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2 years ago

13

Tlaloc pumped up 56 balls. He pumped up 5 dodge balls for every 9 other balls.

1 answer:

borishaifa [10]2 years ago

6 0

Answer:

What do you need from this question? Its not asking anything.

Step-by-step explanation:

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Daniel is drawing a plan for a new animal pen for his farm. The rectangle plotted in the coordinate plane represents the pen, me

Nat2105 [25]

Answer:

22ft

Step-by-step explanation:

I just did the edge.

3 0

2 years ago

Which statement comparing these measures is true? A microgram is 1,000 times greater than a milligram.

ololo11 [35]

The statement is false.

The statement below is true.
A milligram is 1000 times greater than a microgram.

3 0

3 years ago

In a class 20 of students, 7 have a cat and

blsea [12.9K]

Answer:

$\frac{3}{20}$

Step-by-step explanation:

Since there are 20 students, this will be the denominator.

Now, to find how many students have a dog and a cat, you need to subtract all the numbers with 20. The leftover number will be the amount of students.

7+8+8=23

23-20=3

So, there are 3 students who have a dog and a cat.

So, the fraction will be $\frac{3}{20}$

8 0

2 years ago

Eliza volunteers at a nearby aquarium

pishuonlain [190]

It is 3٫100 miles right

8 0

2 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

2 years ago