All categories

2 years ago

NEED HELP ASAP WILL MARK BRAINLIEST IF RIGHT!!!

Mathematics

1 answer:

Zarrin [17]2 years ago

6 0

Answer: Area = x² + 6x + 8

Step-by-step explanation:

Area = length x width

Area = (x+2)(x+4)

Area = (x)(x) + (x)(4) + (2)(x) + (2)(4)

Area = x² + 4x + 2x + 8

Area = x² + 6x + 8

You might be interested in

If f(x)=2x+1 and g(x)=x-3 then g(f(x))= ?x + ?

Thepotemich [5.8K]

Answer:

g(f(x)) = 2x - 2

Step-by-step explanation:

to evaluate g(f(x)) substitute x = f(x) into g(x)

g(f(x)) = g(2x + 1) = 2x + 1 - 3 = 2x - 2

6 0

3 years ago

5(x+3)-7(6-3)+29=50-3(8-x)

Sveta_85 [38]

Answer:

$\quad x=\frac{3}{2}$

Step-by-step explanation:

$7\left(6-3\right)=21$ , $\mathrm{Expand\:}5\left(x+3\right)-21+29:\quad5x+23$ , $\mathrm{Expand\:}50-3\left(8-x\right):\quad 3x+26$ , Subtract 23 and 3x from both sides and simplify: $2x=3$ , Divide both sides by 2 and simplify: $x=\frac{3}{2}$

<em>Hope this helps!!!</em>

6 0

2 years ago

Determine whether the following sequence is geometric. If so, find the common ratio.

Debora [2.8K]

The sequence is geometric and the common ratio is 2

7 0

2 years ago

How to know if a function is periodic without graphing it ?

zhenek [66]

A function $f(t)$ is periodic if there is some constant $k$ such that $f(t+k)=f(k)$ for all $t$ in the domain of $f(t)$ . Then $k$ is the "period" of $f(t)$ .

Example:

If $f(x)=\sin x$ , then we have $\sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x$ , and so $\sin x$ is periodic with period $2\pi$ .

It gets a bit more complicated for a function like yours. We're looking for $k$ such that

$\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5$

Expanding on the left, you have

$\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2$

and

$1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5$

It follows that the following must be satisfied:

$\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}$

The first two equations are satisfied whenever $k\in\{0,\pm4,\pm8,\ldots\}$ , or more generally, when $k=4n$ and $n\in\mathbb Z$ (i.e. any multiple of 4).

The second two are satisfied whenever $k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}$ , and more generally when $k=\dfrac{10n}7$ with $n\in\mathbb Z$ (any multiple of 10/7).

It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when $k$ is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

$\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2$

$\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5$

More generally, it can be shown that

$f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))$

is periodic with period $\mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n)$ .

4 0

2 years ago

you are allowed to work a total of no more than 30 hours each week at your two jobs. Lawn mowing pays $5 per hour and babysittin

timama [110]

Answer:

5X + 8Y >= 300; intersection at (-20, 50)

Step-by-step explanation:

let t = work hours

0 < t < 30

X = time lawn mowing

Y = time babysitting.

X + Y < 30

5X + 8Y >= 300

We could solve...

X < 30 - Y

5(30 - Y) + 8Y >=300

150 - 5Y + 8Y >= 300

3Y >=150

Y >=50

then X < -20

intersection at (-20, 50)

4 0

2 years ago

Read 2 more answers

NEED HELP ASAP WILL MARK BRAINLIEST IF RIGHT!!!​

NEED HELP ASAP WILL MARK BRAINLIEST IF RIGHT!!!