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3 years ago

Adding fractions with unlike denominators

Mathematics

1 answer:

hodyreva [135]3 years ago

8 0

Multiply the 3 times 4. Then do the same on the top.

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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!

3 0

1 year ago

Match the equation to the graph where n is the Number of Burgers and C is the cost.

Softa [21]

Answer:

How did life during the Industrial Revolution differ in the North and the South?

In what ways did the Industrial Revolution make life better?

How did it make life more difficult? How did life during the Industrial Revolution differ in the North and the South?

In what ways did the Industrial Revolution make life better?

How did it make life more difficult?

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the length of side BC if AC is 5 square root 2 m?

Jet001 [13]

This is a right triangle, so AC is a hypotenuse. The other angles are each 45 degrees, so the legs AB and BC are equal length. a^2 + b^2 = c^2 a = b 2a^2 = (5sqrt2)^2 = 50 a^2 = 25 a = 5 Answer: 5

4 0

2 years ago

Solve the equation. 2x squared-3x+1=0

natali 33 [55]

$2x^2-3x+1=0\\\\2x^2-2x-x+1=0\\\\2x(x-1)-1(x-1)=0\\\\(x-1)(2x-1)=0\iff x-1=0\ \vee\ 2x-1=0\ \ \ \ |+1\\\\x=1\ \vee\ 2x=1\ \ \ |:2\\\\\boxed{x=1\ \vee\ x=0.5}$

4 0

3 years ago

Write the subtraction expression as an equivalent addition expression and then evaluate it.

Harrizon [31]

Answer:

Step-by-step explanation:

Given the expression -3-(-5)

This is also expressed as;

-3(-5)

= -3 + 5 (The product of two negative sign gives a positive sign)

= 5 - 3

= 2

Hence the result of the expression is 2

6 0

3 years ago

Adding fractions with unlike denominators​

Adding fractions with unlike denominators