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3 years ago

What is the y intercept 8th math

Mathematics

2 answers:

Mamont248 [21]3 years ago

8 0

Answer:

(0.3) , That's It love hope this works

ruslelena [56]3 years ago

7 0

Answer:

(0,1)

Step-by-step explanation:

the y intercept is when x = 0

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I'LL NAME YOU BRAINIEST!!!! Find the equation for the line which contains the points (0, -3) and (14, 0). Express your answer in

vova2212 [387]

It’s 5+3x 10-1= 21345

5 0

4 years ago

Is it mathematically correct to say that:

Step2247 [10]

Answer: Yes

Step-by-step explanation:

Since the numerator is multiplying square root 13 by square root 2, we can break them apart.

$\displaystyle \frac{\sqrt{13*2}}{\sqrt{13}} =\frac{\sqrt{13}*\sqrt{2} }{\sqrt{13}}$

You are correct, $\sqrt{13}$ over $\sqrt{13}$ is equal to one, so we can "cancel them out" of our expression.

$\displaystyle \frac{\sqrt{13} }{\sqrt{13}} = 1$

This leaves us with: $\sqrt{2}$

We can also test this by using a calculator:

$\frac{\sqrt{13*2}}{\sqrt{13}} \approx 1.41421356\\\sqrt{2} \approx 1.41421356\\1.41421356=1.41421356$

8 0

2 years ago

Hi please help , With solution if it's ok

Snowcat [4.5K]

Answers with Method:-

1) Multiply the length value by 100,000.

$2.5 \: hm = 2.5 \times 100000$

$= >2.5 \: hm = 250000 \: mm$

2) For this, divide the length value by 1000.

$= > \frac{1800}{1000}$

$= > 1.8 \: dam$

3) For finding the approximate value, just multiply the value of length by 1.609.

$= >6450 \: m = (6450 \times 1.609) \: km$

$= > 10380.27 \: km$

4) Multiply the given mass value by 100.

$= > 1.2 \: kg = 1.2 \times 100$

$= > 1200 \: g$

5) Multiply the given value by 10,000.

$4.37 \: dag = 4.37 \times 10000$

$= > 43700 \: mg$

5 0

2 years ago

Find the distance from the point (8, 4) to the line y = x+ 2.

natulia [17]

Answer:

The distance is $3\sqrt{2}\ units$

Step-by-step explanation:

step 1

Find the slope of the give line

we have

y=x+2

the slope m is equal to

m=1

step 2

Find the slope of the perpendicular line to the given line

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal of each other

we have

m=1 -----> slope of the given line

therefore

The slope of the perpendicular line is equal to

m=-1

step 3

With m=-1 and the point (8,4) find the equation of the line

y-y1=m(x-x1)

substitute

y-4=-(x-8)

y=-x+8+4

y=-x+12

step 4

Find the intersection point lines y=x+2 and y=-x+12

y=x+2 -----> equation A

y=-x+12 ----> equation B

Adds equation A and equation B

y+y=2+12

2y=14

y=7

Find the value of x

y=x+2 -----> 7=x+2 -----> x=5

The intersection point is (5,7)

step 5

Find the distance between the point (8,4) and (5,7)

we know that

The distance from the point (8,4) to the line y=x+2 is equal to the distance from the point (8,4) to the point (5,7)

Find the distance AB

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute

$d=\sqrt{(7-4)^{2}+(5-8)^{2}}$

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

$d=\sqrt{18}$

$d=3\sqrt{2}\ units$

see the attached figure to better understand the problem

6 0

4 years ago

Santa Claus loves to drop presents out of his sled during Christmas. The height above the ground of one present after t seconds

Ahat [919]

Answer:

The present was 300 feet above ground 5 seconds after being thrown.

6.47 seconds passed before the present hit the ground.

Step-by-step explanation:

We have the following quadratic function

$h(t) = -16t^{2} - 20t + 800$

Which determines the height of the present.

Determine when the present was 300 feet above the ground.

This is when $h(t) = 300$ . So

$h(t) = -16t^{2} - 20t + 800$

$300 = -16t^{2} - 20t + 800$

$-16t^{2} - 20t + 500 = 0$

This is $t = -6.25$ and $t = 5$ . There are no negative time instants. So the present was 300 feet above ground 5 seconds after being thrown.

Determine how many seconds passed before the present hit the ground.

This is t when $h(t) = 0$

$h(t) = -16t^{2} - 20t + 800$

$-16t^{2} - 20t + 800 = 0$

$t = 6.47$

6.47 seconds passed before the present hit the ground.

5 0

3 years ago