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3 years ago

In triangle abc b is a right angle and a is 20 degrees greater than

Mathematics

1 answer:

MatroZZZ [7]3 years ago

8 0

Answer:

Step-by-step explanation:

let ∠c=x

∠a=x+20

∠a+∠b+∠c=180

x+20+x+90=180

2x=180-110=70

x=35

∠c=35°

∠b=90°

∠a=35+20=55°

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Solve the system of equations Please help me! It would honestly mean so much to me! Thank you !

slega [8]

Answer:

(x, y) = (-3, 5)

Step-by-step explanation:

There are many ways to solve a system of two equations with two unknowns. Almost all of them involve reducing the system to one equation in one unknown. (Graphical solution, as in the attachment, bypasses that algebraic manipulation.)

In general, the first step is to look at the equations to see if ...

one is of the form x = ( ) or y = ( )
the coefficients of one of the variables are opposites
the coefficients of one of the variables are related by a simple number.

If the first condition is true, then the system may be easily solved by "substitution." The expression you have for one of the variables can be substituted for that variable in the other equation.

If the second condition is true, you can add the equations to eliminate the variable with opposite coefficients. (Opposites add to give zero.)

Here, the third condition holds: the coefficient of x in the first equation (4) is simply related to the coefficient of x in the second equation (8) by a factor of 2.

___

So, we can eliminate the x-variable from the system of equations by multiplying the first equation by -2 and adding that result to the second equation:

-2(4x +2y) +(8x +5y) = -2(-2) +(1)

-8x -4y +8x +5y = 4 +1 . . . . eliminate parentheses

y = 5 . . . . . . . . . . . . . . . . . . . collect terms

Now, we can substitute this value into either equation to find the value of x. Using the first equation, we get ...

4x +2(5) = -2

4x = -12 . . . . . . . subtract 10

x = -3 . . . . . . . . . divide by 4

The solution to the system of equations is (x, y) = (-3, 5).

5 0

3 years ago

On the real number line below, with coordinates as labeled, an object moves according to the following set of instructions: From

gavmur [86]

huh Step-by-step explanation:

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3 years ago

What is the sum of 129+3-12×3+15×6

Sladkaya [172]

The sum would be 186

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4 years ago

Find the value of x.

Pani-rosa [81]

Use the proportion 18/3x+6 = 10/3x-6 , and you should be left with your answer as 7.

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4 years ago

A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan

Artyom0805 [142]

Answer:

1) $\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) $y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let $y$ represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: $\frac{dy}{dt}=rate\:in-rate\:out$

The amount coming in is $0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}$

The rate going out depends on the concentration of salt in the tank at time t.

If there is $y(t)$ pounds of salt and there are $100+2t$ gallons at time t, then the concentration is: $\frac{y(t)}{2t+100}$

The rate of liquid leaving is is 3gal\min, so rate out is $=\frac{3y(t)}{2t+100}$

The required differential equation becomes:

$\frac{dy}{dt}=2.5-\frac{3y}{2t+100}$

2) We rewrite to obtain:

$\frac{dy}{dt}+\frac{3}{2t+100}y=2.5$

We multiply through by the integrating factor: $e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }$

to get:

$(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }$

This gives us:

$((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }$

We integrate both sides with respect to t to get:

$(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C$

Multiply through by: $(50+t)^{-\frac{3}{2}}$ to get:

$y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }$

$y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}$

We apply the initial condition: $y(0)=0$

$0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}$

$C=-12500\sqrt{2}$

The amount of salt in the tank at time t is:

$y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}$

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

$y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}$

$y(50)=98.23$

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

$\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})$

As $t\to \infty$ , $50+t\to \infty$ and $\frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0$

This implies that:

$\lim_{t \to \infty}y(t)=\infty- 0=\infty$

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

6 0

3 years ago