8x42+3x59-62=n. What does "n" equal?

8 + 3 + 2a = -1 + 3a + 4

otez555 [7]

Answer:

a = 8

Step-by-step explanation:

8 + 3 + 1 - 4 = 3a - 2a

8 = a

a = 8

I hope this helps a little bit.

3 0

3 years ago

Read 2 more answers

If angle bisector EG is drawn find angle EGF

Gnesinka [82]

Answer:

90°

Step-by-step explanation:

because by bisecting the angle, you are going through the centre in this triangle. Splitting to triangle in half gives two right-angled triangles :)

sorry for bad drawing

6 0

3 years ago

The admission fee at a zoo is $2 for children and $4 for adults. If 1800 people enter the zoo and $6000 is collected, how many a

ipn [44]

Answer:

The number of adults is $1,200$

Step-by-step explanation:

Let

x------> the number of children

y------> the number of adults

we know that

$x+y=1800$ -------> equation A

$2x+4y=6000$ -------> equation B

Multiply equation A by $-2$

$-2(x+y)=-2*1800$ --------> $-2x-2y=-3600$ ------> equation C

Adds equation B and equation C

$2x+4y=6000\\-2x-2y=-3600\\---------\\4y-2y=6000-3600\\2y=2400\\y=2400/2\\y=1200$

4 0

3 years ago

Read 2 more answers

Katt and sanjeev are collecting money for charity. For every £2 that kat collects, sanjeev collects £3. They collect £255 in tot

il63 [147K]

Answer: Kate = £102

Sanjeev = £153

Step-by-step explanation:

From the question, we are informed that for every £2 that kat collects, sanjeev collects £3. This csn be represented in a fraction as:

Kat = 2/5

Sanjeev = 3/5

The amount that each collects will be:

Kat = 2/5 × £255

= 2 × £51

= £102

Sanjeev = 3/5 × £255

= 3 × £51

= £153

6 0

3 years ago

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$

3 0

3 years ago