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1 year ago

A dust particle weighs 7.42 × 10-10 kilograms. What is the weight of 5 × 106 dust particles represented in scientific notation?

Mathematics

2 answers:

vova2212 [387]1 year ago

7 0

5 x 106 = 530 particles

530 x 7.42x10^-10 = 3.93x10^-7

elena55 [62]1 year ago

5 0

3.71 × 10-3 kilograms

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If mBC = (9x-53) and mCD = (2x + 45) find mBAD

Rom4ik [11]

Answer:

$m\angle BAD=73^o$

Step-by-step explanation:

The picture of the question in the attached figure

step 1

Find the value of x

Let

O ----> the center of the circle

we know that

Triangle BOC≅Triangle COD

$m\angle BOC=arc\ BC$ ----> by central angle

$m\angle COD=arc\ CD$ ----> by central angle

$m\angle BOC=m\angle COD$

therefore

$arc\ BC=arc\ CD$

substitute the given values

$(9x-53)^o=(2x+45)^o$

solve for x

$9x-2x=45+53\\7x=98\\x=14$

step 2

Find the measure of angle BAD

we know that

The inscribed angle is half that of the arc it comprises.

$m\angle BAD=\frac{1}{2} [arc\ BC+arc\ CD]$

$arc\ BC=9(14)-53=73^o$

$arc\ CD=2(14)+45=73^o$

substitute

$m\angle BAD=\frac{1}{2} [73^o+73^o]=73^o$

7 0

1 year ago

What is 4/ 15 divided by 10/ 13

Alja [10]

Answer:

26/75

Step-by-step explanation:

4 / 15 / 10 / 13

= 4/15 * 13/10

= 52 / 150

= 26/75 answer

5 0

1 year ago

Read 2 more answers

Select the correct answer. Which transformation will always map a parallelogram onto itself? A. a 90° rotation about its center

prisoha [69]

Answer:

C. a 180° rotation about its center

Step-by-step explanation:

The question is on rotation of an object about the origin through 180°

When a point B, (h,k) is rotated about the origin 0 through 180° both clockwise and anticlockwise directions , the new position is B'(-h,-k)

Example, if we have point K (5,8) rotated through 180°, clockwise or anti-clockwise about the origin the new position will be; K'(-5,-8)

7 0

1 year ago

Suppose you roll a pair of honest dice. If you roll a total of 7 you win $22, if you roll a total of 11 you win $66, if you roll

JulijaS [17]

Answer:

The expected payoff for this game is -$1.22.

Step-by-step explanation:

It is given that a pair of honest dice is rolled.

Possible outcomes for a dice = 1,2,3,4,5,6

Two dices are rolled then the total number of outcomes = 6 × 6 = 36.

$\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),\\(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

The possible ways of getting a total of 7,

{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

Number of favorable outcomes = 7

Formula for probability:

$Probability=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

So, the possibility of getting a total of 7 = $\frac{6}{36}=\frac{1}{6}$

The possible ways of getting a total of 11,

{(5,6), (6,5)}

So, the probability of getting a total of 11 = $\frac{2}{36}$ = $\frac{1}{18}$

Now, other possible rolls = 36 - 6 - 2 = 36 - 8 = 28,

So, the probability of getting the sum of numbers other than 7 or 11 = $\frac{28}{36}$ = $\frac{7}{9}$

Since, for the sum of 7, $ 22 will earn, for the sum of 11, $ 66 will earn while for any other total loss is $11,

Hence, the expected value for this game is

$\frac{1}{6}\times 22+\frac{1}{18}\times 66-\frac{7}{9}\times 11$

$\frac{11}{3}+\frac{11}{3}-\frac{77}{9}$

$\frac{22}{3}-\frac{77}{9}$

$\frac{66-77}{9}$

$-\frac{11}{9}$

$-1.22$

Therefore the expected payoff for this game is -$1.22.

4 0

1 year ago

Two cross sections of a right hexagonal pyramid are obtained by cutting the pyramid with planes parallel to the hexagonal base.

Tanya [424]

Answer:

The larger cross section is 24 meters away from the apex.

Step-by-step explanation:

The cross section of a right hexagonal pyramid is a hexagon; therefore, let us first get some things clear about a hexagon.

The length of the side of the hexagon is equal to the radius of the circle that inscribes it.

The area is

$A=\frac{3\sqrt{3} }{2} r^2$

Where $r$ is the radius of the inscribing circle (or the length of side of the hexagon).

Now we are given the areas of the two cross sections of the right hexagonal pyramid: $A_1=216\:ft^2\: \:\:\:A_2=486\:ft^2$

From these areas we find the radius of the hexagons:

$r_1=\sqrt{\frac{2A_1}{3\sqrt{3} } } =\sqrt{\frac{2*216}{3\sqrt{3} } }=\boxed{9.12ft}$

$r_2=\sqrt{\frac{2A_2}{3\sqrt{3} } } =\sqrt{\frac{2*486}{3\sqrt{3} } }=\boxed{13.68ft}$

Now when we look at the right hexagonal pyramid from the sides ( as shown in the figure attached ), we see that $r_1$ $r_2$ form similar triangles with length $H$

Therefore we have:

$\frac{H-8}{r_1} =\frac{H}{r_2}$

We put in the numerical values of $r_1$ , $r_2$ and solve for $H$ :

$\boxed{H=\frac{8r_2}{r_2-r_1} =\frac{8*13.677}{13.68-9.12} =24\:feet.}$

8 0

2 years ago