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miskamm [114]
3 years ago
5

A survey is given to every 4th grade student that walks into the school. What type of survey is this?

Mathematics
1 answer:
katrin [286]3 years ago
3 0
Pop quiz? I at least think
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What is the measure of the hypotenuse?
SOVA2 [1]
<h2><em>Answer:</em></h2><h3><em>1</em><em>3</em></h3>

<em>solution \\ hypotenuse = x \\ base = 12 \\ perpendicular = 5 \\ using \: pythagoras \: theorem \\  {h}^{2}  =  {p}^{2}  +  {b}^{2}  \\ or \:  {x}^{2}  =  {5}^{2}  +  {12}^{2}  \\ or \:  {x}^{2}  = 25  + 144 \\ or \:  {x}^{2}  = 169 \\ or \: x =  \sqrt{169}  \\ x = 13 \\</em>

<em>hope </em><em>it</em><em> helps</em><em>.</em><em>.</em><em>.</em><em>.</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

3 0
3 years ago
Read 2 more answers
Bài 1: Tính giá trị của biểu thức sau:
Alex Ar [27]

Answer:

Lời giải có ở trong ảnh bạn nhé

8 0
3 years ago
This figure is made up of a quadrilateral and a semicircle.
vodomira [7]

Here the figure is made up of a quadrilateral and a semi circle.

ABCD is the quadrilateral here. We will find the sides of the quadrilateral by using the distance formula.

If (x₁, y₁) and (x₂, y₂) are two points given, then the distance between two points by using distance formula is,

d =\sqrt({ x_{1}-x_{2})^2  +( y_{1} -y_{2}  )^2}

The co-ordinate of A is (-1,2) and co-ordinate of B is (-2,-1).

So the length of side AB = \sqrt{(-1-(-2))^2+(2-(-1))^2}

=\sqrt{(-1+2)^2+(2+1)^2}(As negative times negative is positive)

= \sqrt{(1)^2+(3)^2} =\sqrt{1+9}  =\sqrt{10}

The co-ordinate of C is (4,-3) and D is (5,0)

The length of side CD

= \sqrt(4-5)^2+(-3-0)^2}

= \sqrt{(-1)^2+(-3)^2}

= \sqrt{1+9} =\sqrt{10}

So the sides AB and CD are equal.

The length of side AD

= \sqrt{(-1-5)^2+(2-0)^2}

= \sqrt{(-6)^2+(2)^2} =\sqrt{36+4} =\sqrt{40}

The length of side BC

= \sqrt{(-2-4)^2+(-1-(-3))^2}

= \sqrt{(-2-4)^2+(-1+3)^2}

= \sqrt{(-6)^2+(2)^2}=\sqrt{36+4} =\sqrt{40}

So the lengths of the sides AD and BC are equal.

So the quadrilateral is a rectangle whose length is \sqrt{40} and width is \sqrt{10}.

Area of a rectangle = length × width

= (\sqrt{40}) (\sqrt{10})

= \sqrt{(40)(10)}=\sqrt{400}

= 20 unit^2

Now the diameter of the semicircle is the side AD = \sqrt{40}

So, the radius of the semi-circle = \frac{\sqrt{40}}{2}

= \frac{\sqrt{(4)(10)}}{2}

= \frac{(\sqrt{4})(\sqrt{10})}{2}

= \frac{2\sqrt{10}}{2} = \sqrt{10}

Area of semi-circle = \frac{1}{2} \pi r^2, where r is the radius.

= \frac{1}{2} \pi  (\sqrt{10})^2

= \frac{1}{2} \pi   (10)

= \frac{(\pi)(10)}{2}

= \frac{10\pi}{2}   = 5\pi = 15.7 unit^2 ( Approximately taken to the nearest tenth)

Total area of the figure = (20+15.7) unit^2 = 35.7 unit^2

We have got the required answer.

Option a is correct here.

3 0
3 years ago
Read 2 more answers
Need help plss! ASAP
olasank [31]
The answer i think is a because those segments map onto each other, a-e, b-c, c-d
4 0
3 years ago
Which of the following ordered pairs are solutions to the system of
Elenna [48]
I’m pretty sure the answer is C and b
8 0
2 years ago
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