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grigory [225]
3 years ago
13

Idk what to do can someone help?

Mathematics
1 answer:
dolphi86 [110]3 years ago
4 0
She will rent 12 cages
You might be interested in
Find the common difference for the sequence below. 1/4, 5/16, 3/8,..
vredina [299]

Answer:

the common difference of the sequence is (\frac{1}{16})

Step-by-step explanation:

A sequence of numbers is said to be in arithmetic progression of each term of the sequence has same common difference.

Now, here first term a1 = 1/4

Second term a2 = 5/16

Third term a3 = 3/8

Now, Common difference (d) = a2 - a1 = a3 - a2

Now, a2 - a1 = \frac{5}{16}  - \frac{1}{4}  = \frac{5 - 4}{16}  = \frac{1}{16}

Now, a3 - a2 =  \frac{3}{8}  - \frac{5}{16}  = \frac{6 - 5}{16}  = \frac{1}{16}

Hence, the common difference of the sequence is (\frac{1}{16})

5 0
3 years ago
Which statement correctly finds the interquartile range for the set of data represented by the box plot?
gizmo_the_mogwai [7]

Answer:

<em>B. 11 - 6 = 5</em>

Step-by-step explanation:

Median: 9.5

Lower Quartile: 6

Upper Quartile: 11

Interquartile Range: upper quartile - lower quartile = answer

6 0
3 years ago
What is the largest prime factor of 38
miskamm [114]

Answer:

The greatest prime factor of 38 is 19

5 0
3 years ago
Read 2 more answers
4. Use the two patterns below. What are the first four ordered pairs formed from
kobusy [5.1K]

Answer:

0,2,4,8

Step-by-step explanation:

:)

7 0
3 years ago
Roberto rowed 20 miles downstream in 2.5 hours. The trip back, however, took him 5 hours. Find the rate that Roberto rows in sti
almond37 [142]

Answer:

<em>Roberto's speed in still water is 6 miles/hour and the river speed is 2 miles/hour</em>

Step-by-step explanation:

<u>Relative Speed</u>

When a body is moving at a constant speed v, the distance traveled in a time t is:

d=v.t

When Roberto rows downstream, his speed in still water is added to the speed of the water, making it easier to travel the required distance.

When Roberto rows upstream, his speed in still water is affected by the speed of the water, both are subtracted and the required distance is covered in more time.

Let's call

x = Roberto's rowing speed in still water

y = Speed of the river current

The speed when rowing downstream is x+y, thus the distance traveled is

d=(x+y).t_1

Where t1=2.5 hours. Substituting values:

20=(x+y)*2.5

Rearranging, we find the downstream equation:

2.5x+2.5y=20\qquad[1]

The speed when rowing upstream is x-y, and the distance traveled is

d=(x-y).t_2

Where t2=5 hours. Substituting values:

20=(x-y)*5

Rearranging, we find the upstream equation:

5x-5y=20\qquad[2]

Multiplying [1] by 2:

5x+5y=40

Adding this equation to [2]:

10x=60

Solving:

x=60/10=6

Dividing [2] by 5:

x-y=4

Solving for y

y=x-4=6-4=2

Thus, Roberto's speed in still water is 6 miles/hour and the river speed is 2 miles/hour

3 0
3 years ago
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