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2 years ago

Messi kicks a soccer ball from a position that is 3 meters from a sideline and 4 meters from a goal

Mathematics

1 answer:

Katarina [22]2 years ago

6 0

Answer:

$( 4.8 ,7.6)$

Step-by-step explanation:

We want to find the point, B(x,y) that divides the directed line segment from Messi, M(3,4) to S(6,10) in the ratio m:n=3:2

We use the section formula:

$( \frac{mx_2+nx_1}{m + n} ,\frac{my_2+ny_1}{m + n})$

We substitute the coordinates and the ratio into the section formula to get:

$( \frac{3 \times 6+2 \times 3}{3 + 2} ,\frac{3 \times 10+2 \times 4}{3+ 2})$

We simplify to get:

$( \frac{18+6}{5} ,\frac{30+8}{5})$

This gives us:

$( \frac{24}{5} ,\frac{38}{5})$

$( 4.8 ,7.6)$

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Elena wants to make a scale drawing of her bedroom. Her bedroom is a rectangle with length 5 m and width 3 m. She decides on a s

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We are given dimensions of rectangular bedroom = length of 5m and width 3 m.

Also scale drawing of factor 1/50 of the original dimension.

In order to find the dimensions of scale drawing, we need to multiply scale factor by each dimension.

Therefore, length of the scale drawing is 5 * 1/50 = 1/10 m and

Let us convert it in centimeter now.

1 m = 100 cm

1/10 m = 100 * 1/10 = 10cm.

Width of the scale drawing = 3 * 1/50 = 3/50 m.

100*3/50 = 6 cm.

Therefore, the dimensions of a scale drawing of Elenas bedroom is length 10cm and width 6cm.

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Answer:

25%

Step-by-step explanation:

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2 years ago

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In a competition, two people will be selected from four finalists to receive the first and second prizes. The prize winners will

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Answer:

1. JG (Jim gets first prize, George gets second prize)

2. JH (Jim gets first prize, Helen gets second prize)

3. JM (Jim gets first prize, Maggie gets second prize)

4. GH (George gets first prize, Helen gets second prize)

5. GJ (George gets first prize, Jim gets second prize)

6. GM (George gets first prize, Maggie gets second prize)

7. MJ (Maggie gets first prize, Jim gets second prize)

8. MG (Maggie gets first prize, George gets second prize)

9. MH (Maggie gets first prize, Helen gets second prize)

Step-by-step explanation:

The question asks for the list of outcomes in the event "Not A". We are looking for the reverse or negative of Event A.

The above given list is the list of outcomes in the event where Helen DOES NOT get first prize.

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2 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

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2 years ago

Which data set is least Likely to resemble a normal distribution?<br> Look at picture

kicyunya [14]

Answer: B) The heights of girls who live on a certain street in the city of Buffalo

Every answer choice starts with "the heights of all 14-year-old girls who", so we can ignore that part. Choice A describes the largest population while choice B describes the smallest population. In other words, choice A is very general and broad, while choice B is very specific and narrow. The more specific you get and the smaller the population is, the less likely its going to be normally distributed.

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2 years ago