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3 years ago

A parallelogram has an area of 100 square units. Its perimeter is

Mathematics

1 answer:

Nonamiya [84]3 years ago

4 0

good morning ☕️

___________________

Step-by-step explanation:

—Dimension1 =25 and dimension2 =4

Area=25*4=100

Perimeter=2(25+4)=2(29)=58.

—-Dimension1 =12.5 and dimension2 =8

Area=12.5*8=100

Perimeter=2(12.5 + 8)=2(20.5)=41

:).

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Harry reads that a particular element has an atom with a mass of 0.000000000012 grams. What is the weight of the atom expressed

Arlecino [84]

1.2 e⁻¹¹

Because you need a number between 1 and 10 for scientific notation. So you move the decimal over 11 places. Since you move it to the right, it becomes a negative exponent.

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2 years ago

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julia-pushkina [17]

Number 3 is D, number 4 is B, number 5 is B

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3 years ago

Find the standard equation of a sphere that has diameter with the end points given below. (3,-2,4) (7,12,4)

DiKsa [7]

Answer:

The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Let the first endpoint be represented as $(x_{1}, y_{1}, z_{1})$ and the second endpoint be $(x_{2}, y_{2}, z_{2})$ .

Hence,

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Midpoint = $(\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})$

Midpoint = $(\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\$

Midpoint = $(5, 5, 4)$

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

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

$d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}$

$d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}$

$d = \sqrt{16 +196 + 0$

$d =\sqrt{212}$

$d = 2\sqrt{53}$

This is the diameter

To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is $(5, 5, 4)$

Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

Hence, the equation of the sphere of radius $\sqrt{53}$ and center $(5, 5, 4)$ is

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}$

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

This is the standard equation of the sphere

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2 years ago

Identify the variables in the following.<br> A) 4r- 3<br> B) x + 7

Doss [256]

A). 4r − 3

There are no like terms.

Answer:

= 4r − 3

B). x + 7

There are no like terms.

Answer:

= x + 7

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2 years ago

Whats the ninth term of 25 20 15 10 5

Kaylis [27]

Given:

The sequence is 25, 20, 15, 10, 5.

To find:

The ninth term of the given sequence.

Solution:

We have,

25, 20, 15, 10, 5

It is an AP because the difference between two consecutive terms are same.

Here,

First term (a) = 25

Common difference (d) = 20-25

= -5

The nth terms of an AP is

$a_n=a+(n-1)d$

Where, a is the first term and d is the common difference.

Putting a=25, n=9 and d=-5 to get the 9th term.

$a_9=25+(9-1)(-5)$

$a_9=25+(8)(-5)$

$a_9=25-40$

$a_9=-15$

Therefore, the ninth term of the given sequence is -15.

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2 years ago