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seraphim [82]
3 years ago
7

Help quickly plz!!

Mathematics
1 answer:
miskamm [114]3 years ago
3 0
<h2>Explanation:</h2><h2 /><h3><u>PART 1. Using an algorithm:</u></h3>

<u>First expression:</u>

3 \times \frac{3}{4}

Step 1: Multiply numerators and denominators.

3 \times \frac{3}{4}=\frac{9}{4}

<u>Second expression:</u>

\frac{1}{4} \times \frac{2}{3}=\frac{2}{12}

Step 1: Multiply numerators and denominators.

3 \times \frac{3}{4}=\frac{9}{4}

<u>Third expression:</u>

1\frac{1}{3} \times 2

Step 1: Convert mixed fraction into improper fraction:

1\frac{1}{3} \\ \\ \\ Add \ whole \ and \ fractional \ part: \\ \\ 1+\frac{1}{3}=\frac{4}{3}

Step 2: Replace the improper form into the expression

\frac{4}{3}\times 2

Step 2: Multiply numerators and denominators.

\frac{4}{3}\times 2=\frac{8}{3}

<h3><u>PART 2. Using diagram:</u></h3><h3 />

The diagram is shown below. Basically the steps we follow are the same as if it were using algorithms.

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4 0
3 years ago
Let f(x)= 8/(1+3e^−0.7x)
Nataly [62]

Answer:

  0.31

Step-by-step explanation:

This is a good question to ask your calculator. Be sure to use appropriate parentheses. You need them around the exponent of e as well as around the denominator.

8 0
3 years ago
Select all the correct locations on the table. Select all of the complex numbers in the given table.
goblinko [34]

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answer in  picture

Step-by-step explanation:

4 0
4 years ago
Can anyone help me solve this? Please and thank you!
AveGali [126]

It's the sum of a geometric sequence.

Let's rewrite it a bit:

\displaystyle\\\sum_{n=1}^{10}8\left(\dfrac{1}{4}\right)^{n-1}=8\sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n\cdot \left(\dfrac{1}{4}\right)^{-1}=8\sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n\cdot 4=32\sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n

And now let's calculate this sum \displaystyle \sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n:

S_n=\dfrac{a(1-r^n)}{1-r}\\\\a=\dfrac{1}{4}\\n=10\\r=\dfrac{1}{4}\\\\S_{10}=\dfrac{\dfrac{1}{4}\cdot\left(1-\left(\dfrac{1}{4}\right)^{10}\right)}{1-\dfrac{1}{4}}=\dfrac{\dfrac{1}{4}\cdot\left(1-\dfrac{1}{1048576}\right)}{\dfrac{3}{4}}=\dfrac{\dfrac{1048575}{1048576}}{3}=\dfrac{1048575}{3145728}=\\=\dfrac{349525}{1048576}

Now let's calculate the initial sum:

\displaystyle 32\cdot  \sum_{n=1}^{10}\left(\dfrac{1}{4}\right)^n=32\cdot \dfrac{349525}{1048576}=\dfrac{349525}{32768}\approx10.67

8 0
2 years ago
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