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aniked [119]
3 years ago
10

I have been trying to figure this out no luck.​

Mathematics
2 answers:
algol [13]3 years ago
8 0

Answer:

Step-by-step explanation:

sergiy2304 [10]3 years ago
5 0

Since a number over the same number (ex: 1/1, 2/2, etc) is always 1, and since the two expressions are completely identical, the answer is 1.

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Can someone help me understand these questions?
Soloha48 [4]
<span>kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other.

a square does fir this definition since it has 4 equal sides, hence 2 pairs and hte equal sides are adjacent to each other.
so yes a square is a kite

the definition of a rectangle is </span>
<span>a figure with four straight sides and four right angles, </span>

a parallelogram with 4 right angle woudl be a rectangle but any without would not

so it can be, but generally is not
7 0
3 years ago
Read 2 more answers
Help Literal Equations
Rom4ik [11]
If you are trying to find x it is x=15-y
8 0
3 years ago
Erik drives 19.2 miles in 16 minutes.
Kipish [7]
Bruh it clearly says 65 mph so that’s the answer
7 0
3 years ago
Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8
Kitty [74]
Substitute z=\ln x, so that

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)

Then the ODE becomes


x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0
\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0

which has the characteristic equation r^2+1=0 with roots at r=\pm i. This means the characteristic solution for y(z) is

y_C(z)=C_1\cos z+C_2\sin z

and in terms of y(x), this is

y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)

From the given initial conditions, we find

y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1
y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8

so the particular solution to the IVP is

y(x)=\cos(\ln x)+8\sin(\ln x)
4 0
3 years ago
Consider the expression (3x+7)(3x+2)
Nastasia [14]

Answer:

a. shown below

b. 9x+14

Step-by-step explanation:

a. (on the bottom)

b. (3x+7)(3x+2)

multiply the terms:

9x+14

thats the area

6 0
3 years ago
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