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Nonamiya [84]
3 years ago
13

Help me please !!!!!!!! Thank you .

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
7 0

Answer:

You made this up didn't you

Step-by-step explanation:

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A newborn baby weighs 3.136 times 10 superscript 3 units. which unit of measure was used? grams ounces pounds tons
lorasvet [3.4K]

Answer:

its A

Step-by-step explanation:

have you smiled today? its very important:)

8 0
2 years ago
Solve for X.<br> - 4x + 10 = 50
erastovalidia [21]

Answer: -10

Step-by-step explanation:

10 x 4 is 40 so with 2 negatives, it turns positive, and 40 plus 10 = 50

5 0
2 years ago
Read 2 more answers
The difference between an expression and an_________ is that an expression does not contain an equal sign.
kvv77 [185]
Fill in the blank with equation.

An equation is nearly the same thing as an expression. However, the dofference is that equations have an = sign while expression do not.
8 0
3 years ago
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
Scientists discover two planets orbiting a distant star. The average distance from the star to Planet A is 4 AU, and it takes 43
Kitty [74]

Answer:

Option C - 9 AU

Step-by-step explanation:

To find : What is the average distance from Planet B to the star?

Solution :

According to kepler's law,

The squares of the sidereal periods (of revolution) of the planets are directly proportional to the cubes of their mean distances from the Sun.

i.e. P^2\propto S^3

We have given,

The average distance from the star to Planet A is S_1=4 AU.

It takes 432 Earth days for Planet A to orbit the star i.e. P_1=432

It takes 1,460 days for Planet B to complete an orbit i.e.  P_2=1460

Substitute the values in (\frac{P_1}{P_2})^2=(\frac{S_1}{S_2})^3

(\frac{432}{1460})^2=(\frac{4}{S_2})^3

0.0875=(\frac{4}{S_2})^3

Taking root cube both side,

\sqrt[3]{0.0875}=\frac{4}{S_2}

0.444=\frac{4}{S_2}

S_2=\frac{4}{0.444}

S_2=9.00

The average distance from Planet B to the star is 9 AU.

Therefore, Option C is correct.

5 0
3 years ago
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