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3 years ago

The ratio of Mark’s age to Reeta’s age is 3 : 5 Mark’s age is 24 years. Work out Reeta's age.

1 answer:

6 0

The ratio of Reeta's age and Site's age is 3:5 and the sum of their ages is 80 years. The ratio of their ages after 10 years will be

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X-5 over (3x+9)(x-4)

STatiana [176]

Answer:

Simplified equation: $\frac{x-5}{3(x+3) (x-4)}$

<em>Brainliest Pls</em>

6 0

3 years ago

Please find the exact length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2). Thank y

I am Lyosha [343]

Answer:

the exact length of the midsegment of trapezoid JKLM = $\mathbf{ = 3 \sqrt{5} }$ i.e 6.708 units on the graph

Step-by-step explanation:

From the diagram attached below; we can see a graphical representation showing the mid-segment of the trapezoid JKLM. The mid-segment is located at the line parallel to the sides of the trapezoid. However; these mid-segments are X and Y found on the line JK and LM respectively from the graph.

Using the expression for midpoints between two points to determine the exact length of the mid-segment ; we have:

$\mathbf{ YX = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }$

$\mathbf{ YX = \sqrt{(8-5)^2+(8-2)^2} }$

$\mathbf{ YX = \sqrt{(3)^2+(6)^2} }$

$\mathbf{ YX = \sqrt{9+36} }$

$\mathbf{ YX = \sqrt{45} }$

$\mathbf{ YX = \sqrt{9*5} }$

$\mathbf{ YX = 3 \sqrt{5} }$

Thus; the exact length of the midsegment of trapezoid JKLM = $\mathbf{ = 3 \sqrt{5} }$ i.e 6.708 units on the graph

8 0

3 years ago

Bob rides his bike from his house to a friend's house. After riding 9 miles, he realizes he has traveled 3/4 of the way. How far

o-na [289]

Answer:

12 miles

Step-by-step explanation:

sorry if im wrong

8 0

3 years ago

What is 15 liters increased by 60%

Alexeev081 [22]

It is 24 liters
100%=15
increase by 60%
100 + 60 = 160
160% of 15
15 x 1.60 =24

6 0

3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

3 years ago