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3 years ago

What 2 numbers is √14 between

Mathematics

2 answers:

k0ka [10]3 years ago

8 0

It is between 3 and 4. hope this helps :D

lesya692 [45]3 years ago

7 0

3 and 4 because 3 squared is 9 and 4 squared is 16

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which number is the largest 1.2x10 to the forth power,4.5x10 to the negitive third power,3.8x10 to the third power,7.5x10 to the

Dennis_Churaev [7]

Answer:

2 × 10 to the fourth power. (160,000)

Step-by-step explanation:

We know that if it is a negative exponent that is also an odd number, it will not be the greatest/largest. 2 times 10 is 20 and then 20 to the fourth power is 160,000. 3.8 times 10 equals 38, then we put it to the third power and we will get 54,872. Finally, 7.5 times 10 equals 75 and when we put that to the third power, it becomes 421,875. Therefore, 2 times 10 to the fourth power is the greatest/largest.

4 0

3 years ago

Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes

Ivan

Answer:

14.86 knots.

Step-by-step explanation:

<em>Given that:</em>

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

At 2 pm:

Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

Distance traveled by boat 1 = 24 + 9= 33 units due east

Distance traveled by boat 2 = 16+8 = 24 units due south

Now, speed of boat 1 changes to 8+7 = 15 knots

At 5 pm:

Distance traveled by boat 1 = 33 + 2 $\times$ 9= 51 units due east

Distance traveled by boat 2 = 24 + 2 $\times$ 15 = 54 units due south

Now, the situation of distance traveled can be seen by the attached right angled $\triangle AOB$ .

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units$

so, the total distance between the two boats is 74.28 units.

Change in distance per hour = $\dfrac{Total\ distance}{Total\ time}$

$\Rightarrow \dfrac{74.28}{5} = 14.86\ knots$

6 0

3 years ago

HELP PLEASE! Will give the Brainliest if shown work!!

scZoUnD [109]

Answer:

$\large\boxed{Q1.\ A_{\triangle PQR}=90\ cm^2}\\\\\boxed{Q2.\ V\approx321\ m^3}$

Step-by-step explanation:

$Q1.\\\\\text{If}\ \triangle ABC\sim\triangle PQR\ \text{then the quotient of the areas is equal}\\\text{the square of the similarity scale}\ k.\\\\\text{The sides}\ AB\ \text{and}\ QP\ \text{are corresponding}.\ \text{Calculate the scale:}\\\\k=\dfrac{4}{6}=\dfrac{4:2}{6:2}=\dfrac{2}{3}\\\\\text{The area of }\ \triangle ABC=40\ cm^2.\\\\\text{Let the area of}\ \triangle PQR=x,\ \text{then}\\\\\dfrac{40}{x}=\left(\dfrac{2}{3}\right)^2\\\\\dfrac{40}{x}=\dfrac{4}{9}\qquad\text{cross multiply}\\\\4x=(9)(40)\qquad\text{divide both sides by 4}\\\\x=(9)(10)\\\\x=90\ cm^2$

$Q2.\\\\\text{The formula of a volume of a sphere:}\\\\V=\dfrac{4}{3}\pi R^3\\\\R-radius\\\\\text{We have the diameter}\ 2R=8.5\ m\to R=\dfrac{8.5}{2}\ m=4.25\ m.\\\\\text{Substitute:}\\\\V=\dfrac{4}{3}\pi(4.25)^3=\dfrac{4}{3}\pi(76.765625)=(4)(25.588541)\pi=102.354164\pi\\\\\pi\approx3.14\\\\V=(102.354164)(3.14)\approx321\ m^3$