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Svetllana [295]
3 years ago
5

The dishes have been cleaned by us into active voice​

Mathematics
2 answers:
KIM [24]3 years ago
6 0

Answer:

the dishes sparkled after we washed them.

Step-by-step explanation:

klasskru [66]3 years ago
4 0
The dishes are clean once you wash them duh
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PLEASE HELP I WILL GIVE BRAINILEST
stira [4]

for A you should take each plotted point and change the sign of the x values. for example (-1,3) should become (1,3) and (-4,4) should become (4,4)

for B take each point and flip the variables and x's sign. (-1,3) should become (3,1)

for C take every x value and subtract 2. (-1,3) should become (-3,3)

for D take each point and flip the variables and y's sign. (-1,3) should become (-3,-1)

5 0
3 years ago
Suppose a and b are both non zero real numbers. Find real numbers c and d such that 1/a+ib= c+id
Thepotemich [5.8K]

\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}

Explanation

\frac{1}{a+bi}=c+di

Step 1

multiplicate by the conjugate

\begin{gathered} \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(bi)^2} \\ \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(-b^2)}=\frac{a-bi}{a^2+b^2} \end{gathered}

notice that

\begin{gathered} \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i \\ \frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di \\ so \\  \end{gathered}\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}

I hope this helsp you

6 0
1 year ago
14x to the power of -2 , for x=7
OverLord2011 [107]

Answer:

:5:5&&'4&4&4&:&;4"33*2**223$&4&4&55&

6 0
2 years ago
Read 2 more answers
I NEED THE ANSWER RIGHT NOW PLEASE!!!
strojnjashka [21]

Answer:

Second : 1/6 x + 47

Third : 1 2/3 x + 35 - 1 1/2 x + 12

Fourth : 5 (1/3x) + (5) (7) - (3) (1/2x) + (3) (4)

Step-by-step explanation:

I would start by expanding the expression, showing each step. Then, simplify by collecting like terms.

To expand, you multiply the term outside of the brackets by each term inside the brackets.

5(1/3x + 7) - 3(1/2x - 4)

= 5(1/3x) + (5)(7) - (3)(1/2x) + (3)(4)       Showing how to expand

= 5/3x + 35 - 3/2x + 12           Simplified above step by multiplying

= 5/3x - 3/2x + 47            Collected like terms (35 + 12 = 47)

= 10/6x - 9/6x + 47          Change fractions to the common denominator 6

= \frac{10-9}{6}x + 47   Combine fractions with common denominator

= 1/6x + 47              Fully expanded

See which options match the steps I took to expand the original expression.

The second step is the same as option 4.

The last step is the same as option 2.

I bolded those steps that are the same.

Some options use mixed fractions (whole numbers and fractions).

From the third step, I will convert the improper fractions to mixed fractions.

5/3x + 35 - 3/2x + 12

= 1 2/3x + 35 - 1 1/2x + 12

This is the same as option 3.

\frac{5}{3}x = 1\frac{2}{3}x    On the left side, 5 goes into 3 ONE time. The ONE becomes the whole number. After 5 goes into 3, there is 2 left, which becomes the numerator.

\frac{3}{2}x = 1\frac{1}{2}x   On the left side, 3 goes into 2 ONE time. The ONE becomes the whole number. The remainder is 1, which becomes the numerator.

6 0
3 years ago
Read 2 more answers
What is the justification for the first step in proving the formula for factoring the sum of cubes?
Anna11 [10]

Explanation:

The formula isnt correctly written, it should state:

a^3+b^3 = (a+b)(a^2-ab+b^2)

You have to start from (a+b)(a^2-ab+b^2)  and end in a³+b³. On your first step, you need to use the distributive property.

(a+b)(a^2-ab+b^2) = a*(a^2-ab+b^2) + b*(a^2-ab+b^2)

This is equal to

a*a^2-a*(ab) + a*b^2 + b*a^2-b*(ab) + b*b^2 = a^3 - a^2b + ab^2 +ba^2 -b^2a +b^3

Note that the second term, -a²b, is cancelled by the fourth term, ba², and the third term, ab², is cancelled by the fifht term, -b²a. Therefore, the final result is a³+b³, as we wanted to.

5 0
3 years ago
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