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emmasim [6.3K]
3 years ago
9

How can you use functions to solve real world problems ??

Mathematics
1 answer:
Ber [7]3 years ago
5 0
You can use functions to slove real world problems by saying if some thing cost 2 dollar you would have 1 thing

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I need to know The value of f(1/2)
Lostsunrise [7]

\huge\boxed{\textsf{a.)}\ 10}

Substitute \frac{1}{2} for x.

f\left(\frac{1}{2}\right)=5\cdot4^\frac{1}{2}

Use a^\frac{m}{n}=\sqrt[n]{a^m}.

\begin{aligned}f\left(\frac{1}{2}\right)&=5\cdot\sqrt[2]{4^1}\\&=5\cdot\sqrt[2]{4}\\&=5\cdot2\\&=\boxed{10}\end{aligned}

7 0
1 year ago
Does (-2)^n/( 3^n+1) converge or diverge?
hodyreva [135]
\left|\dfrac{(-2)^n}{3^n+1}\right|=\dfrac{2^n}{3^n+1}

As n\to\infty, the sequence a_n=\left(\dfrac23\right)^n converges to zero.

If you're talking about the infinite series

\displaystyle\sum_{n\ge0}\dfrac{(-2)^n}{3^n+1}

well we've shown by comparison that this series must also converge because we know any geometric series \sum\limits_n r^n will converge as long as |r|.
3 0
3 years ago
Help please and thank you ​
Dominik [7]

Answer:

  D

Step-by-step explanation:

We assume the rotation R is <em>counterclockwise</em> 60°.

__

The exponent on R is the number of times it is applied. That is, R² = R(R(figure)). So, the composition is equivalent to R^(2-4) = R^-2.

When the exponent of R is negative, it is essentially the inverse function. That is, applying the function R to the result will give the figure you started with. Equivalently, it is rotation in the other direction.

  (\mathcal{R}^2\circ\mathcal{R}^{-4})(\text{figure})=\mathcal{R}^{-2}(\text{figure})=\text{figure rotated $120^{\circ}$ CW}

The point 120° clockwise from B is D.

The desired image point is D.

3 0
3 years ago
Is 12:6 more than 3:2
Dahasolnce [82]
Yes. Think of it as a fraction -
12/6 reduces to 4/2.
4/2 > 3/2.
4 0
3 years ago
Read 2 more answers
X is directly proportional with the square root of Y. If X=128 and Y=16 find: X when Y=36 B) Y when X=48 ( HELP ASAP!)
Tems11 [23]

x = sqrt{y}k

K is the constant of proportionality.

128 = sqrt{16}k

128 = 4k

128/4 = k

32 = k

x = sqrt{y}k

x = sqrt{36}•32

x = 6 • 32

x = 192

Find y when x = 48.

x = sqrt{y}k

48 = sqrt{y}32

48/32 = sqrt{y}

(48/32)^2 = [sqrt{y}]^2

2.25 = y

7 0
3 years ago
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