simplify the radical expression the square root of 63x to the 15th power y to the 9th power divided by 7xy^11

Which table represents a linear function?

Digiron [165]

It would be the first one because it’s going up by a constant rate

5 0

3 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

3 years ago

Two dice are tossed what is the probability of obtaining a sum greater than 6

katrin2010 [14]

7/12 because it would come out to be 21/36, and the simplified version of that is 7/12<span />

5 0

3 years ago

Find the equation of the ellipse with the following properties:

steposvetlana [31]

Answer: The equation of an ellipse:

(

x

−

h

)

2

a

2

+

(

y

−

k

)

2

b

2

=

1

;

a

>

b

Has vertices at

(

h

±

a

,

k

)

Has foci at

(

h

±

√

a

2

−

b

2

,

k

)

Use the vertices to write 3 equations:

k

=

4

[1]

h

−

a

=

−

6

[2]

h

+

a

=

10

[3]

Use equations [2] and [3] to solve for h and a:

2

h

=

4

h

=

2

a

=

8

Use the focus to write another equation:

8

=

h

+

√

a

2

−

b

2

Substitute values for h and a:

8

=

2

+

√

8

2

−

b

2

6

=

√

64

−

b

2

36

=

64

−

b

2

b

2

=

64

−

36

b

2

=

28

b

=

√

28

Substitute the values into the standard form:

(

x

−

2

)

2

8

2

+

(

y

−

4

)

2

(

√

28

)

2

=

1

5 0

3 years ago

Two cones have their heights in the ratio 1 : 3 and the radius of their bases in the ratio 3 : 1, show that their volumes are in

Nuetrik [128]

<h2> $\large\bold{\underline{\underline{Question:-}}}$ </h2>

Two cones have their heights in the ratio 1 : 3 and the radius of their bases in the ratio 3 : 1 show that their volumes are in the ratio 3 : 1.

<h2> $\large\bold{\underline{\underline{Solution:-}}}$ </h2>

⇒ Ratio of heights of two cones = 1 : 3

⇒ Ratio of radius of their bases = 3 : 1

⇒ We know that V = 1/3πr²h

⇒ Ratio of their volume = V1 : V2

$\underline{ \underline{ \sf{Let's \: find \: the \: ratio \: of \: their \: volumes:}}}$

${ \implies{ \sf {V _{1} : V_{2}}}}$

${ \implies{ \sf{ \frac{1}{3} \pi \: { r_{1}}^{2} h_{1} : \frac{1}{3} \pi \: { r_{2} }^{2} h_{2}}}}$

${ \implies{ \sf{ {r_{1}}^{2} h_{1} : { r_{2}}^{2} h_{2}}}}$

${ \implies{ \sf{ \frac{ { r_{1} }^{2} }{ { r_{2} }^{2} } : \frac{ h_{2} }{ h_{1} } }}} \\$

${ \implies{ \sf{ \frac{ {3}^{2} }{ {1}^{2} } : \frac{3}{1} }}} \\$

${ \implies{ \sf{ \frac{9}{1} : \frac{3}{1} }}} \\$

${ \implies{ \sf{3 : 1}}} \\$

${ \therefore{ \sf{ \green{Ratio \: of \: their \: volumes = 3 : 1}}}}$

4 0

3 years ago