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3 years ago

Find the distance between the points (10,10) and (4,10).

1 answer:

3 0

The formula for distance is :

d = √(x₂-x₁)²+(y₂-y₁)²

Let the point (10;10) be (x₂; y₂) , and let the point (4;10) be (x₁;y₁).

So it would look like this -

d = √(10-4)² + (10-10)²

d = √(6)²+ (0)

d = √36

d = 6 units

Hope this helped~

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The ratio of boys to girls in Coach Sheppard's class is 3 to 5. There are 20 girls in the

kondaur [170]

Answer:

32 Students

Step-by-step explanation:

3/5 = 12/20

This means you add the boys and girls to get 32.

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3 years ago

Read 2 more answers

The general solution of 2 y ln(x)y' = (y^2 + 4)/x is

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Replace $y'$ with $\dfrac{\mathrm dy}{\mathrm dx}$ to see that this ODE is separable:

$2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\mathrm dx}{x\ln x}$

Integrate both sides; on the left, set $u=y^2+4$ so that $\mathrm du=2y\,\mathrm dy$ ; on the right, set $v=\ln x$ so that $\mathrm dv=\dfrac{\mathrm dx}x$ . Then

$\displaystyle\int\frac{2y}{y^2+4}\,\mathrm dy=\int\dfrac{\mathrm dx}{x\ln x}\iff\int\frac{\mathrm du}u=\int\dfrac{\mathrm dv}v$

$\implies\ln|u|=\ln|v|+C$

$\implies\ln(y^2+4)=\ln|\ln x|+C$

$\implies y^2+4=e^{\ln|\ln x|+C}$

$\implies y^2=C|\ln x|-4$

$\implies y=\pm\sqrt{C|\ln x|-4}$

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3 years ago

Find the length x to the nearest whole number. A right triangle has a vertical leg of length x units, a base leg with a length o

wel

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The value of x is $x = 274 \ unites$

Step-by-step explanation:

From the question we are told that

The length of the vertical length is $x$

The length of the base leg is L = 330 + d units

The length of the bisecting line segment is h

The base angle is $\theta = 28^o$

The angle between d and line segment is $\theta_1 = 56^o$

For the first angle

$Tan \theta_1 = \frac{x}{d}$

=> $d = \frac{x}{Tan \theta _1}$

For the whole big triangle

$Tan \theta = \frac{x}{330 + d}$

=> $d = \frac{x}{Tan \theta } -330$

So equating the both d

$\frac{x}{Tan \theta _1} = \frac{x}{Tan \theta } -330$

Substitute values

$\frac{x}{Tan (56)} = \frac{x}{Tan (28) } -330$

$0.6745 x = 1.880x -330$

$1.20549 x = 330$

$x = 274 \ unites$

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3 years ago

PLZZZZZZZZ HELP IM VERY STUCK ON THIS

Neporo4naja [7]

Answer:

B= 15 adults tickets

C= 35 student tickets

Step-by-step explanation:

x+y=50

solve for y

y=50-x

5x+3y=180

5x+3(50-x)=180

5x+150-3x=180

2x=30

x=15

y=50-15

y=35

6 0

3 years ago

A family has two cars. The first car has a fuel efficiency of 15 miles per gallon of gas and the second has a fuel efficiency of

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The fuel consumption of each of the cars during the week are as follows;

The first car consumes 20 gallons of fuel.

The second car consumes 30 gallons of fuel.

Reasons:

Fuel efficiency of the first car = 15 miles per gallon

Fuel efficiency of the second car = 35 miles per gallon

Combined distance traveled by the two cars in a week = 1,350 miles

The total gas consumption during the week = 50 gallons

Let x represent the number of gallons consumed by the first car, and let y

represent the number of gallons consumed by the second car, we get the

following system of simultaneous equations;

15·x + 35·y = 1,350...(1)

x + y = 50...(2)

Therefore;

y = 50 - x

Which gives;

15·x + 35 × (50 - x) = 1,350

15·x + 1,750 - 35·x = 1,350

1,750 - 1,350 = 35·x - 15·x = 20·x

400 = 20·x

x = 400 ÷ 20 = 20

The number of gallons consumed by the first car, x = 20 gallons

From equation (2), we have;

x + y = 50

y = 50 - x

Therefore;

y = 50 - 20 = 30

The number of gallons consumed by the second car, y = 30 gallons

Learn more about word problems that lead to simultaneous equations here:

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2 years ago