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alukav5142 [94]
2 years ago
7

Which statements about the hyperbola are true? Check all that apply.

Mathematics
2 answers:
mariarad [96]2 years ago
8 0

Answer:

True options: 1, 2 and 5

Step-by-step explanation:

From the given diagram, you can see that the center of the hyperbola is placed at the origin, so first option is true (see attached diagram for definition of center, vertices, foci, i.e.)

There are two vertices of the hyperbola, they are placed at (-6,0) and (6,0), so second option is true.

The transverse axis is the segment connecting vertices, this segment is horizontal, so option 3 is false.

The foci are not placed within the rectangular reference box, so this option is false.

The directrices are vertical lines with equations x=\pm \dfrac{a}{e}, so this option is true.

dusya [7]2 years ago
6 0

Answer:1 2 5

Step-by-step explanation:

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FWML is a parallelogram. Find the values of x and y. Solve for the value of z, if z=x−y
Dima020 [189]

Answer:

x=5

y=8

z=-3

Step-by-step explanation:

We have been given a parallelogram. We are asked to solve for the values of x and y.

We know that opposite sides of parallelogram are equal, so we can set equation as:

3x-3=x+7

3x-x-3=x-x+7

2x-3=7

2x-3+3=7+3

2x=10

\frac{2x}{2}=\frac{10}{2}

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Similarly, we will solve for y.

2y-6=y+2

2y-y-6=y-y+2

y-6=2

y-6+6=2+6

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To solve for z, we will subtract y from x as:

z=x-y\\z=5-8\\z=-3

Therefore, the value of z is negative 3.

4 0
2 years ago
Sydney needs to earn $35 so she can buy her father a birthday present. her mom said she can make $6 per hour cleaning around the
Dmitry_Shevchenko [17]

Answer:

if i can be brainliest that would be great

f(x) = x^2+3x-10

f(x+5) = (x+5)^2+3(x+5)-10 ... replace every x with x+5

f(x+5) = (x^2+10x+25)+3(x+5)-10

f(x+5) = x^2+10x+25+3x+15-10

f(x+5) = x^2+13x+30

Compare this with x^2+kx+30 and we see that k = 13

Factor and solve the equation below

x^2+13x+30 = 0

(x+10)(x+3) = 0

x+10 = 0 or x+3 = 0

x = -10 or x = -3

The smallest zero is x = -10 as its the left-most value on a number line.

4 0
2 years ago
∆ABC has vertices at A(12, 8), B(4, 8), and C(4, 14). ∆XYZ has vertices at X(6, 6), Y(4, 12), and Z(10, 14). ∆MNO has vertices a
koban [17]
To get the length of each side,
use the distance formula with equation: 
Distance = ((x2-x1)^2+(y2-y1)^2)^0.5 
Solving
<span>AB = 8 units   BC = 6 units AC = 10 units
</span><span>MN =8units    NO = 6 units MO = 10 units
</span><span>XY = 6.32 units   YZ = 6.32 units XZ = 8.94 units
</span>JK = 4.47 units    KL = 4.47 units JL = 6 units 

1 The right answer is the letter b) triangle ABC and Triangle MNO are Congruent <span>triangles
ABC and MNO  are triangles that have the same lengths of its three sides.

</span>2 The answer is the letter c) rotation
There is a rotation of 90º about ABC and MNO the origin is B=N
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C----------O
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3 0
2 years ago
A circle with radius of \greenD{6\,\text{cm}}6cmstart color #1fab54, 6, start text, c, m, end text, end color #1fab54 sits insid
Diano4ka-milaya [45]

Answer: 141 cm^{2}

Step-by-step explanation:

We have two circles:

Cirlce 1, with a radius r=6 cm and area A_{1}:

A_{1}=\pi r^{2}

And Cicle 2, with a radius R=9 cm and area A_{2}:

A_{1}=\pi R^{2}

Since Circle 1 is inside Circle 2 and assuming the area of the shaded region is the shown in the attached image, its area is:

A=A_{2}-A_{1}=\pi R^{2}-\pi r^{2}

A=\pi (R^{2}- r^{2})

A=\pi ((9cm)^{2}- (6cm)^{2})

Finally:

A=141.37 cm^{2} \approx 141 cm^{2}

3 0
2 years ago
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Answer:

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Step-by-step explanation:

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5 0
2 years ago
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