The ratio of boys to girls in the sixth grade is 2:3. If there are 80 boys, how many are girls? Solve using unit rates.

Julia has read her favorite book 6 times,This is 3 times more than her Kami.How many times has Kami read the book?

motikmotik

Answer: The answer is: " 2 " .

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→ Kami has read the book " 2 (two) " times.

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Step-by-step explanation:

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Let "J" represent the number of times julia has read the book.

Let "K" represent the number of times Kami has read the book.

J = 6 ;

J = 3K ;

Solve for "K" ;

J = 6 = 3K ;

→ 6 = 3K ;

↔ 3K = 6 ;

Divide each side of the equation by "3" ;

to isolate "K" on one side of the equation; & to solve for "K" :

→ 3K / 3 = 6 / 3 ;

to get:

K = 2 .

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The answer is: " 2 " .

→ Kami has read the book " 2 (two) " times.

_______________________________________________

8 0

3 years ago

Compare the decimals 18.10 and 18.1

Arte-miy333 [17]

Both are equal since 18.1 we can add to the right zero
then 18.10=18.1

4 0

4 years ago

Read 2 more answers

Use mathematical induction to prove that if L is a linear transformation from V to W, then L (α1v1 + α2v2 +· · ·+αnvn)= α1L (v1)

e-lub [12.9K]

Answer:

The proof is shown in the explanation below.

Step-by-step explanation:

Analysis:

The proof by induction focuses on n. In this case, let n = 1, and $L^{1}$ will be a linear operator since $L^{1} = L$

The exercise will show that $L^{n}$ is a linear operator on V and that $L^{n+1}$ is also a linear operator on V. This, follows that:

$L^{n+1} (av) = L(L^{m}(v_{1}+v_{2})\\ = L(L^{m} (v_{1} + L^{m}v_{2})\\ = L(L^{m}v_{1} + L(L^{m}v_{2})\\ = L^{m+1}(v_{1}) + L^{m+1}(v_{2})$

6 0

3 years ago

Read 2 more answers

The table below represents the relationship of the amount of snowfall

RUDIKE [14]

Answer:

yes in 4 of the five countries the snowfall in inches is double the amount of time in hours

7 0

3 years ago

PLEASE HELP ME If 0 < z ≤ 90 and sin(9z − 1) = cos(6z + 1), what is the value of z? z = 3 z = 4 z = 5 z = 6

Burka [1]

Answer:

z = 6

Step-by-step explanation:

We know that ...

sin(x) = cos(90 -x)

Substituting (9z-1) for x, this is ...

sin(9z -1) = cos(90 -(9z -1))

But we also are given ...

sin(9z -1) = cos(6z +1)

Equating the arguments of the cosine function, we have ...

90 -(9z -1) = 6z +1

90 = 15z . . . . . . . . . add (9z-1) to both sides

6 = z . . . . . . . . . . . . divide by 15

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Comment on the graph

The attached graph shows 5 solutions in the domain of interest. These come from the fact that the relation we used is actually ...

sin(x) = cos(90 +360k -x) . . . . . for any integer k

Then the above equation becomes ...

90 +360k = 15z

6 +24k = z . . . . . . . . . for any integer k

The sine and cosine functions also enjoy the relation ...

sin(x) = cos(x -90)

sin(9z -1) = cos(9z -1 -90) = cos(6z +1)

3z = 92 . . . . . equating arguments of cos( ) and adding 91-6z

z = 30 2/3

6 0

3 years ago