3 years ago

Is (1, 5) ( 2, 2) (3, 1) (4, 2) (3, 5) a function explain

Mathematics

1 answer:

3241004551 [841]3 years ago

8 0

No it’s NOT a function
a function is a relation

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Kenny wants to wrap his mom's present. Wrapping paper is three cents per square inch. How much will the wrapping paper cost Kenn

Lostsunrise [7]

Answer:

There is no answer because we dont know how many square inches his moms present is.

8 0

2 years ago

8 1/3 miles 9 4/5miles 11 5/6miles 12 miles 14 7/8 miles. how many miles would that be that in a week run.

Kaylis [27]

Answer:

8 1/3+9 4/5+ 11 5/6+12+14 7/8=<u>58 61/120</u>

Step-by-step explanation:

First, you would add all of the whole numbers: 8+9+11+1+2+14=54

Second, you would add all of the fractions 1/3+4/5+5/6+7/8=4 61/120

Lastly, 54+4 61/120=<u>58 61/120</u>

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3 years ago

Sketch the curve with the given vector equation. Indicate with an arrow the direction in which t increases. r(t) = 4 cos(t)i + 4

slega [8]

Answer:

The equation to this question is " $x^2+y^2=16$ "

Step-by-step explanation:

Given value:

$\to r(t) = 4 \cos (t)i + 4 sin(t)j + k\\where \\\\\to x(t)= 4 \cos t\\\to y(t) = 4 \sin t\\\to z(t)=1\\$

$\to x^2+y^2= 16 \cos^2 t +16 \sin^2 t\\\\\to x^2+y^2= 16 (\cos^2 t +\sin^2 t)\\\\\therefore (\cos^2 t +\sin^2 t =1)\\\\\to x^2+y^2= 16 (1)\\\\\to x^2+y^2= 16\\\\\to z=1$

So, the curve $r(t) = 4 \cos (t)i + 4 sin(t)j + k$ is basically a circle, which is defined in the attachment file.

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3 years ago

Solve: XC - XV. Show your answer as a Roman Numeral. Standard Roman Number Numeral 1 1 5 V 10 Х 50 L 100 С C 500 D 1,000 M OA) V

Keith_Richards [23]

Answer:

5,6,7

Step-by-step explanation:

SEE THE IMAGE FOR SOLUTION

HOPE IT HELPS

HAVE A GREAT DAY

6 0

2 years ago

What is the area of the isosceles trapezoid shown?

balandron [24]

If you want the area of the entire trapeziod.

The formula for area of a trapeziod is:

1/2h(b₁+b₂)

So the equation applied to the trapeziod is:

2.5(20+12)

20+12 is 32. 32 multiplied by 2.5 is 80.

<h3><u><em>Ｙｏｕｒ　ａｎｓｗｅｒ　ｉｓ　８０.</em></u></h3>

4 0

3 years ago

Read 2 more answers