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erastova [34]
3 years ago
8

PQRS is a parallelogram with PQ=26cm and QR=20cm . If the distance between the longer-sides is 12.5cm , Find​

Mathematics
1 answer:
Kay [80]3 years ago
6 0

Answer:

Below.

Step-by-step explanation:

I am guessing you want the area of the parallelogram.

Take the longer side to be the base.

Area = base * distance between base and opposite side

=  26 * 12.5

= 325 cm^2.

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alexdok [17]

Answer:

Step-by-step explanation:

A linear function is a polynomial function of the first degree that has the form:

f (x) = m*x + b   or  y=m*x + b

where y is the dependent variable, x is the independent variable, m is the slope of the line and b is the intercept with the Y axis.

The slope m measures the inclination of the line with respect to the abscissa axis, that is, the x axis.  According to the value of the slope m, the linear function can be increasing if m> 0, decreasing if m <0 or constant if m = 0.

You know that the number of sea turtle deaths per year is modeled by f(x) = 13.42x + 109.118.   Then the value of the slope is 13.42. In this scenario, the slope indicates that the number of deaths of sea turtles grows in a proportion of 13.42 with respect to the pollution index of the bay.

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3 years ago
Which series of transformations shows that triangle A is congruent to triangle B?
Vilka [71]
I would say A
hope that helps!!!
5 0
3 years ago
Read 2 more answers
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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Answer:

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3 years ago
A fair die is rolled repeatedly. find the expected number of rolls until all 6 faces appear.
Makovka662 [10]
Each face has a chance of 1/6
So you beed 6 rolls
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3 years ago
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