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blondinia [14]
3 years ago
14

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

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer:

The analysis of your expresson is given in the images below.

Please see attached pictures.

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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
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