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3 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Tatiana [17]3 years ago

6 0

First you are going to look at the smallest box and figure out what the area is once you do that you will need to find the area of the bigger box once you figure out both of the areas you will then find out what they are you will then add up the amounts to get the area of both squares together

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Alybia was making 115 cookies for a Valentine’s Day party. They were each either red or purple. Forty percent of the cookies wer

NeTakaya

Answer:

69 purple cookies!

Step-by-step explanation:

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B80%7D%20" id="TexFormula1" title=" \sqrt[3]{80} " alt=" \sqrt[3]{80} " ali

Marianna [84]

Start by decomposing the number inside the root into primes

Then group the terms into cubes if possible

$\begin{gathered} 80=2\cdot2\cdot2\cdot2\cdot5 \\ 80=2^3\cdot2\cdot5 \\ 80=10\cdot2^3 \end{gathered}$

rewrite the root

$\sqrt[3]{80}=\sqrt[3]{10\cdot2^3}$

then cancel the terms that are cubes and bring them out of the root

$\sqrt[3]{80}=2\sqrt[3]{10}$

7 0

1 year 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

3 years ago

TanA+tanB+tanAtanB=1 If A+B=45 degree

djyliett [7]

Answer:

See below

Step-by-step explanation:

$\tan A+\tan B + \tan A.\tan B = 1\\\\ \implies \: \tan A + \tan B = 1 - \tan A.\tan B \\ \\ \implies \: \frac{\tan A + \tan B}{ 1 - \tan A.\tan B } = 1\\ \\ \implies \: \tan (A + B) = 1 \\ \\ \implies \: \tan (A + B) =\tan 45 \degree \\ \\ \implies \: A + B = 45 \degree \\ \\ thus \: proved$

8 0

3 years ago

Determine the unit price of the candy using the graph below (I’m confused on this)

LiRa [457]

Brand A costs $2 per ounce, B is $1, C is $0.50(?) I think

4 0

3 years ago