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3 years ago

3. WEDNESDAY (5/20): The minimum of the graph of a quadratic function is located at (-1,2). The point (2, 2015

Mathematics

1 answer:

Vadim26 [7]3 years ago

8 0

Answer:

C.) f(x) = 2x^2 + 4x + 4

Step-by-step explanation:

The equation of the parabola with vertex (h,k) is y=a(−h+x)^2+k.

Thus, the equation of the parabola is y=a(x+1)^2+2.

To find a, use the fact that the parabola passes through the point (2,20): 20=9a+2.

Solving this equation, we get that a=2.

Thus, the equation of the parabola is y=2(x+1)^2+2.

TO STANDARD FORM

= 2*(x^2+2x+1)+2

=(2x^2+4x+2)+2

= 2x^2+4x+2+2

= 2x^2+4x+4

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Y=4x^2-1 find the inverse. I cant it’s to difficult.

choli [55]

The answer is x equals the square root of y+1/4

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3 years ago

A force of 60 N is used to stretch two springs that are initially the same length. Spring A has a spring constant of 4 N/m, and

goblinko [34]

Answer:

D:Spring A is 3 m longer than spring B because 15 – 12 = 3.

Step-by-step explanation:

In this question, you should remember the Hooke's Law in physics.

The Hooke's Law simply explains that the extension that occurs on a spring is directly proportional to the load applied on it.

The mathematical expression for this law is

$F=-kx$

where;

F= force applied on the spring

x = the extension on the spring

k= the spring constant which varies in spring.

The question will need you to calculate the extension on the springs A and B then compare the values obtained.

In spring A

Force, F=60N and spring constant ,k=4 N/m

To find the extension x apply the expression;

$F=-kx\\\\60=-4*x\\\\60=-4x\\\\\frac{60}{-4} =\frac{-4x}{-4} \\\\\\-15=x$

Here the spring extension is 15 m

In spring B

Force, F=60N and spring constant , k=5N/m

To find the extension x apply the same expression

$F=-kx\\\\60=-5*x\\\\60=-5x\\\\\\\frac{60}{-5} =\frac{-5x}{-5} \\\\\\-12=x$

Here the extension on the spring is 12 m

Compare

The extension on spring A is 3 m longer than that in spring B because when you subtract the value of spring B from that in spring A you get 3m

$=15m-12m=3m$

5 0

3 years ago

Find the function of which is the solution of 36y"-48y'-48y=0 with initial conditions y1(0)=1. y1'(0)=0

lilavasa [31]

Answer:

$y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }$

Step-by-step explanation:

Let, $D=\frac{d}{dx}$

Now, us simplify the given differential equation and write it in terms of D,

$36y''-48y'-48y=0$

or, $3y''-4y'-4y=0$

or, $(3D^2-4D-4)y=0$

We have our auxiliary equation:

$3D^2-4D-4=0$

or, $(3D+2)(D-2)=0$

or, $D=2, - \frac{2}{3}$

Therefore our solution is,

$y=Ae^{2x}+Be^{- \frac{2}{3}x}$

and, $y'=2Ae^{2x}-\frac{2}{3}Be^{\frac{2}{3}x }$

Applying the boundary conditions, we get,

$A+B=1$

$2A-\frac{2}{3}B=0$

Solving them gives us,

$A=\frac{1}{4} ,B=\frac{3}{4}$

Hence,

$y=\frac{1}{4} e^{2x}+\frac{3}{4} e^{-\frac{2}{3}x }$

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3 years ago

What is value of the digit 5 in the number 75

ycow [4]

First, think of your places. You have the ones places, tens places, hundreds places, and so on.

The first number starting from the right is the ones, and as you keep going left, the value of each given digit becomes higher.

Since 5 is in the ones place, its value would be just 5. If it were in the tens place, it would be 50. If it were in the hundreds place, it would be 500, and so on.

Think of it this way;

Ones is just one. If a number is in the 'ones' place, its value would be a single digit. If it were in the tens place, its value would be two digits.
That's how it would be for each place going left.
Every number you move to the left, its value gains a one.

So here's an example:
5555

The value of 5 in the ones place "5555" is simply 5.

In the tens place, you end up adding one zero, so the value of the second five to the left would be, "50"

So with that said, the value of the digit 5 in the number 75 is 5.

Haha, hope this cleared up any confusion, and have a wonderful day! :)

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ruslelena [56]

Answer:14+14=28

Step-by-step explanation:

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