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2 years ago

Solve for yyy. \dfrac{6}{11} = \dfrac{y}{3} 11 6 = 3 y

Mathematics

1 answer:

Volgvan2 years ago

6 0

<em>Multiply both sides by 3.</em>

<em>Express 6/11 × 3 as a single fraction.</em>

<em>Multiply 6 and 3 to get 18.</em>

<em>Swap the sides so that all the terms of the variables are on the left side.</em>

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There are eight teams in a league. I’m how many ways can the teams finish first, second, and third?

jolli1 [7]

Answer:

There are 336 ways can the teams finish first, second, and third

Step-by-step explanation:

Total no. of teams = 8

First prize is given to one of the 8 teams

So, no. of teams left for second prize = 8-1 =7

Second prize is given to one of the remaining 7 teams

So, no. of teams left for third prize = 7-1=6

So,the teams finish first, second, and third in no. of ways = $8 \times 7 \times 6$

The teams finish first, second, and third in no. of ways = 336

Hence there are 336 ways can the teams finish first, second, and third

3 0

3 years ago

I don't understand can someone help please.

Roman55 [17]

Answer: The point where the two graphed lines cross is the solution to the system of equations. (1, -1)

Step-by-step explanation:

The second equation is already in slope-intercept form y = x - 2

the slope is +1 (invisible coefficient of x) and the y-intercept is -2

y = mx +b

"m" is the slope (the coefficient of x) Positive slopes go up from left to right

"b" is the y-intercept, where the graphed line crosses the x-axis

Rewrite the first equation in slope-intercept form.

6x + y = 5 subtract 6x from both sides

<em>-6x</em> + 6x +y = -<em>6x</em> + 5 .( left side 6x + 6x =0 so "cancel")

y = -6x + 5

Then you know the slope and the intercepts

b = 5 so start with a point at +5 in the y-axis

m = -6 so from there go down 6 and over to the right 1 square and plot another point. Draw a straight line through the two points.

The point where the two graphed lines cross is the solution to the system of equations.

Your graph should look like the screenshot below.

4 0

4 years ago

Someone please help asap! Thank you!

nekit [7.7K]

Answer:

The answer to your question is:

Perimeter = 26.5 units

Area = 22.5 units²

Step-by-step explanation:

Data

A (5, 9)

B (8, 0)

C (0, 9)

Formula

d = $\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2} }$

dAB = $\sqrt{(8 - 5)^{2} + (0 - 9)^{2} }$

dAB = $\sqrt{3^{2} + 9^{2} }$

dAB = $\sqrt{90}$

dBC = $\sqrt{(0 - 8)^{2} + (9 - 0)^{2} }$

dBC = $\sqrt{8^{2} + 9^{2} }$

dBC = $\sqrt{145}$

dAC = $\sqrt{(0 - 5)^{2} + (9 - 9)^{2} }$

dAC = $\sqrt{5^{2} + 0^{2} } \\$

dAC = $\sqrt{25} = 5$

Perimeter = $\sqrt{90} + \sqrt{145} + 5$

Perimeter = 26.5 units

Area = 1/2 [ ( 5x0) + (8x9) + (0 x 9) - (5x 9) - (0x 0) - (8x9)]

= 1/2 [ 0 + 72 + 0 - 45 - 0 - 72]

= 1/2 [-45]

= 45/2

= 22.5 units²

4 0

4 years ago

What is the answer for these two?

Vesna [10]

Answer: 1 is 2

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Estimate the size of a crowd walking in a charity fundraising march that occupies a rectangular space with dimensions of 10 feet

jolli1 [7]

Answer:

4000 is size of a crowd walking in a charity fundraising March.

Step-by-step explanation:

Given:

Dimensions of rectangular space 10 feet by 12000 feet

So we can say that,

Length = 1200 ft

Width = 10 ft

Now we will calculate the area of rectangular space which is given by

Hence Area will be = $length\times width= 10\ ft \times 1200 \ ft= 12000 \ ft^2$

Now we know that 40 people occupy a rectangle measuring 10 feet by 12 feet.

Length = 12 ft

Width = 10 ft

Area = $length\times width= 10\ ft \times 12 \ ft= 120 \ ft^2$

It says that 40 people are occupied in 120 $ft^2$

So how many people will be there in 12000 $ft^2$

By using unitary method we get,

Number of people = $\frac{40\times 12000 \ ft^2}{120 \ ft^2} = 4000 \ peoples$

Size of a crowd walking in a charity fundraising March is 4000.

8 0

4 years ago

Read 2 more answers

Solve for yyy. \dfrac{6}{11} = \dfrac{y}{3} 11 6 ​ = 3 y ​

Solve for yyy. \dfrac{6}{11} = \dfrac{y}{3} 11 6 = 3 y