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2 years ago

Urgent, pls help me!!!

Mathematics

1 answer:

BlackZzzverrR [31]2 years ago

8 0

Answer:

Radius: 2

Center: (-3, 1)

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Find the midpoint of the line segment joining the points (7,1) and (-1,-1)

chubhunter [2.5K]

= x1 + x2/2 , y1 + y2/2
= 7+-1/2 , 1+(-1)/2
Midpoint = (3,0)

6 0

2 years ago

PLEASE HURRY

s2008m [1.1K]

Answer:

1.25p

Step-by-step explanation:

p = amount raised last year

amount raised this year = p +(25% of p)

=1 p+0.25p

=p(1+0.25)

=1.25p

8 0

3 years ago

Jim will rent a car for the weekend. He can choose one of two plans. The first plan has an initial fee of $ 59 and costs an addi

AveGali [126]

Answer:This problem is about linear equations. We assume Dale drive X miles, and the total cost is $Y, then we can get:

Plan I: Y=38+0.11X

Plan II: Y=49+0.07X

When both plans cost the same, 38+0.11X=49+0.07X. We will get X = 275miles, and Y=$68.25

Step-by-step explanation:

6 0

2 years ago

5. Find the general solution to y'''-y''+4y'-4y = 0

CaHeK987 [17]

For any equation,

$a_ny^(n)+\dots+a_1y'+a_0y=0$

assume solution of a form, $e^{yt}$

Which leads to,

$(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0$

Simplify to,

$e^{yt}(y^3-y^2+4y-4)=0$

Then find solutions,

$\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}$

For non repeated real root y, we have a form of,

$y_1=c_1e^t$

Following up,

For two non repeated complex roots $y_2\neq y_3$ where,

$y_2=a+bi$

and,

$y_3=a-bi$

the general solution has a form of,

$y=e^{at}(c_2\cos(bt)+c_3\sin(bt))$

Or in this case,

$y=e^0(c_2\cos(2t)+c_3\sin(2t))$

Now we just refine and get,

$\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}$

Hope this helps.

r3t40

5 0

3 years ago

Suppose it takes 4 hours for a certain strain of bacteria to reproduce by dividing in half. If 70 bacteria are present to begin

Sauron [17]

Answer:

1 day: 4480

2 days: 8960

3 days: 13440

$f^{-1}(x) = \sqrt{\frac{x -4}{4}}$ ---- function inverse

Step-by-step explanation:

Given

$f(x) = 70 \cdot 64x$

Solving (a): The amount present after 1 day.

Here, $x =1$

So:

$f(1) = 70 \cdot 64*1= 4480$

Solving (b): The amount present after 2 days.

Here, $x =2$

So:

$f(2) = 70 \cdot 64*2= 8960$

Solving (c): The amount present after 3 days.

Here, $x = 3$

So:

$f(3) = 70 \cdot 64*2= 13440$

Solving (d): The inverse function of:

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

Replace f(x) with y

$y= 4x^2 + 4$

Swap x and y

$x= 4y^2 + 4$

Rewrite as:

$4y^2 = x -4$

Divide by 4

$y^2 = \frac{x -4}{4}$

Take square roots

$y = \sqrt{\frac{x -4}{4}}$

Replace y with function inverse

$f^{-1}(x) = \sqrt{\frac{x -4}{4}}$

6 0

2 years ago

Urgent, pls help me!!!​

Urgent, pls help me!!!