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Gnoma [55]
3 years ago
13

Bryan bought 10 daisies and 5 sunflowers. Each daisy cost $2 and each sunflower cost $5. Bryan had $12 left after he made his pu

rchase.
How much money did Bryan spend on the flowers? Which info isnt needed to answer this question?
Mathematics
1 answer:
sasho [114]3 years ago
3 0

Answer:

Bryan spent $45 on flowers. The last sentence isn't needed to answer this question.

Step-by-step explanation:

If you multiply 10 by 2, then 5 by 5, and add it at the end you get 45.

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Find the GCE.<br> 10rs, 35rst
lukranit [14]
The GCE of 10 and 35 is 5
7 0
3 years ago
-4x - 6x = -20 =?<br><br> -3(2x - 3) = 33 = ?<br><br> 4x + 3x + 2x = 180 = ?
postnew [5]

Answer:

-4x-6x=-20=

x=2

-3(2x-3)=33=

x=-4

4x+3x+2x=180

x=20

Step-by-step explanation:

-4x-6x=-20                                

-4+(-6)=-10x

-10x=20 (divide both by 10)

x=-2 (divide by -1 to get positive)

x=2

-3(2x-3)=33

-6x+9=33

     -9 .  -9

-6x=24 (divide both by -6)

x=-4 (divide by -1 to make positive)

4x+3x+2x=180

(combine like terms)

9x=180 (divide both sides by 9)

x=20

5 0
2 years ago
Please help with imaginary numbers worksheet!
mihalych1998 [28]

Problem 5

<h3>Answer:   6i</h3>

------------------

Explanation:

We have these four identities

  • i^0 = 1
  • i^1 = i
  • i^2 = -1
  • i^3 = -i

Notice how computing i^4 leads us back to 1. So i^4 = i^0. The pattern repeats every 4 terms. So we divide the exponent by 4 and look at the remainder. We ignore the quotient entirely. We can see that 28/4 = 7 remainder 0. Meaning that i^28 = i^0 = 1.

We can think of it like this if you wanted

i^28 = (i^4)^7 = 1^7 = 1

Then the sqrt(-36) becomes 6i

So overall, we end up with the final answer of 6i

=============================================

Problem 6

<h3>Answer:  -3i</h3>

------------------

Explanation:

We'll use the ideas mentioned in problem 5

46/4 = 11 remainder 2

i^46 = i^2 = -1

sqrt(-9) = 3i

The two outside negative signs cancel out, but there's still a negative from -1 we found earlier. So we end up with -3i

In other words, here is one way you could write out the steps

-i^{46}*-\sqrt{-9}\\\\-i^{2}*-3i\\\\i^{2}*3i\\\\-1*3i\\\\-3i\\\\

=============================================

Problem 7

<h3>Answer:   -1</h3>

------------------

Work Shown:

i^10 = i^2 because 10/4 = 2 remainder 2

i^19 = i^3 because 19/4 = 4 remainder 3

i^7 = i^3 because 7/4 = 1 remainder 3

Again, all we care about are the remainders.

i^{10}+i^{19} - i^{7}\\\\i^{2}+i^{3} - i^{3}\\\\i^{2}\\\\-1

=============================================

Problem 8

<h3>Answer:    -1 + i</h3>

------------------

Work Shown:

i^22*i^6 = i^(22+6) = i^28

Earlier in problem 5, we found that i^28 = i^0 = 1

So,

i^1 - \left(i^{22}*i^{6}\right)\\\\i^1 - i^{28}\\\\i^1 - i^{0}\\\\i - 1\\\\-1+i

8 0
2 years ago
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$68 net price

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$80 x 15% (discount) = $12 discount
$80- $12 = $68 net price
4 0
2 years ago
Solve the inequality (29/289)z&lt;(16/161)z
stellarik [79]

is that what you looking for

4 0
2 years ago
Read 2 more answers
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