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3 years ago

Hurry please thank you!

Mathematics

1 answer:

charle [14.2K]3 years ago

3 0

The answer is C good luck

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Use the figure above of Quadrilateral ABCD to answer the following: Use the check list to determine which properties fit. You wi

prohojiy [21]

Explanation:

First, let's draw the quadrilateral. So:

Then, the distance d and slope m between two points with coordinates (x1, y1) and (x2, y2) can be calculated as:

$\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ m=\frac{y_2-y_1}{x_2-x_1} \end{gathered}$

So, the distance and slope of AB where A is (-5,3) and B is (0, 6) are:

$\begin{gathered} d=\sqrt[]{(0-(-5))^2+(6-3)^2} \\ d=\sqrt[]{(0+5)^2+3^2} \\ d=\sqrt[]{34} \\ m=\frac{6-3}{0-(-5)}=\frac{3}{0+5_{}}=\frac{3}{5} \end{gathered}$

The distance and slope of BC where B is (0,6) and C is (5, 3) are:

$\begin{gathered} d=\sqrt[]{(5-0)^2+(3-6)^2}=\sqrt[]{34} \\ m=\frac{3-6}{5-0}=-\frac{3}{5} \end{gathered}$

The distance and slope of CD where C is (5,3) and D is (0, 0) is:

$\begin{gathered} d=\sqrt[]{(0-5)^2+(0-3)^2}=\sqrt[]{34} \\ m=\frac{0-3}{0-5}=\frac{-3}{-5}=\frac{3}{5} \end{gathered}$

The distance and slope of AD where A is (-5,3) and D is (0, 0) are:

$\begin{gathered} d=\sqrt[]{(0-(-5))^2+(0-3)^2} \\ d=\sqrt[]{(0+5)^2+(3)^2}=\sqrt[]{34} \\ m=\frac{0-3}{0-(-5)}=-\frac{3}{5} \end{gathered}$

Therefore, the correct answers are:

Option 1 : Opposite sides (AB, CD, and BC, AD) have equal slopes making them parallel to each other, making ABCD a parallelogram.

Option 4: The distance of the sides AB, BC, CD, and AD are all congruent

3 0

11 months ago

The point (3, 6) is on the graph of y= 5f(2(x+3))-4 . Find the original point on the graph of y=f(x).

Sonja [21]

Answer:

(12, 2) is the original point on the graph of $y=f(x)$ .

Step-by-step explanation:

Given:

$y= 5f(2(x+3))-4$ has a point (3, 6) on its graph.

To find:

Original point on graph $y=f(x)$ = ?.

Solution:

We are given that The point (3, 6) is on the graph of $y= 5f(2(x+3))-4$

If we put x = 3 and y = 6 in $y= 5f(2(x+3))-4$ , it will satisfy the equation.

Let us the put the values and observe:

$6= 5f(2(3+3))-4\\\Rightarrow 6= 5f(2(6))-4\\\Rightarrow 6= 5f(12)-4\\\Rightarrow 6+4=5f(12)\\\Rightarrow 5f(12)= 6+4\\\Rightarrow 5f(12)= 10\\\Rightarrow f(12)= \dfrac{10}{5}\\\Rightarrow f(12)= 2\\OR\\\Rightarrow 2=f(12)$

Now, let us compare the above with the following:

$y=f(x)$

we get y = 2 and x = 12

So, the original point on graph of $y=f(x)$ is (12, 2).

7 0

3 years ago

Consider the following theorem. Theorem If f is integrable on [a, b], then b a f(x) dx = lim n→[infinity] n i = 1 f(xi)Δx where

mel-nik [20]

Split up the interval [1, 9] into n subintervals of equal length (9 - 1)/n = 8/n :

[1, 1 + 8/n], [1 + 8/n, 1 + 16/n], [1 + 16/n, 1 + 24/n], …, [1 + 8 (n - 1)/n, 9]

It should be clear that the left endpoint of each subinterval make up an arithmetic sequence, so that the i-th subinterval has left endpoint

1 + 8/n (i - 1)

Then we approximate the definite integral by the sum of the areas of n rectangles with length 8/n and height $f(x_i)$ :

$\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx \approx \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right)$

Take the limit as n approaches infinity and the approximation becomes exact. So we have

$\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx = \lim_{n\to\infty} \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right) \\\\ = \lim_{n\to\infty} \frac8n \sum_{i=1}^n \left(1+\frac{16}n(i-1)+\frac{64}{n^2}(i-1)^2-4-\frac{32}n(i-1)+6\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=1}^n \left(64(i-1)^2-16n(i-1)+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=0}^{n-1} \left(64i^2-16ni+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(64\sum_{i=0}^{n-1}i^2 - 16n\sum_{i=0}^{n-1}i + 3n^2\sum{i=0}^{n-1}1\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{64(2n-1)n(n-1)}{6} - \frac{16n^2(n-1)}{2} + 3n^3\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{49n^3}3-24n^2+\frac{32n}3\right) \\\\= \lim_{n\to\infty} \frac{8\left(49n^2-72n+32\right)}{3n^2} = \boxed{\frac{392}3}$