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2 years ago

Evaluate c/2 - 3 + 6/d when c=14 and d=13???

Mathematics

1 answer:

Kazeer [188]2 years ago

5 0

c/2-3+6/d

substitute c=14 and d=13:

14/2-3+6/13

simplify:

7-3+6/13

4+6/13

multiply 4 by 13/13 to group like terms (for a fraction):

52/13+6/13

combine:

58/13

Therefore, the answer is 58/13 or around 4.462

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Anyone know how to do this?

Lera25 [3.4K]

Answer:

Part A) 25

Part B) 225

Step-by-step explanation:

Part A

The slope of the graph gives us the mile per gallon of the car.

Recall that the slope of the graph is rise over run.

$m = \frac{200}{8} = 25$

Therefore car gets 25 miles per gallon

Part B) Since we know the slope of the direct variation line, we substitute into y=kx to get:

$y = 25x$

To find the number of miles the car travels using 9 gallons, we substitute x=9 to get:

$y = 25 \times 9$

This gives us 225 miles

8 0

2 years ago

Perimeter of an isosceles triangle is 90 m. If the length of the two equal sides is three fourth of the unequal side , find the

Sveta_85 [38]

Answer:

The length of the equal side is 27 meters each and that of the unequal side is 36 meters

Step-by-step explanation:

An isosceles triangle is a triangle with two sides being equal (also two equal base angles).

Let us assume the length of the two equal sides to be x meters each, and the length of the unequal side to be y meter. Since the perimeter of the triangle is 90 m, it can be expressed as:

x + x + y = 90

2x + y = 90

But the length of the equal side is three fourth of the unequal side, i.e x = 3/4y

Therefore:

2(3/4y) + y = 90

3/2y + y = 90

2.5y = 90

y = 90/2.5

y = 36 meters

Also x = 3/4 * 36 = 27 meters

The length of the equal side is 27 meters each and that of the unequal side is 36 meters

5 0

2 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

2 years ago

Please help me with this. I will give brainliest.

Temka [501]

Answer:

do you still need this answered?

Step-by-step explanation:

7 0

2 years ago

Which equation should you use to solve for x?

oee [108]

Answer:

1st Option

Step-by-step explanation:

All angles in a triangle must add up to 180°.

Step 1: Set up equation

40° + 60° + 16x = 180°

Step 2: Rewrite

60° + 16x + 40° = 180°

8 0

3 years ago

Read 2 more answers