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2 years ago

12

Do you know the answer to this

1 answer:

ludmilkaskok [199]2 years ago

8 0

Answer:

$4x^2y\sqrt[4]{3y}$

Step-by-step explanation:

$\sqrt[4]{768x^8y^5} =$

$= \sqrt[4]{256 \times 3(x^2)^4y^4y}$

$= \sqrt[4]{4^4 \times 3(x^2)^4y^4y}$

$= 4x^2y\sqrt[4]{3y}$

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Work out the following questions

Likurg_2 [28]

Answer:

a)8/21

b)1/9

Step-by-step explanation:

because when we divide fraction we resprocal the divisor

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2 years ago

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14% of the towns population is over the age 65. If there are 320 residents over the age of 65, approximately what is the towns p

Anestetic [448]

Answer:

about 315

Step-by-step explanation:

2,250×0.14

hope this helps!

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3 years ago

Solve the inequality: 12p + 7 > 139

lara31 [8.8K]

12p+7>139 Start
12p>132 Subtract 7
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4 years ago

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Caleb gets newsletters by e-mail. He gets one for sports every 5 days, one for model railroads every 10 days, and one for music

valentina_108 [34]

He would have to wait 40 more days

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