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3 years ago

Name the minor arc and find it's measure.

Mathematics

2 answers:

Vladimir79 [104]3 years ago

6 0

Based on the given figure, the minor arc is AB and its measure is 115°.

Minor arc is the arc that is less than half of the circle or less than 180°.
Major arc is the arc that is more than 180°. In this case, the major arc is ADB. Its measure is the difference of 360° and the minor arc, 245°

choli [55]3 years ago

5 0

Answer:

1.c. 72cm^2

2.c. m<1=36,m<2=72

3.b.7/2 and 49/4

4.c. 247.7 in^2

5.b. AB;115

6.a. 30 in

7.b. 10.89m^2

8.d. 11.7 m^2

9.(270pie+81/3)m^2

10.d. 20.73

Step-by-step explanation:

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on january 1,1999,the price of gasoline was$1.39 per gallon .if the price of gasoline increased by 0.5%per month ,what was the w

arlik [135]

$P(t)=1.39.(1.005)^t \ \ \ \ where t \ is \ time \ in \ months\\
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P(1)=1.39.(1.005)^1\\
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P(1)=1.39*1.005\\
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P(1)=\$ \ 1.40$

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3 years ago

If p percent = 30 percent ,CP =RS1000, SP =? profit amount =?

lubasha [3.4K]

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2 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 11z on the curve of intersection of the plane x − y + z =

Taya2010 [7]

Answer:

$\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}$

<em>Maximum value of f=2.41</em>

Step-by-step explanation:

<u>Lagrange Multipliers</u>

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

$\bigtriangledown f=\lambda \bigtriangledown g$

for some scalar $\lambda$ called the Lagrange multiplier.

For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is

$\bigtriangledown f=\lambda \bigtriangledown g+\mu \bigtriangledown h$

The gradient of f is

$\bigtriangledown f=$

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in $x,y,z,\lambda,\mu$ .

We have

$f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0$

Let's compute the partial derivatives

$f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0$

The Lagrange condition leads to

$1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)$

Operating and simplifying

$1=\lambda+2x\mu\\2=-\lambda +2y\mu \\\lambda=11$

Replacing the value of $\lambda$ in the two first equations, we get

$1=11+2x\mu\\2=-11 +2y\mu$

From the first equation

$\displaystyle 2\mu=\frac{-10}{x}$

Replacing into the second

$\displaystyle 13=y\frac{-10}{x}$

Or, equivalently

$13x=-10y$

Squaring

$169x^2=100y^2$

To solve, we use the restriction h

$x^2 + y^2 = 1$

Multiplying by 100

$100x^2 + 100y^2 = 100$

Replacing the above condition

$100x^2 + 169x^2 = 100$

Solving for x

$\displaystyle x=\pm \frac{10}{\sqrt{269}}$

We compute the values of y by solving

$13x=-10y$

$\displaystyle y=-\frac{13x}{10}$

For

$\displaystyle x= \frac{10}{\sqrt{269}}$

$\displaystyle y= -\frac{13}{\sqrt{269}}$

And for

$\displaystyle x= -\frac{10}{\sqrt{269}}$

$\displaystyle y= \frac{13}{\sqrt{269}}$

Finally, we get z using the other restriction

$x - y + z = 1$

Or:

$z = 1-x+y$

The first solution yields to

$\displaystyle z = 1-\frac{10}{\sqrt{269}}-\frac{13}{\sqrt{269}}$

$\displaystyle z = \frac{-23\sqrt{269}+269}{269}$

And the second solution gives us

$\displaystyle z = 1+\frac{10}{\sqrt{269}}+\frac{13}{\sqrt{269}}$

$\displaystyle z = \frac{23\sqrt{269}+269}{269}$

Complete first solution:

$\displaystyle x= \frac{10}{\sqrt{269}}\\\\\displaystyle y= -\frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{-23\sqrt{269}+269}{269}$

Replacing into f, we get

f(x,y,z)=-0.4

Complete second solution:

$\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}$

Replacing into f, we get

f(x,y,z)=2.4

The second solution maximizes f to 2.4

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3 years ago

Which equations represent the line that is parallel to 3x − 4y = 7 and passes through the point (−4, −2)? Check all that apply.

DIA [1.3K]

Reta a:

$3x-4y=7 \\ -4y=-3x+7 \\ 4y=3x-7 \\ y= \frac{3}{4}x- \frac{7}{4}$

reta x:

$y=ax+b$

Retas paralelas possuem coeficientes angulares iguais ⇒ mₐ=mₓ

$y= \frac{3}{4}x+b$

(-4,-2)

$\frac{3}{4}.(-4)+b=-2 \\ -3+b=-2 \\ b=-2+3 \\ b=1$

∴ y=3/4x+1

8 0

3 years ago

The pie chart shows the favourite lesson of a group of students.

anastassius [24]

Answer:

Step-by-step explanation:

hope that helped

3 0

2 years ago