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3 years ago

Describe and correct the error in finding the measure of the angle

Mathematics

1 answer:

DerKrebs [107]3 years ago

5 0

Sorry, it's late, and I'm a bad explainer.

The error is adding (2x-12) with x and 30. This is wrong because you are adding the angles inside the triangle and you are assuming that (2x - 12) is the unlabeled angle INSIDE the triangle, when it is the exterior angle/outside of the triangle.

A straight line is also 180°.

(2x - 12) + ? = 180

30 + x + ? = 180

If you look at the equations, and put parentheses around 30 + x, (30 + x) and (2x - 12) should be the SAME NUMBER. So you could set them equal to each other to find x. (or you could also look at the picture and see that they both need/are missing the same angle)

2x - 12 = 30 + x

x = 42

Now you plug 42 into the exterior angle equation

2(42) - 12 = 84 - 12 = 72°

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Need help please !!!

blsea [12.9K]

Answer:

option C is the correct answer

Step-by-step explanation:

The trigonometric equation you are using has a general form

y = A* tan w*(x - r)

Where

A is the amplitude of the function

w is the frequency rad/s

r is the phase shift

In your case

A = -2

The frequency is

w = (2*pi)/period

period = pi/4

w = 8

r = -pi/2

y = -2* cos 8*(x + pi/2)

7 0

3 years ago

Which algebraic expression is equivalent to

poizon [28]

Hey there :)

( 2x⁵ - 3x⁴ + x² + 5x + 7 ) - ( - 4x⁵ - x⁴ - 3x² - 5x + 2 )

Let us combine like-terms
2x⁵ - ( - 4x⁵ ) - 3x⁴ - ( - x⁴ ) + x² - ( - 3x² ) + 5x - ( - 5x ) + 7 - 2

Minus and minus becomes plus
2x⁵ + 4x⁵ - 3x⁴ + x⁴ + x² + 3x² + 5x + 5x + 7 -2

6x⁵ - 2x⁴ + 4x² + 10x + 5

Your answer will be option A) 6x⁵ - 2x⁴ + 4x² + 10x + 5

8 0

3 years ago

Read 2 more answers

Please help me on this one?

blondinia [14]

3x + 4y = 16 Subtract 4y from both sides. Put brackets around the 2 terms on the right.

3x + 4y - 4y = (16 - 4y)

3x = (16 - 4y) Divide by 3

$\text{ x =}\dfrac{(16 - 4y)}{3} = \dfrac{16}{3} - \dfrac{4y}{3}$

Answer That's B.

3 0

3 years ago

Read 2 more answers

34− 6 × 3+ 78 pemdas

Mazyrski [523]

Answer:

Step-by-step explanation:

PEDMAS

$34 - 6 \times 3 + 78 \\ 34 - 18 + 78 \\ 34 + 60 \\ = 94$

Step 1 : Multiplication :

-6×3 =-18

Step 2: Addition: -18+78 = +60

Step 3: Addition : 35+60

= 94

8 0

3 years ago

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Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

aliya0001 [1]

Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}$

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, $C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}$

$R_{out}$ =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}$

Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{250}$

We solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$

Recall that when t=0, A(t)=50 (our initial condition)

$50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}$

4 0

3 years ago