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3 years ago

How did newton know that 1%of 80 is 0.8

1 answer:

8 0

So,

Newton knows that 1% of anything really just means moving the decimal point two places to the left.

Examples:
1% of 10 = 0.1

1% of 389,092,192 = 3,890,921.92

1% of 3.1415926 = 0.031415926

Knowing this, Newton figures that all he has to do is move the decimal place two places to the left.

1% of 80 = 0.8

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The following data points represent the number of classes that each teacher at Broxin High School teaches.

nika2105 [10]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Tonya and Pearl each completed a separate proof to show that alternate interior angles AKL and FLK are congruent. Who completed

zalisa [80]

Answer:

Tonya proof is correct
and Pearl proof is wrong

Step-by-step explanation:

AKL and GKB are obviously congruent. But the reasons given are different, One of the justification is Vertical Angles Theorem, and the other is Adjacent Angles.

But the correct justification is the Vertical Angles Theorem, because the angles are Vertically away from each other, or the angles are at opposite sides of each other. Which is when Vertical Angles Theorem is applied.
Making Tonya correct since that was the answer he given.

The definition of Adjacent Angles is incorrect, this Theorem is used when the angles are Adjacent to each other. When the angles are on the same line as the other angle, is when the definition of Adjacent Angles is applied. Not in this situation

8 0

2 years ago

Which of the following functions are an example of exponential growth ?

yanalaym [24]

Answer:

i don't know maybe D. 2 and 3

8 0

3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

Read 2 more answers

An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each oth

dedylja [7]

Answer:

(a) D(t) = 250t -500 miles

(b) Controller has 2 hours, but including time for pilots to divert course or altitude.

Step-by-step explanation:

Given:

two planes at same altitude heading in a collision course.

Plane A at 400 miles from collision point at 200 mph

Plane B at 300 miles from collision point at 150 mph.

Theoretical collision happens in

t = 400/200 = 300/150 = 2 hours

Distance ya of plane A from collision point as a function of time in hours

ya(t) = 400 -200t

Distance yb of plane B from collision point as a function of time in hours

yb(t) = 300-150t

(a) Distance between two planes,

Since the two planes are on courses perpendicular to each other, will need using pythagorean theorem

D(t) = sqrt(ya(t)^2+yb(t)^2)

= sqrt((400-200t)^2+(300-150t)^2)

= 250(t-2)

D(t) = 250t -500 miles

b. time available

Time until D(t) = 0

solve D(t) = 0

D(t) = 0

250(t-2) = 0

t = 2 (two hours)

3 0

3 years ago