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Irina-Kira [14]
3 years ago
5

Help please. I don't know what a vector is in this context

Mathematics
1 answer:
ale4655 [162]3 years ago
8 0

Answer: The vector relies on how the objects of where the magnitude and directions should lay on.

Step-by-step explanation:

So, I’ll help you with these 4 problems. So, for the vector 2/4, it would 2 units up and 4 units to the right. The second would be 5 units down and 6 units to the right. The third one would be 3 units up and 1 unit to the left. The last one would be 10 units downs and 12 units to the left. A simplified way of thinking of this is to just look at the signs of the number and see where the vector’s magnitude and directions would go in that point of view when you visual the graph or count it in your head.

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8.49

Step-by-step explanation:

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Find the area of the composite figure.
Roman55 [17]

Answer:

The Area of the composite figure would be 76.26 in^2

Step-by-step explanation:

<u>According to the Figure Given:</u>

Total Horizontal Distance = 14 in

Length = 6 in

<u>To Find :</u>

The Area of the composite figure

<u>Solution:</u>

Firstly we need to find the area of Rectangular part.

So We know that,

\boxed{ \rm \: Area  \:  of \:  Rectangle = Length×Breadth}

Here, Length is 6 in but the breadth is unknown.

To Find out the breadth, we’ll use this formula:

\boxed{\rm \: Breadth = total  \: distance - Radius}

According to the Figure, we can see one side of a rectangle and radius of the circle are common, hence,

\longrightarrow\rm \: Length \:  of \:  the  \: circle = Radius

  • Since Length = 6 in ;

\longrightarrow \rm \: 6 \: in   = radius

Hence Radius is 6 in.

So Substitute the value of Total distance and Radius:

  • Total Horizontal Distance= 14
  • Radius = 6

\longrightarrow\rm \: Breadth = 14-6

\longrightarrow\rm \: Breadth = 8 \: in

Hence, the Breadth is 8 in.

Then, Substitute the values of Length and Breadth in the formula of Rectangle :

  • Length = 6
  • Breadth = 8

\longrightarrow\rm \: Area \:  of  \: Rectangle = 6 \times 8

\longrightarrow \rm \: Area \:  of  \: Rectangle = 48 \: in {}^{2}

Then, We need to find the area of Quarter circle :

We know that,

\boxed{\rm Area_{(Quarter \; Circle) }  = \cfrac{\pi{r} {}^{2} }{4}}

Now Substitute their values:

  • r = radius = 6
  • π = 3.14

\longrightarrow\rm Area_{(Quarter \; Circle) } =  \cfrac{3.14 \times 6 {}^{2} }{4}

Solve it.

\longrightarrow\rm Area_{(Quarter \; Circle) } =  \cfrac{3.14 \times 36}{4}

\longrightarrow\rm Area_{(Quarter \; Circle) } =  \cfrac{3.14 \times \cancel{{36} } \: ^{9} }{ \cancel4}

\longrightarrow\rm Area_{(Quarter \; Circle)} =3.14 \times 9

\longrightarrow\rm Area_{(Quarter \; Circle) } = 28.26 \:  {in}^{2}

Now we can Find out the total Area of composite figure:

We know that,

\boxed{ \rm \: Area_{(Composite Figure)} =Area_{(rectangle)}+ Area_{ (Quarter Circle)}}

So Substitute their values:

  • \rm Area_{(rectangle)} = 48
  • \rm Area_{(Quarter Circle)} = 28.26

\longrightarrow \rm \: Area_{(Composite Figure)} =48 + 28 .26

Solve it.

\longrightarrow \rm \: Area_{(Composite Figure)} =\boxed{\tt 76.26 \:\rm in {}^{2}}

Hence, the area of the composite figure would be 76.26 in² or 76.26 sq. in.

\rule{225pt}{2pt}

I hope this helps!

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Quick algebra 1 question for 10 points!
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