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4 years ago

Solve 2a² and 3b²-c³ if A=3 ,B=2 ,C=1

1 answer:

6 0

Since we know a = 3, substitute this into the expression:

2a² = 2 x 3²

3² = 9

So 2a² = 2 x 9

2a² = 18

We know b = 2 and c = 1, so substitute these values into the expression:

3b² - c³ = 3 x 2² - 1³

2² = 4
1³ = 1

So:

3 x 4 - 1

Using the order of operations we multiply first:

3 x 4 = 12

Take off the 1:

Final answer:

3b² - c³ = 11

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What are the sum of the two pink angles?

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45 degrees

Step-by-step explanation:

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3 years ago

Maya has 5 over 6 cup of ice cream. How many 1 over 4 -cup servings are in 5 over 6 cup of ice cream?

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The answer is c because it’s not a or d and maybe not b so ye c

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3 years ago

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A recipe calls for 2/3 cup of flour for every 4 teaspoons of vanilla. How much vanilla should be used for every 1 cups of flour?

tiny-mole [99]

Answer: For 1 cup of flour, there should be 6 teaspoons of vanilla.

Step-by-step explanation:

What I did is that I found what 1 ÷ 2/3 equals. To find that, I used the reciprocal so the new equation would be 1 × 3/2 which equals 3/2. So since 2/3 × 3/2 equals 1, we have to do the same to 4 teaspoons. 4 × 3/2 equals 6. Therefore, there should be 6 teaspoons of vanilla for every one cup of flour.

Hope this helps! Don't hesitate to ask for clarification.

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3 years ago

Delilah finds her answer is 90, but she copied the expression incorrectly from the board. If her reasoning and answer for the ex

USPshnik [31]

Question:

Three students were working on 630,000÷700 in math class.

a. Desiree finds the quotient is 9,000 and Micah finds the quotient is 900.

Determine which student found the correct quotient.

b. Delilah finds her answer is 90, but she copied the expression incorrectly from the board. If her reasoning and answer for the expression she copied are correct, what expression might she have written?

Answer:

a. Micah is correct

$Expression = 63000 \div 700$

$Expression = 630000 \div 7000$

Step-by-step explanation:

Given

$Expression = 630000 \div 700$

$Desiree = 9000$

$Micah = 900$

Solving (a): Who is correct?

To do this, we first calculate the quotient of the expression.

$Expression = 630000 \div 700$

Divide 630000 by 700

$Quotient = 900$

Hence:

Micah is correct

Solving (b): What Delilah must have copied?

We have:

$Delilah = 90$

There are no options to pick from. So, we just need to work out expressions that equals 0.

Possible expressions are:

$Expression = 63000 \div 700$

$Expression = 630000 \div 7000$

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3 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,

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

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3 years ago

Read 2 more answers

Solve 2a² and 3b²-c³ if A=3 ,B=2 ,C=1​

Solve 2a² and 3b²-c³ if A=3 ,B=2 ,C=1