How do you put -26/3 on a number line

Mathematics

1 answer:

earnstyle [38]3 years ago

5 0

U make a big number line

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a cartoon holds 1.248 liters of fruit juice. what is this amount rounded to the nearest tenth? use the number line below.

mestny [16]

amount rounded to the nearest tenth of 1.248 should be 1.2
Answer is 1.2

6 0

3 years ago

Read 2 more answers

A baseball team had 75 players show up for tryouts last year and this year had 90 players show up for tryouts. Find the percent

olga_2 [115]

Answer: 33.3%

Step-by-step explanation:

First find the difference between the players in try-outs over the years.

= 90 - 75

= 25

Then use this difference to find the percent increase:

= Difference / Players last years * 100%

= 25 / 75 * 100%

= 33.3%

3 0

3 years ago

A wire that is 22 feet long connects the top of a pole to the ground. The wire is attached to the ground at a point that is 10 f

krok68 [10]

✰ Concept Used :-

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In this question, we can clearly observer that the diagram shows a right angled triangle. And, we have been provided with the value of base, and the value of hypotenuse, using the pythagoras theorem, now we can easily find out the value of the perpendicular i.e. the value of the side h. According to the pythagoras theorem, square of hypotenuse is equal to the sum of square of perpendicular and square of side respectively. Therefore, square of side is equal to the difference of square of hypotenuse and square of perpendicular.

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✰ Given Information :-

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Hypotenuse = 22 ft.
Base = 10 ft.

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✰ To Find :-

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The value of side or the perpendicular

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✰ Formula Used :-

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$\star \: \underline{ \boxed{ \purple { \sf {Side}^{2} = {Hypotenuse}^{2} - {Base}^{2} }}} \: \star$

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✰ Solution :-

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$\sf \longrightarrow {Side}^{2} = {(22 \: ft)}^{2} - {(10 \: ft)}^{2} \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{2} = {484 \: ft}^{2} - {100 \: ft}^{2} \: \: \: \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{2} = {384 \: ft}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{} = \sqrt{ {384 \: ft}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \sf \longrightarrow {Side}^{} = \underline{ \boxed{ \frak{ \green{19.60 \: ft}}}} \: \star \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\$