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3 years ago

4m3/3m2n x 9n2/30m2n2

Mathematics

1 answer:

Brut [27]3 years ago

5 0

4*m*3/3*m*2*n*x*9*n*2/30*m*2*n*2 evaluates to 96/5n^3m^3mx and that is the final answer.

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Numbers 4,5,6 !! Please help

aleksandrvk [35]

Answer:

4) VW= 23

5) VW= 12

6) VW= 75

Step-by-step explanation:

VU= 31

WU= 8

subtract 8 from 31

VW= 23

UW= 18

UV= 6

subtract 6 from 18

VW= 12

UV= 30

WU=45

add 30 and 45

VW= 75

7 0

3 years ago

The double number line shows how many bananas are needed to make 111 loaf of banana bread.

skelet666 [1.2K]

Answer:

Chart

Number of loaves Number of Bananas

1 2 1/2

2 5

4 10

Step-by-step explanation:

My friend helped me with this and since I saw no answer here I wanted to help YOU!!!!

So the first one was easy because it already said it in the number line.

The second one you simply multiply 2x 2 1/2

The third one you multiply 4x 2 1/2.

Hope it helps! Stay safe and dont go out during Quarantine!!!!!!!!!!

5 0

3 years ago

34x−5=4 x = <br> im confused

IRINA_888 [86]

Answer:

9/34

Step-by-step explanation:

Let us move all the numerical values to one side:

34x= 4+5

34x = 9

x=9/34

<em>I hope this helps! :)</em>

8 0

3 years ago

Lizzie rolls two dice. What is the probability that the sum of the dice is:

zhenek [66]

Answer:

$A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\$

Step-by-step explanation:

Total outcomes possible: 36

A. Divisible by 3

Possible options are:

3, 6, 9 and 12.

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

Possible outcomes for 12 are: {(6,6)} Count 1

Total count = 2 + 5 + 4 + 1 = 12

Probability of an event E can be formulated as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(A) = \dfrac{12}{36} = \dfrac{1}{3}$

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)} Count 1

Possible cases for sum 3: {(1,2), (2,1)} Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)} Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)} Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

$P(B) = \dfrac{15}{36} = \dfrac{5}{12}$

C. Divisible by 3 and less than 7:

$P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}$

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

$P(A \cap B) = \dfrac{7}{\text{36}}$

5 0

2 years ago

In the xy-plane , the graph of y=x^2 and the circle with center ( 0,1) and the radius 3 have how many points of intersection ?

Nikitich [7]

There are 2 points of intersection

5 0

3 years ago