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3 years ago

Plzzzz helpppppppp...........

Mathematics

1 answer:

Flura [38]3 years ago

8 0

Answer:

Why isn't there a picture-

Step-by-step explanation:

HEHEH BOIS

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Solve by completing the square x^2-6x=1

Dmitry [639]

(x-3)^2=10
I hope this helps you.

4 0

3 years ago

A golfer hits an errant tee shot that lands in the rough. A marker in the center of the fairway is 150 yards from the center of

Gwar [14]

$\bold{\huge{\underline{ Solution}}}$

<h3>Given :-</h3>

A marker in the center of the fairway is 150 yards away from the centre of the green
While standing on the marker and facing the green, the golfer turns 100° towards his ball
Then he peces off 30 yards to his ball

We have to find the distance between the golf ball and the center of the green .

<h3>Let's Begin :-</h3>

Let assume that the distance between the golf ball and central of green is x

Here, 

Distance between marker and centre of green is 150 yards
That is, Height = 150 yards
For facing the green , The golfer turns 100° towards his ball
That is, Angle = 100°
The golfer peces off 30 yards to his ball
That is, Base = 30 yards

According to the law of cosine :-

$\bold{\red{ a^{2} = b^{2} + c^{2} - 2ABcos}}{\bold{\red{\theta}}}$

Here, a = perpendicular height
b = base
c = hypotenuse
cos theta = Angle of cosine

So, For Hypotenuse law of cosine will be :-

$\sf{ c^{2} = a^{2} + b^{2} - 2ABcos}{\sf{\theta}}$

Subsitute the required values,

$\sf{ x^{2} = (150)^{2} + (30)^{2} - 2(150)(30)cos}{\sf{100°}}$

$\sf{ x^{2} = 22500 + 900 - 900cos}{\sf{\times{\dfrac{5π}{9}}}}$

$\sf{ x^{2} = 22500 + 900 - 900( - 0.174)}$

$\sf{ x^{2} = 22500 + 900 + 156.6}$

$\sf{ x^{2} = 23556.6}$

$\bold{ x = 153.48\: yards }$

Hence, The distance between the ball and the center of green is 153.48 or 153.5 yards

5 0

3 years ago

Using the Law of Cosines, in triangle DEF, if e=18yd, d=10yd, f=22yd, find measurement of angle D

Ivahew [28]

Check the picture below.

$\bf \textit{Law of Cosines}\\ \quad \\
c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies 
c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}
\\\\\\
\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}=cos(C)\implies cos^{-1}\left(\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}\right)=\measuredangle C\\\\
-------------------------------\\\\
\begin{cases}
d=10\\
e=18\\
f=22
\end{cases}\implies cos^{-1}\left(\cfrac{{{ 18}}^2+{{ 22}}^2-10^2}{2(18)(22)}\right)=\measuredangle D$

$\bf \implies cos^{-1}\left(0.893\overline{93}\right)=\measuredangle D\implies 26.63^o\approx \measuredangle D$

7 0

3 years ago

Equations

sergij07 [2.7K]

Answer:

2x+12=24

first you subtract 12 from both sides

2x+12=24

-12. -12

The 12-12 should cancel itself, the rest of the equation you bring down to get

2x=12 (because 24-12=12)

Now you have 2x=12.
you then divide 2x by both sides.

2x=12

/2x=/2x

The 2x/2x cancels itself out so you then solve for 12/2x.

For this you just divide 12/2 which is 6!

x= 6 is your final answer.
to check this equation you can plug your number back into x to see if it is true! 2(6)+12=24.
6 times 2 is 12 and 12+12 is 24 so your answer (6) is true!

hope this helps! :D

4 0

3 years ago

Read 2 more answers

Find a solution of x dy dx = y2 − y that passes through the indicated points. (a) (0, 1) y = (b) (0, 0) y = (c) 1 6 , 1 6 y = (d

Leni [432]

Answers:

(a) $y = \frac{1}{1 - Cx}$ , for any constant C

(b) Solution does not exist

(c) $y = \frac{256}{256 - 15x}$

(d) $y = \frac{64}{64 - 15x}$

Explanations:

(a) To solve the differential equation in the problem, we need to manipulate the equation such that the expression that involves y is on the left side of the equation and the expression that involves x is on the right side equation.

Note that

$x\frac{dy}{dx} = y^2 - y
\\
\\ \indent xdy = \left ( y^2 - y \right )dx
\\
\\ \indent \frac{dy}{y^2 - y} = \frac{dx}{x}
\\
\\ \indent \int {\frac{dy}{y^2 - y}} = \int {\frac{dx}{x}} 
\\
\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln x + C_1}$ (1)

Now, we need to evaluate the indefinite integral on the left side of equation (1). Note that the denominator y² - y = y(y - 1). So, the denominator can be written as product of two polynomials. In this case, we can solve the indefinite integral using partial fractions.

Using partial fractions:

$\frac{1}{y^2 - y} = \frac{1}{y(y - 1)} = \frac{A}{y - 1} + \frac{B}{y}
\\
\\ \indent \Rightarrow \frac{1}{y^2 - y} = \frac{Ay + B(y-1)}{y(y - 1)} 
\\
\\ \indent \Rightarrow \boxed{\frac{1}{y^2 - y} = \frac{(A+B)y - B}{y^2 - y} }$ (2)

Since equation (2) has the same denominator, the numerator has to be equal. So,

$1 = (A+B)y - B
\\
\\ \indent \Rightarrow (A+B)y - B = 0y + 1
\\
\\ \indent \Rightarrow \begin{cases}
 A + B = 0
& \text{(3)}\\-B = 1
 & \text{(4)} \end{cases}$

Based on equation (4), B = -1. By replacing this value to equation (3), we have

A + B = 0
A + (-1) = 0
A + (-1) + 1 = 0 + 1
A = 1

Hence,

$\frac{1}{y^2 - y} = \frac{1}{y - 1} - \frac{1}{y}$

So,

$\int {\frac{dy}{y^2 - y}} = \int {\frac{dy}{y - 1}} - \int {\frac{dy}{y}} 
\\
\\ \indent \indent \indent \indent = \ln (y-1) - \ln y
\\
\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln \left ( \frac{y-1}{y} \right ) + C_2}$

Now, equation (1) becomes

$\ln \left ( \frac{y-1}{y} \right ) + C_2 = \ln x + C_1
\\
\\ \indent \ln \left ( \frac{y-1}{y} \right ) = \ln x + C_1 - C_2
\\
\\ \indent \frac{y-1}{y} = e^{C_1 - C_2}x
\\
\\ \indent \frac{y-1}{y} = Cx, \text{ where } C = e^{C_1 - C_2}
\\
\\ \indent 1 - \frac{1}{y} = Cx
\\
\\ \indent \frac{1}{y} = 1 - Cx
\\
\\ \indent \boxed{y = \frac{1}{1 - Cx}}
$ (5)

At point (0, 1), x = 0, y = 1. Replacing these values in (5), we have

$y = \frac{1}{1 - Cx}
\\
\\ \indent 1 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1

$

Hence, for any constant C, the following solution will pass thru (0, 1):

$\boxed{y = \frac{1}{1 - Cx}}$

(b) Using equation (5) in problem (a),

$y = \frac{1}{1 - Cx}$ (6)

for any constant C.

Note that equation (6) is called the general solution. So, we just replace values of x and y in the equation and solve for constant C.

At point (0,0), x = 0, y =0. Then, we replace these values in equation (6) so that

$y = \frac{1}{1 - Cx}
\\
\\ \indent 0 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1$

Note that 0 = 1 is false. Hence, for any constant C, the solution that passes thru (0,0) does not exist.

(c) We use equation (6) in problem (b) and because equation (6) is the general solution, we just need to plug in the value of x and y to the equation and solve for constant C.

At point (16, 16), x = 16, y = 16 and by replacing these values to the general solution, we have

$y = \frac{1}{1 - Cx}
\\
\\ \indent 16 = \frac{1}{1 - C(16)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 16C}
\\
\\ \indent 16(1 - 16C) = 1
\\ \indent 16 - 256C = 1
\\ \indent - 256C = -15
\\ \indent \boxed{C = \frac{15}{256}}


$

By replacing this value of C, the general solution becomes

$y = \frac{1}{1 - Cx}
\\
\\ \indent y = \frac{1}{1 - \frac{15}{256}x} 
\\ 
\\ \indent y = \frac{1}{\frac{256 - 15x}{256}}
\\
\\
\\ \indent \boxed{y = \frac{256}{256 - 15x}}



$

This solution passes thru (16,16).

(d) We do the following steps that we did in problem (c):
- Substitute the values of x and y to the general solution.
- Solve for constant C

At point (4, 16), x = 4, y = 16. First, we replace x and y using these values so that

$y = \frac{1}{1 - Cx} 
\\ 
\\ \indent 16 = \frac{1}{1 - C(4)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 4C} 
\\ 
\\ \indent 16(1 - 4C) = 1 
\\ \indent 16 - 64C = 1 
\\ \indent - 64C = -15 
\\ \indent \boxed{C = \frac{15}{64}}$

Now, we replace C using the derived value in the general solution. Then,

$y = \frac{1}{1 - Cx} \\ \\ \indent y = \frac{1}{1 - \frac{15}{64}x} \\ \\ \indent y = \frac{1}{\frac{64 - 15x}{64}} \\ \\ \\ \indent \boxed{y = \frac{64}{64 - 15x}}$

5 0

3 years ago