value of the conditional statement ~p → ~q is <span>T T → T.</span>
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That probability is about 6.7%.

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Solve this equation with substitution method :-<br> x+5 = 3 (y+5)<br> x-5 = 7 (y-5)
Damm [24]
From the first equation,
x+5 = 3(y+5)
x = 3y + 15 - 5
Now substitue x in the second equation with (3y +15 - 5).
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3y +5 = 7y - 35
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Since y is 10, and x is (3y +15 - 5),
x = 30 + 15 - 5 = 40
5 0
2 years ago
When each side of an equation has been simplified, equations that have BLANK coefficients and BLANK constants on each side have
kobusy [5.1K]

Answer:

The answer is below

Step-by-step explanation:

An equation shows the relationship between two or more variables. An equation is a statement that shows the equality between expressions. An equation with infinitely many solution is when all numbers are solutions, that is there is no one solution. Example is: x + 3 = x + 3.

When each side of an equation has been simplified, equations that have the same coefficients and the same constants on each side have infinitely many solutions

7 0
2 years ago
What is the equation of the line that passes through the points (15, 9) and (Negative 2, 9)? y =
Bad White [126]

\bf (\stackrel{x_1}{15}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{9}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{9}-\stackrel{y1}{9}}}{\underset{run} {\underset{x_2}{-2}-\underset{x_1}{15}}}\implies \cfrac{0}{-17}\implies 0 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{9}=\stackrel{m}{0}(x-\stackrel{x_1}{15})\implies y-9=0\implies y=9

3 0
3 years ago
Read 2 more answers
7,,,,,,,,,,,,,,,,,,,,please help
Goryan [66]

Answer:

1

Step-by-step explanation:

The given expression is

\frac{2^{\frac{11}{2} }\times 2^{-7}\times 2^{-5}}{2^3\times 2^{\frac{1}{2} }\times 2^{-10}}

Recall and use

a^m\times a^n=a^{m+n}

\frac{2^{\frac{11}{2} -7-5}}{ 2^{3+\frac{1}{2} -10}}

\frac{2^{-\frac{13}{2}}}{ 2^{-\frac{13}{2}}}

Same bases are dividing, so we write one base and subtract the exponent.

2^{-\frac{13}{2}+\frac{13}{2}}

2^0=1

8 0
3 years ago
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