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Tamiku [17]
3 years ago
9

Help me please it’s due tonight. Need this asap

Mathematics
1 answer:
liraira [26]3 years ago
3 0

Answer:

Repost your picture; it is super blurry.

Step-by-step explanation:

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Find the value of the given variable (-4, r) (-8, 3) m=5
UNO [17]
R=23 because its slope the formula is Y-Y/X-X


6 0
3 years ago
I dont understand how to answer this question, help please?
kakasveta [241]
It is A. ....... i wish you had a whole chart

8 0
3 years ago
Kallisto built a rectangular sign that measured 2 and 3/4. The width is 1 and 1/4 shorter than the length. To find the area, mul
prisoha [69]

Answer:

<em>l</em> = 1 5/16

w = 1/16

A = 21/16 x 1/16 = 21/256

Step-by-step explanation:

Perimeter = 11/4

<em>l</em> = length

w = <em>l</em> - 5/4

2<em>l</em> + 2(<em>l</em> - 5/4) = 11/4

2<em>l</em> + 2<em>l</em> - -5/2 = 1/4

multiply each side by 4

8<em>l</em> + 8<em>l </em>- 10 = 11<em> </em>

16<em>l</em> = 21

<em>l</em> = 21/16 or 1 5/16

width = 21/16 - 20/16 = 1/16

8 0
3 years ago
Isaiah has $7.15 in Dimes and Nickels. He has a total of 91 coins. How many of each coin does he have? Group of answer choices d
yarga [219]

Answer:

Dimes = x = 52

Nickels = y = 39

Step-by-step explanation:

Let

Dimes = x

Nickels = y

x + y = 91 (1)

0.10x + 0.05y = 7.15 (2)

From (1)

x = 91 - y

Substitute x = 91 - y into (2)

0.10x + 0.05y = 7.15

0.10(91 - y) + 0.05y = 7.15

9.10 - 0.10y + 0.05y = 7.15

- 0.10y + 0.05y = 7.15 - 9.10

-0.05y = -1.95

y = -1.95 / 0.05

= 39

Substitute y = 39 into (1)

x + y = 91

x + 39 = 91

x = 91 - 39

x = 52

Dimes = x = 52

Nickels = y = 39

7 0
3 years ago
To find the interquartile range you need to ____the upper and lower quartile values.​
Blababa [14]

Answer:

The correct options are: Interquartile ranges are not significantly impacted by outliers. Lower and upper quartiles are needed to find the interquartile range. The data values should be listed in order before trying to find the interquartile range. The option Subtract the lowest and highest values to find the interquartile range is incorrect because the difference between lowest and highest values will give us range. The option A small interquartile range means the data is spread far away from the median is incorrect because a small interquartile means data is nor spread far away from the median

8 0
3 years ago
Read 2 more answers
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