I need help with this math homework

Suppose 12 quarts of pure antifreeze is mixed with 8 quarts of a solution that is 72% antifreeze. Answer the questions below. Do

ra1l [238]

(a)

12+0.72(8)

12+5.76

17.76 qts of antifreeze

(b)

100(12+8(.72))/(12+8)

100(12+5.76)/20

100(17.76)/20

5(17.76)

88.8%

4 0

3 years ago

how do the graphs of f(x) = sin x and g(x) = sin2x+3 compare. Must make two selections of the following that apply.

Elena L [17]

Answer:

The correct options are:

g(x) is shifted three units higher than f(x).

g(x) has a period that is half the period of f(x).

Step-by-step explanation:

We have to compare the graphs of the function:

$f(x)=\sin x$

and $g(x)=\sin 2x+3$

We have to select the correct options among the following:

As we know that the period of sine function is 2π.

i.e. Period of function f(x) is: 2π.

The period of sin(2 x) is π.

Hence, the period of the function g(x) function is π.

Hence, the period of g(x) is half the period of f(x).

Also we could observe that g(x) is shifted 3 units upward.

3 0

3 years ago

Read 2 more answers

Two numbers less than 8 that have 3 as their GCF

Anarel [89]

Answer:

3 and 6

Step-by-step explanation:

GCF's or greatest common factors of numbers 1 to 8

1 : 1

2 : 1, 2

3: 1 , 3

4 : 1, 2, 4

5 : 1, 5

6 : 1, 2, 3, 6

7 : 1, 7

8 : 1, 2, 4, 8

3 and 6 have 3 as their GCF

4 0

2 years ago

According to the Holy Bible, Jesus begins to perform healings. There were many healers during Jesus’ time. What made Jesus’ heal

Gekata [30.6K]

Healing was essential to the ministry of Jesus because He envisioned healing as a physical symbol of forgiveness. He guaranteed the ultimate glory of the human body through His personal resurrection, but forecast that restoration by healing twisted, shrunken, blinded limbs and organs. The paralytic's restoration is but one of many such examples (Mark 2:1-12).

Though there were many healers, Jesus was able to even raise up the dead, and also there was one instance where the Canaanite woman struggled through His disciples' desire to dismiss her, and His own initial, courteous refusal, to get what she knew she could trust Him to grant (Matthew 16:28). The crowds "begged him to let the sick just touch the edge of his cloak," (Matthew 14:36), for "all who touched him were healed."

Healers were not able to do that, and Jesus also claimed that he was God's son, so performing miracles like this was like proving the point. Also, healers were not always able to heal the person, but Jesus was able to do so all the time (for free even!) so many people traveled to meet him so he could heal them.

The woman with a hemorrhage crept silently through the crowd to merely touch His clothes (Mark 5:28). She also claimed that she went to many healers, but she didn't get healed, in fact, she got worse! So that's an instance that proves the point.

Thank you :D

3 0

3 years ago

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:

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