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3 years ago

10

Between which pair of consecutive integers does square root 21 lie? 10 and 11 5 and 6 20 and 22 4 and 5

1 answer:

Zolol [24]3 years ago

8 0

square root of 11. 3 & 4 2. square root of 33. 5 & 6 3. square root of 8. 2 & 3 4. cube root of 220. 6 & 7 5. cube root of -80. 4 & 5 6. fourth root of 20. 2 & 3 7. square root of 300. 17 & 18 8. fifth root of -35. 2 & 3 9. square root of 150. 12 & 13 10. cube root of -312. 6 & 7

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Suppose that you want to find out if more Canadians prefer eating toast or cereal in the morning. List 4 ways you could find sec

Rina8888 [55]

Information collected through censuses or government departments like housing, social security, electoral statistics, tax records.
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7 0

3 years ago

Joshua's Math textbook weighs 4.21 ounces. His Social Studies book weighs 3.99 ounces. What is the difference in weight between

pochemuha

Answer:

0.22

Step-by-step explanation:

4.21-3.99= 0.22

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3 years ago

Which of the following is equivalent to the expression

nata0808 [166]

A.) i is the answer

5 0

3 years ago

Drag point P around the unit circle to find the values of the trigonometric functions.

DanielleElmas [232]

Answer:

See the attachment

Step-by-step explanation:

A unit circle is one whose radius is 1

sin(90) will have coordinates (0,1)

sin(40)= 0.643

sin(40)= P/H= 0.643/1

cos(40)= B/H= 0.766

Similarly for cos(245)= cos(-65)= -0.423/1

sin(245)= sin(-65)= -0.906

8 0

4 years ago

Read 2 more answers

The coordinates of the vertices of ∆PQR are P(-2,5), Q(-1,1), and R(7,3). Determine whether ∆PQR is a right triangle. Show your

Mashutka [201]

Given

∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

Determine whether ∆PQR is a right triangle

To proof

As given ∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

First find out the sides of triangle

FORMULA

Distance formula between two points

$D^{2}= (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

Distance between two points P(-2,5) and Q(-1,1)

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

$PR = \sqrt{17}$

Distance between two points Q(-1,1)and R(7,3)

$QR = \sqrt{(7+1)^{2} +(3-1)^{2} }$

$QR =\sqrt{68}$

Distance between two points R(7,3) and P(-2,5)

$RP =\sqrt{(-2-7)^{2} + (5-3)^{2} }$

$RP=\sqrt{85}$

now show that ∆PQR is a right triangle

$RP^{2} = PQ^{2} +QR^{2}$

Putting the value given above

$(\sqrt{85}) ^{2} = \sqrt{17} ^{2} +\sqrt{68} ^{2}$

85 = 17 +68

85 =85

In the right triangle

HYPOTENUSE² = BASE² + PERPENDICULAR²

This is prove above

Hence ∆PQR is a right triangle

Hence proved

7 0

4 years ago