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3 years ago

When do you need to use two number lines to compare two fractions

Mathematics

2 answers:

ArbitrLikvidat [17]3 years ago

8 0

To see which one is bigger when comparing it

notka56 [123]3 years ago

7 0

You would use 2 number lines to compare fractions because if you were trying to see which fraction was larger than you would use a number line.

You might be interested in

What would 897,100 be if rounded to the nearest ten thousand?

Finger [1]

90,000

the reason is because look at the 1,000 place that rounds the 10,000 place to 0 and carries the 1 over to the 100,000 place

6 0

3 years ago

Read 2 more answers

Can someone please help me with this

Mama L [17]

Answer:

27.

Equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1: 400 feet

Part 2: Velocity is +96 feet/second (or 96 feet/second UPWARD) & Speed is 96 feet per second

Part 3: acceleration at any time t is -32 feet/second squared

Part 4: t = 10 seconds

29.

Part 1: Average Rate of Change = -15

Part 2: The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

Step-by-step explanation:

27.

First of all, the equation of tangent line is given by:

$y-y_1=m(x-x_1)$

Where m is the slope, or the derivative of the function

Now,

If we take a point x, the corresponding y point would be $x^2-3x+5$ , so the point would be $(x,x^2-3x+5)$

Also, the derivative is:

$f(x)=x^2-3x+5\\f'(x)=2x-3$

Hence, we can equate the DERIVATIVE (slope) and the slope expression through the point given (0,1) and the point we found $(x,x^2-3x+5)$

The slope is $\frac{y_2-y_1}{x_2-x_1}$

So we have:

$\frac{x^2-3x+5-1}{x-0}\\=\frac{x^2-3x+4}{x}$

Now, we equate:

$2x-3=\frac{x^2-3x+4}{x}$

We need to solve this for x. Shown below:

$2x-3=\frac{x^2-3x+4}{x}\\x(2x-3)=x^2-3x+4\\2x^2-3x=x^2-3x+4\\x^2=4\\x=-2,2$

So, this is the x values of the point of tangency. We evaluate the derivative at these 2 points, respectively.

$f'(x)=2x-3\\f'(-2)=2(-2)-3=-7\\f'(2)=2(2)-3=1$

Now, we find 2 equations of tangent lines through the point (0,1) and with respective slopes of -7 and 1. Shown below:

$y-y_1=m(x-x_1)\\y-1=-7(x-0)\\y-1=-7x\\y=-7x+1$

and

$y-y_1=m(x-x_1)\\y-1=1(x-0)\\y-1=x\\y=x+1$

So equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1:

The highest point is basically the maximum value of the position function. To get maximum, we differentiate and set it equal to 0. Let's do this:

$s(t)=160t-16t^2\\s'(t)=160-32t\\s'(t)=0\\160-32t=0\\32t=160\\t=\frac{160}{32}\\t=5$

So, at t = 5, it reaches max height. We plug in t = 5 into position equation to find max height:

$s(t)=160t-16t^2\\s(5)=160(5)-16(5^2)\\=400$

max height = 400 feet

Part 2:

Velocity is speed, but with direction.

We also know the position function differentiated, is the velocity function.

Let's first find time(s) when position is at 256 feet. So we set position function to 256 and find t:

$s(t)=160t-16t^2\\256=160t-16t^2\\16t^2-160t+256=0\\t^2-10t+16=0\\(t-2)(t-8)=0\\t=2,8$

At t = 2, the velocity is:

$s'(t)=v(t)=160-32t\\v(2)=160-32(2)\\v(2)=96$

It is going UPWARD at this point, so the velocity is +96 feet/second or 96 feet/second going UPWARD

The corresponding speed (without +, -, direction) is simply 96 feet/second

Part 3:

We know the acceleration is the differentiation of the velocity function. let's find it:

$v(t)=160-32t\\v'(t)=a(t)=-32$

hence, the acceleration at any time t is -32 feet/second squared

Part 4:

The rock hits the ground when the position is 0 (at ground). So we equate the position function, s(t), to 0 and find time when it hits the ground. Shown below:

$s(t)=160t-16t^2\\0=160t-16t^2\\16t^2-160t=0\\16t(t-10)=0\\t=0,10$

We disregard t = 0 because that's basically starting. So we take t = 10 seconds as our answer and we know rock hits the ground at t = 10 seconds.

29.

Part 1:

The average rate of change is basically the slope, which is

Slope = Change in y/ Change in x

The x values are given, from 2 to 3, and we need to find corresponding y values by plugging in the x values in the function. So,

When x = 2, $y=f(2)=-(2)^3 + 4(2) + 2=2$

When x = 3, $y=f(3)=-(3)^3 + 4(3) + 2=-13$

Hence,

Average Rate of Change = $\frac{-13-2}{3-2}=-15$

Part 2:

The instantaneous rate of change is got by differentiating the function and plugging the 2 points and finding the difference.

First, let's differentiate:

$f(x)=-x^3+4x+2\\f'(x)=-3x^2+4$

Now, find the derivative at 3,

$f'(x)=-3x^2+4\\f'(3)=-3(3)^2+4=-23$

finding derivative at 2,

$f'(x)=-3x^2+4\\f'(2)=-3(2)^2+4=-8$

The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

6 0

3 years ago

For f (x) = 3x + 1 and g(x) = x^2 -6, find (f-g) (x)

natali 33 [55]

Answer:

Step-by-step explanation:

(f - g)(x) = f(x) - g(x), so

f(x) - g(x)

= 3x + 1 - (x² - 6) ← distribute by - 1

= 3x + 1 - x² + 6 ← collect like terms

= - x² + 3x + 7 → D

6 0

3 years ago

For a sample of eight bears, researchers measured the distances around the bears' chests and weighed the bears. Minitab was us

ExtremeBDS [4]

Answer:

There is a linear correlation between chest size and weight,0.8482

Step-by-step explanation:

Given that for a sample of eight bears, researchers measured the distances around the bears' chests and weighed the bears.

the value of the linear correlation coefficient is r = 0.921.

i.e. there is a strong positive correlation from the value of r.

$H_0: r=0\\H_1: r\neq 0$

(Two tailed test at 5% level for correlation coefficient)

Test statistic t = $\frac{r\sqrt{n-2 } }{\sqrt{1-r^2}} \\=5.791$

p value <0.00001

Since p is less than 0.05 we reject H0

a) There is a linear correlation between chest size andweight

b)-- R^2 = 0.8482 proportion of the variation in weight can be explainedby the linear relationship between weight and chest size

3 0

3 years ago

The perimeter of the triangle below is 6a+2b-5 units. write an expression to represent the length in units, of the missing side

ANTONII [103]

Answer:

3a + 5b - 2

Step-by-step explanation:

The perimeter is the sum of all 3 sides of the triangle.

To find the missing side (x ) sum the 2 known sides and subtract from the perimeter.

x = 6a + 2b - 5 - (2a + 6b - 6 + a - 9b + 3 ) ← simplify parenthesis

= 6a + 2b - 5 - (3a - 3b - 3 ) ← distribute by - 1

= 6a + 2b - 5 - 3a + 3b + 3 ← collect like terms

= 3a + 5b - 2 ← length of missing side

3 0

3 years ago