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My name is Ann [436]
3 years ago
6

A bookmark is shaped like a rectangle with a semicircle attached at both ends. The rectangle is 10 cm long and 4 cm wide. The di

ameter of each circle is the width of the rectangle. What is the area of the bookmark? Use 3.14 for π.
The area of the bookmark is cm2.

Plzz help me

Mathematics
1 answer:
Naddika [18.5K]3 years ago
3 0

Answer:

Area of the bookmark = 52.56 cm²

Step-by-step explanation:

The two semicircles attached at both end of the rectangle will give us a full circle.

Therefore,

Area of the bookmark = area of rectangle + Area of a circle

Area of the bookmark = L × W + πr²

Length (L) = 10 cm

Width (W) = 4 cm

radius (r) = ½(4) = 2 cm

π = 3.14

Plug in the values into the equation

Area of bookmark = 10 × 4 + 3.14 × 2²

= 40 + 12.56

= 52.56 cm²

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Answer:

Converting the equation x^2-20x+13=0 into completing the square method we get: \mathbf{(x-10)^2=87}

Step-by-step explanation:

we are given quadratic equation: x^2-20x+13=0

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Completing the square method is of form: a^2-2ab+b^2=(a-b)^2

Looking at the given equation x^2-20x+13=0

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then we have middle term 20x that can be written in form of 2ab So, we have a=x and b=? Multiplying 10 with 2 we get 20 so, we can say that b = 20

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So, we need to add and subtract (10)^2 on both sides

x^2-20x+13=0\\x^2-2(x)(10)+(10)^2-(10)^2+13=0\\(x^2-2(x)(10)+(10)^2) \:can\: be\: written\: as\: (x-10)^2 \\(x-10)^2-100+13=0\\(x-10)^2-87=0\\(x-10)^2=87

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Find the area of this triangle..
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d

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Masteriza [31]

\huge\text{Hey there!}

\large\textsf{y = 4x - 1}

\large\text{If x= -3, then substitute it into the given equation}

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\large\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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