MATHEMATICS HIGH SCHOOL

When writing an equation in slope-intercept form, you need the values of the

Answers

Answer 1

Answer:

When writing an equation in slope-intercept form, there will be need for slope and y- intercept.

What is the slope intercept form of an equation ?

When any equation is represented in the y = mx +c form, it is called the slope intercept form of the equation.

Here m is slope and c is constant.

To writing the slope intercept form of the equation,

There will be requirement of slope,

and y intercept, it will be in form of (0, a), here a will be value of y coordinate.

To know more about Slope intercept form :

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Answer 2

Answer:

Answer:

slope, m

y-intercept

Step-by-step explanation:

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How do you " find, to the nearest tenth of an inch, the diagonal of a square whose perimeter is 28 inches ?"

Answers

Well if the perimeter is 28 inches, each side of the square is 7 inches. 28/4 = 7 because all sides of a square are the same.

If we split the square in half to create 2 triangles we can find the length of the diagonal. The two sides are the bases of 7 inches.

Use the Pythagorean theorem to solve for the hypotenuse or the diagonal.

7^2 + 7^2 = diagonal ^ 2

Therefore, the diagonal equals 9.9 inches.

2. Find the least no. by which 294 must be multiplied to make it a perfect square?

Answers

$294|2\n147|3\n.49|7\n.\ 7|7\n.\ 1|\n\n294=2*3*7^2=6*7^2\n\nAnswer:this\ number\ is\ 6\ becaue\ 6*294=6*6*7^2=6^2*7^2=(6*7)^2$

What is the distance between points (3,8) and (-9,-8)5
10
15
20

what is the distance between points (2,6) and (5,2)
3
4
5

what is the distance between points (21,16) and (9,11)
13
12
22
24

Answers

It is better to graph these points, to easily understand that the distance of the two points form the hypotenuse of the right triangle. We need to use the Pythagorean theorem to solve for the distance of the two points.

a² + b² = c² ; a = x value ; b = y value

1) (3,8) and (-9,-8)
a = 3 - (-9) = 3 + 9 = 12 ⇒ 12² = 144
b = 8 - (-8) = 8 + 8 = 16 ⇒ 16² = 256 ⇒⇒ 144 + 256 = 400 ; c = √400 = 20

2) (2,6) and (5,2)
a = 2 - 5 = 3 ⇒ 3² = 9
b = 6 - 2 = 4 ⇒ 4² = 16 ⇒⇒ 9 + 16 = 25 ; c = √25 = 5

3) (21,16) and (9,11)
a = 21 - 9 = 12 ⇒ 12² = 144
b = 16 - 11 = 5 ⇒ 5² = 25 ⇒⇒ 144 + 25 = 169 ; c = √169 = 13

9 x (2 + 7) = (? x 2) + (? x 7)

Answers

i believe the answer is 9x ( 2 + 7) = +9 x 2) + (9 x 7)

In a survey, 44% of the people said they voted for Mr. Johnson, while 2/5 of the people said they voted for Ms. Smith. Which group is larger? Explain.

Answers

Since 2/5 is equivalent to 4/10 = 40%, then the group which voted for Mr. Johnson is the larger group since they are 44%.

44% Is Bigger Because 2/5 is Only 40%

Consider the 1-to10^19 scale on which the disk of the Milky Way Galaxy fits on a football field. On this scale, how far is it from the sun to the alpa centauri (real distance:4.4 light-years) How big is the sun itself on this scale? Compare the sun's size on this scale to the actual size of a typical atom (about 10^-10 m in diameter).

Answers

Answer:

Sun is $0.4162708$ cm away from alpha century.

Sun is $1.391016*10^(-10)$ m.

Sun is 1.391016 time the size of an atom at this scale.

Step-by-step explanation:

Light year is a measure of distance. It is the distance light travels in an year.

Light year = $9.4607*10^(12)$ km

So 4.4 light years = $4.4*9.4607*10^(12)$ km

$41.62708*10^(12)$ km

Lets scale this down to the level of $1*10^(-19)$

$41.62708*10^(12)*1*10^(-19)$ km

= $41.62708*10^(-7)$ km

Change the units to centimeters:

$41.62708*10^(-7)*1*10^(5)$ cm

= $41.62708*10^(-2)$ cm

= $0.4162708$ cm

Therefore on the new scale sun is $0.4162708$ cm away from alpha century.

Diameter of the sun is 1.391016 million km

Lets change Sun's diameter to the new scale:

$1.391016*10^(6) *10^(-19)$ km

= $1.391016*10^(-13)$ km

Lets change kilometers in to meters:

$1.391016*10^(-13)*10^(3)$ m

= $1.391016*10^(-10)$ m

Therefore, sun is $1.391016*10^(-10)$ m

and an atom is $1*10^(-10)$

Therefore the sun is 1.391016 time the size of an atom at this scale.

Final answer:

On a 1-to-10^19 scale, the distance from the Sun to the Alpha Centauri is about 44 cm. On this same scale, the Sun itself would have a diameter of about 150 picometers, which is larger than a typical atom.

Explanation:

The 1-to-10^19 scale means for every actual meter in space, we represent 10^19 meters on our model. The Alpha Centauri is 4.4 light-years away from the sun. Considering 1 light-year equals to approximately 9.46x10^15 meters, the real distance from the sun to Alpha Centauri is about 4.16x10^16 meters. So, on the scale, this is about 0.44 meters or 44 cm.

The Sun's real size, with a diameter of 1.5 million kilometers or 1.5x10^9 meters, is represented as 1.5x10^-10 meters or 150 picometers on the scale. This is much bigger than an actual atom, which has a diameter of 0.1 to 0.5 nanometers or 100 to 500 picometers. Hence, on this scale, the Sun would be larger than a typical atom.

Learn more about Scale Representation here:

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