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Juli2301 [7.4K]
3 years ago
7

Find the vertices and locate the foci for the hyperbola whose equation is given. 49x2 - 100y2 = 4900

Mathematics
1 answer:
Fiesta28 [93]3 years ago
5 0
<h2>Answer:</h2>

A hyperbola is the set of all points in a plane such that  the difference of whose  distances from two distinct fixed points called foci is a positive constant. In this problem, we have the following equation:

49x^2-100y^2=4900

What if we divide the whole equation by 49 \times 100=4900? Well the result is:

\frac{1}{4900}(49x^2-100y^2=4900) \\ \\ \frac{49x^2}{4900}-\frac{100y^2}{4900}=\frac{4900}{4900} \\ \\ \frac{x^2}{100}-\frac{y^2}{49}=1 \\ \\ \boxed{\frac{x^2}{10^2}-\frac{y^2}{7^2}=1}

The standard form of the equation of the hyperbola given that the vertex lies on the origin is:

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

SO THE VERTICES ARE:

\boxed{(a,0)=(10,0) \ and \ (-a,0)=(-10,0)}

Calculating the foci:

We \ know \ that \ foci \ are: \\ \\ (-c,0) \ and \ (c.0) \\ \\ Also: \\ \\ c=\sqrt{a^2+b^2} \\ \\ c=\sqrt{100+49} \therefore c=\sqrt{149}

SO THE FOCI ARE:

\boxed{(-\sqrt{149},0) \ and \ (\sqrt{149},0)}

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IRINA_888 [86]

Answer:

The answer is

x equal -243

4 0
3 years ago
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Habiendose organizado un bingo se ha recaudado $1900. Por la entrada de hombres $15 y mujeres $10 y se ha reportado una asistenc
aev [14]

Answer:

The number of men was 80

The number of women was 70

Step-by-step explanation:

The question in English is

Having organized a bingo has raised $1900. For the entrance of men $15 and women $10 and an attendance of 150 people has been reported. Determine the number of men and the number of women who participated in the bingo.

Let

x----> the number of men

y----> the number of women

we know that

x+y=150

x=150-y -----> equation A

15x+10y=1,900 -----> equation B

substitute equation A in equation B and solve for y

15(150-y)+10y=1,900

2,250-15y+10y=1,900

15y-10y=2,250-1,900

5y=350

y=70

Find the value of x

x=150-70=80

therefore

The number of men was 80

The number of women was 70

3 0
3 years ago
Suppose that S is the set of successful students in a classroom, and that F stands for the set of freshmen students in that clas
lora16 [44]

Answer:

b) 24

Step-by-step explanation:

We solve building the Venn's diagram of these sets.

We have that n(S) is the number of succesful students in a classroom.

n(F) is the number of freshmen student in that classroom.

We have that:

n(S) = n(s) + n(S \cap F)

In which n(s) are those who are succeful but not freshmen and n(S \cap F) are those who are succesful and freshmen.

By the same logic, we also have that:

n(F) = n(f) + n(S \cap F)

The union is:

n(S \cup F) = n(s) + n(f) + n(S \cap F)

In which

n(S \cup F) = 58

n(s) = n(S) - n(S \cap F) = 54 - n(S \cap F)

n(f) = n(F) - n(S \cap F) = 28 - n(S \cap F)

So

n(S \cup F) = n(s) + n(f) + n(S \cap F)

58 = 54 - n(S \cap F) + 28 - n(S \cap F) + n(S \cap F)

n(S \cap F) = 24

So the correct answer is:

b) 24

3 0
3 years ago
9x 2 - 18x - 7 ÷ (3x + 1)
Harman [31]

Answer:

The quotient is: 3x-7

The remainder is: 0

Step-by-step explanation:

We need to divide 9x^2 - 18x - 7 ÷ (3x + 1)

The Division is shown in the figure attached.

The quotient is: 3x-7

The remainder is: 0

8 0
3 years ago
Evaluate:
Anni [7]

Answer:

a) = 4.5

b) = 3.3

Step-by-step explanation:

Before solving our problems given to us let us under stand the rule of cube roots

It says \sqrt[3]{x \times x \times x} = x   -----(A)

Also

\sqrt[3]{x \times x \times x \times y \times y \times y} = x \times y   ---(B)

Now let us see each part one by one

a) we have

\sqrt[3]{64} + \sqrt[3]{0.027} + \sqrt[3]{0.008}

Now 64 = 4 x 4 x 4

0.027 = 0.3 x 0.3 x 0.3

0.008 = 0.2 x 0.2 x 0.2

substituting these values

\sqrt[3]{4 \times 4 \times 4} + \sqrt[3]{0.3 \times 0.3 \times 0.3} + \sqrt[3]{0.2 \times 0.2 \times 0.2}

Applying Rule A in above

=4+0.3+0.2

=4+0.5

4.5

b) we have \sqrt[3]{0.3 \times 0.3 \times 0.3 \times 11 \times 11 \times 11}

Applying the B rule in this

=0.3 \times 11

3.3

4 0
3 years ago
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