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3 years ago

11

Sam has 3/4 of a can of hot chocolate mix. To make a cup of hot chocolate, Sam adds hot water to one scoop of mix. The scoop hol

ds 1/24 of a can. How many cups of hot chocolate can Sam make?

2 answers:

Mashutka [201]3 years ago

3 0

Sam could make 18 cups

Scilla [17]3 years ago

3 0

18 cups would be the answer

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The diameter of circle is 36 miles. What is the circle's area<br><br>Use 3.14 for pi<br><br>

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3.14 on the length. but if you add to a graph send another one to my comment and show me

explanation

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2 years ago

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Number 1 and 2 please (25 points)

Elenna [48]

Answers:

1) $(3x+2)+(4-5x)=-2x+6$

$(3+2i)+(4-5i)=7-3i$

2) $(1-3x)(2+x)=-3x^{2}-5x+2$

$(1-3i)+(2+i)=5-5i$

Step-by-step explanation:

In mathematics there are rules related to complex numbers, specifically in the case of addition and multiplication:

<u>Addition: </u>

If we have two complex numbers written in their binomial form, the sum of both will be a complex number whose real part is the sum of the real parts and whose imaginary part is the sum of the imaginary parts (similarly as the sum of two binomials).

For example, the addition of these two binomials is:

$(3x+2)+(4-5x)=3x-5x+2+4=-2x+6$

Similarly, the addition of two complex numbers is:

$(3+2i)+(4-5i)=3+4+2i-5i=7-3i$ Here the complex part is the number with the $i$

<u>Multiplication: </u>

If we have two complex numbers written in their binomial form, the multiplication of both will be the same as the multiplication (product) of two binomials, taking into account that $i^{2}=-1$ .

For example, the multiplication of these two binomials is:

$(1-3x)(2+x)=2+x-6x-3x^{2}=-3x^{2}-5x+2$

Similarly, the multiplication of two complex numbers is:

$(1-3i)+(2+i)=2+i-6i-3i^{2}=2-5i-3(-1)=5-5i$

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3 years ago

PLEASE HELP ME FIND THE ANSWER

n200080 [17]

B is true.
I think E is also true.

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3 years ago

What is the simplified form of the quantity y−squared minus y minus 12 over the quantity y−squared minus 2y minus 15?

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(y^2-y-12)/(y^2-2y-15) factor both the numerator and denominator...

(y^2-4y+3y-12)/(y^2-5y+3y-15)

(y(y-4)+3(y-4))/(y(y-5)+3(y-5))

((y-4)(y+3))/((y-5)(y+3)) so the (y+3) terms cancel leaving

(y-4)/(y-5)

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2 years ago

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natta225 [31]

Answer:

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Step-by-step explanation:

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3 years ago

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