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2 years ago

What is a true statement comparing the graphs of x^2/6^2 = 1 and x^2/8^2-y^2/6^2=1

Mathematics

1 answer:

Jlenok [28]2 years ago

4 0

I am assuming that the statements are as follows:

The foci of both graphs are the same points.

The lengths of both transverse axes are the same.

The directrices of = 1 are horizontal while the directrices of = 1 are vertical.

The vertices of = 1 are on the y-axis while the vertices of = 1 are on the x-axis.

It should be A because of

Edge 21.

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Can someone please answer. There is a picture. There is only one question. Thank you so much

KatRina [158]

Hello,

Answer A

(x-x0)²+(y-y0)²=r²

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3 years ago

Please help with this simple trig problem- angle of two sectors of a circle

igomit [66]

Answer:

K = (1/2)r^2(sin(θ) +θ)

Step-by-step explanation:

The area of the triangle to the left is ...

A1 = (1/2)r^2·sin(180°-θ) = (1/2)r^2·sin(θ)

The area of the sector to the right is ...

A2 = (1/2)r^2θ

so the total area of the blue shaded region is ...

K = A1 + A2 = (1/2)r^2·sin(θ) + (1/2)r^2·θ

K = (1/2)r^2(sin(θ) +θ)

4 0

3 years ago

What is the value of i=

vaieri [72.5K]

$\sum\limits_{t=1}^3\left(4\cdot\left(\dfrac{1}{2}\right)^{t-1}\right)=4\cdot\left(\dfrac{1}{2}\right)^{1-1}+4\cdot\left(\dfrac{1}{2}\right)^{2-1}+4\cdot\left(\dfrac{1}{2}\right)^{3-1}\\\\=4\cdot\left(\dfrac{1}{2}\right)^0+4\cdot\left(\dfrac{1}{2}\right)^1+4\cdot\left(\dfrac{1}{2}\right)^2=4\cdot1+4\cdot\dfrac{1}{2}+4\cdot\dfrac{1}{4}\\\\=4+2+1=7\\\\Answer:\ \boxed{D.\ 7}$

5 0

3 years ago

Please help me find the answer pls

fiasKO [112]

Answer:

A. Solutions are: x = 2, y = 1.

B. Solutions are: x = 3, y = 2.

1. Inconsistent

2. Inconsistent

3. Consistent

Step-by-step explanation:

A. Solutions of each system of linear equations by substitution method:

Equation 1: 3x - 2y = 4

Equation 2: x = 2y

Step 1: Substitute x = 2y into the Equation 1:

3(2y) - 2y = 4

6y - 2y = 4

4y = 4

Step 2: Divide both sides of the equation by 4 to isolate y:

$\frac{4y}{4} = \frac{4}{4}$

y = 1.

Step 3: For Equation 2, x = 2y, substitute y = 1 into the equation to solve for x:

x = 2y

x = 2(1)

x = 2

Therefore, the solutions are: x = 2, y = 1.

B. Find the solutions of each system of linear equations by elimination method:

Equation 1: 2x + y = 8

Equation 2: x + y = 5

Step 1: Multiply Equation 2 by 2:

2(x + y) = 5(2)

2x + 2y = 10

Step 2: Subtract Equation 1 from the equation derived from Step 1, 2x + 2y = 10:

2x + 2y = 10

- 2x + y = 8

y = 2

Step 3: Plug in y = 2 into Equation 1, 2x + y = 8 to solve for x:

2x + y = 8

2x + (2) = 8

Step 4: subtract both sides of the equation by 2 to isolate x:

2x + 2 - 2 = 8 - 2

2x = 6

Step 5: Divide both sides of the equation by 2 to solve for x:

$\frac{2x}{2} = \frac{6}{2}$

x = 3.

The solutions are: x = 3, y = 2.

1. Inconsistent

2. Inconsistent

3. Consistent (infinitely many solutions)

3 0

3 years ago

Demonstrate how repeated subtraction can be used to show that 22 dividend by 7 = 3 r 1

Rudik [331]

22-7-7-7=1 this will show that you can subtract 7 three times from 22 and have a remainder of 1.

6 0

3 years ago