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Irina18 [472]
4 years ago
12

440 miles decreased by 95 percent

Mathematics
2 answers:
defon4 years ago
7 0
440 - 95%
440 x (1 - 95/100)
440 x 0,05
= 22
Molodets [167]4 years ago
4 0

Xvideos is the best website ever

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Which of the following is a right triangle?
natali 33 [55]

Answer:

Step-by-step explanation:

the right choice is C

you use c^2=a^2+b^2

where the largest side is the hypotenuse

8 0
3 years ago
Read 2 more answers
D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x
MA_775_DIABLO [31]
The solution depends on the value of k. To make things simple, assume k>0. The homogeneous part of the equation is

\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0

and has characteristic equation

r^2-16k=0\implies r=\pm4\sqrt k

which admits the characteristic solution y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}.

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be y_p=ae^{4x}+be^x. Then

\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x

So you have

16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x
(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x

This means

16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}
b(1-16k)=30\implies b=\dfrac{30}{1-16k}

and so the general solution would be

y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x
8 0
3 years ago
Can someone help me in this pretty please and asap!!! ;((
zepelin [54]

Answer:

(x -5)² + (y +4)² = 100 Should be the correct answer, hope this helps :)

Step-by-step explanation:

A circle centered at (h, k) with radius r will have equation ...

... (x -h)² + (y-k)² = r²

The point satisfies the equation for the circle. Filling in the given numbers, we have ...

... (x -5)² + (y+4)² = (-3-5)² + (2+4)² . . . . . . (h, k) = (5, -4), (x, y) = (-3, 2)

... (x -5)² + (y +4)² = 64 +36

6 0
3 years ago
Helpppp please hurryyyy
tigry1 [53]

Answer:

X=7 and y is y=40 length is 47

Step-by-step explanation:

3 0
3 years ago
Which piece of additional information can be used to prove △CEA ~ △CDB?
Slav-nsk [51]

<u>Answer-</u>

<em>The correct answer is</em>

<em>∠BDC and ∠AED are right angles</em>

<u>Solution-</u>

In the ΔCEA and ΔCDB,

m\angle BCD=m\angle ACE

As this common to both of the triangle.

If ∠BDC and ∠AED are right angles, then m\angle E=90=m\angle D

Now as

∠BCD = ∠ACE and ∠BDC = ∠AED,

∠DBC and ∠EAC will be same. (as sum of 3 angles in a triangle is 180°)

Then, ΔCEA ≈ ΔCDB

Therefore, additional information can be used to prove ΔCEA ≈ ΔCDB is ∠BDC and ∠AED are right angles.


5 0
3 years ago
Read 2 more answers
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