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Anna11 [10]
3 years ago
10

What is the following sum in simplest form - see attached.

Mathematics
1 answer:
melomori [17]3 years ago
5 0
You must first simplify the radicals by taking out any perfect squares.

√8  + 3√2  + √32 =
√4 * √2   +  3√2   + √16 * √2 =
2√2  + 3√2  + 4√2 =
9√2
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Determine the coordinate of the point shown. It’s on a (3,-5) (1,-2) (1.5, -2.5) (-2.5, 1.5)
Sveta_85 [38]

Answer:

its a function

slope form:y+5=-1.18·(x-3)

(-2.5,1.5)

Step-by-step explanation:

4 0
2 years ago
In the year 2003, a company made $6.8 million in profit. For each consecutive year after that, their profit increased by 13%. Ho
LenKa [72]

Answer:

$7.7 million

Step-by-step explanation:

Given: $6.8 million profit made by company.

           13% increase in profit every year.

Now, finding the company´s profit in the year of 2006.

As we know there is 13% increase in profit.

∴ Profit increase= \frac{13}{100} \times 6.8= \$ 0.884

Next adding company´s profit with increase in profit to get profit in the year 2006.

∴ 6.8+0.884= \$ 7.684 ≅ $7.7

The company´s profit in the year 2006 is $7.7 million.

6 0
2 years ago
4/7 + 2/7 = ? can yall plzz help me simplify the awnser
noname [10]

Answer:

6/7 is the correct answr

4 0
2 years ago
Read 2 more answers
I wish to have $10,000 at the end of 8 yers in a bank offering a simple interest rate of 7.5% per year. How much should I deposi
kolezko [41]

Answer:

You have to invest  $6,250

Step-by-step explanation:

very simple applying the simple interest formula which is

A = P (1 + rt)

Given data

A, final amount =  $10,000

P, initial principal balance=  ?

r, annual interest rate = 7.5%

t, time (in years)= 8 years

we can substitute our given data to find the principal needed.

10000= P(1+0.075*8)\\\10000= P(1.6)\\\

Divide both sides by 1.6 we have

P= \frac{10000}{1.6} \\\P= 6250

P= $6,250

3 0
3 years ago
CALCULUS: Determine which function is a solution to the differential equation y ' − y = 0.
Montano1993 [528]

C: none of these are solutions to the given equation.

• If<em> y(x)</em> = <em>e</em>², then <em>y</em> is constant and <em>y'</em> = 0. Then <em>y'</em> - <em>y</em> = -<em>e</em>² ≠ 0.

• If <em>y(x)</em> = <em>x</em>, then <em>y'</em> = 1, but <em>y'</em> - <em>y</em> = 1 - <em>x</em> ≠ 0.

The actual solution is easy to find, since this equation is separable.

<em>y'</em> - <em>y</em> = 0

d<em>y</em>/d<em>x</em> = <em>y</em>

d<em>y</em>/<em>y</em> = d<em>x</em>

∫ d<em>y</em>/<em>y</em> = ∫ d<em>x</em>

ln|<em>y</em>| = <em>x</em> + <em>C</em>

<em>y</em> = exp(<em>x</em> + <em>C </em>)

<em>y</em> = <em>C</em> exp(<em>x</em>) = <em>C</em> <em>eˣ</em>

8 0
2 years ago
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