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8090 [49]
3 years ago
5

Solve the equation 7 + 3m = 28

Mathematics
1 answer:
balandron [24]3 years ago
7 0
The answer is D. M=7 



===================
explanation 

7+3(7)=28

7+21=28

28=28

making the D.M=7 as true.
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Caroline bakes brownies to give to her friends. She starts with 48 brownies and gives a brownie to 6 friends each hour until she
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Answer: The  below figure shows the graph of the given situation.

Step-by-step explanation:

Since, at first Caroline has brownies= 48

And, according to the question, it is decreasing by 6 per hour.

Then, the function of the given situation can be written as,

f(x)=48 - 6 x,     ------ (1)

where,  x is the number of hours and f(x) is the number of brownies Caroline has left in x hour.

Thus by substitution x= 0,1,2,3,4,5,6,7,8 in equation (1),

We get, f(x)=48, 42, 36, 30, 24, 18, 12, 6, 0

So, the function has points, (0, 48), (1, 42), (2, 36), (3, 30), (4, 24), (5,18 ),(6,12 ) (7, 6) (8, 0).

With help of the above points we can draw the graph of the given situation.

Note: since at x=8, f(x)=0 therefore after 8 hours she has no brownies left.


8 0
3 years ago
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astra-53 [7]
The numbers are 59 and 61
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Perform the indicated operations. Write the answer in standard form, a+bi.<br> 5-3i / -2-9i
Vsevolod [243]

\huge \boxed{\mathfrak{Answer} \downarrow}

\large \bf\frac { 5 - 3 i } { - 2 - 9 i } \\

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\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)})  \\

Multiplication can be transformed into difference of squares using the rule: \sf\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}})  \\

By definition, i² is -1. Calculate the denominator.

\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85})  \\

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85})  \\

Do the multiplications in \sf5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9\left(-1\right).

\large \bf \: Re(\frac{-10+45i+6i+27}{85})  \\

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\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85})  \\

Do the additions in \sf-10+27+\left(45+6\right)i.

\large \bf Re(\frac{17+51i}{85})  \\

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\large \bf \: Re(\frac{1}{5}+\frac{3}{5}i)  \\

The real part of \sf \frac{1}{5}+\frac{3}{5}i \\ is \sf \frac{1}{5} \\.

\large  \boxed{\bf\frac{1}{5} = 0.2} \\

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