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Nataliya [291]
3 years ago
12

Hours worked / pay 2/$15.00 4/$30.00 6/$45.00 8/$60.00 A/p=7.50h B/ p=15h C/ p=h+15 D/h=7.50p

Mathematics
2 answers:
alex41 [277]3 years ago
7 0
Working for 2 hours pays $15.
Divide $15 by 2 to get $7.50 per hour.
Divide each amount of pay by its hours, and all are $7.50 per hour.
The equation is $7.50 multiplied by the number of hours, or

p = 7.50h
nikitadnepr [17]3 years ago
4 0

Answer:

The answer is D. p=7.50h

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Then, the probability of getting heads on the second toss is  1/2 .

Then, the probability of getting heads on the third toss is  1/2 .

The probability of all three things happening is ...

                 (1/2) x (1/2) x (1/2)  =  1/8  =  0.125  =  12.5% .



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Using a graphical method, solve the simultaneous equations x+y =4 and y=2x+1
adoni [48]

Answer:

(1, 3)

Step-by-step explanation:

Given the expression

x+y = 4 ... 1

y = 2x+1 ....2

Substitute equation 2 into 1

x + (2x+1) = 4

3x + 1 = 4

3x = 4-1

x = 3/3

x = 1

Since y = 2x + 1

y = 2(1) + 1

y = 3

Hence the solution to the equation is (1,3). This means that the coordinate point on the graph where both lines intersect will be at (1, 3)

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Which of the following are congruent angles?
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Letter B. X is a vertical angle, and so is F. Vertical angles are all congruent.
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Solve 73 make sure to also define the limits in the parts a and b
Aleks04 [339]

73.

f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}

a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

b)

Since we can't divide by zero, we need to find when:

x^4-2x^2+144=0

But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

so, for x = -3:

\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty

For x = 4:

\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty

4 0
1 year ago
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