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solong [7]
4 years ago
12

Fo two disk i problem 3 what is the ratio of linear accerlation of a point on the rim of disk a

Mathematics
1 answer:
Nitella [24]4 years ago
7 0
DdffcfggtttjhsdeYWE8757Y2YU5725YW532R
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What is the product of
nydimaria [60]
\frac{10}{ m^{5} }
3 0
3 years ago
Helppp!!!! please!!!
KatRina [158]

Answer:

Option B is the correct answer

Step-by-step explanation:

Center of the circle lie on point (-1, - 3) = (h, k)

\implies \: h =  - 1 \:  \: k =  - 3 \\radius \: r \:  = 2  \\ equation \: of \: circle \: in \: standard \: form \:  \\ is \: given \: as :  \\  {(x - h)}^{2}  +  {(y - k)}^{2}  =  {r}^{2}  \\  {(x  +  1)}^{2}  +  {(y  +  3)}^{2}  =  {2}^{2}  \\  \purple{ \boxed{ \bold{ {(x  +  1)}^{2}  +  {(y  + 3)}^{2}  = 4}}}

8 0
3 years ago
Which of the following shows the extraneous solution(s) to the logarithmic equation? log4 (x) + log4(x - 3) = log4 (-7x + 21)
Salsk061 [2.6K]
First join the log4 on the left:

log4( x*(x-3) = log4(-7x+21)

Then x = -7, works: -7*(-10)=70 = -7*(-7)+21

x=-3, 18 = 42, does not work

x=3 0=0 works,

However, when one puts x = -7 in the *original* exression, log4(-7) or log4(-10) do not exist (you know why?). So x= -7 is extraneous.

Now x=3 gives log4(0) on the left and right, which does not exist.

So, C is the answer, both are extraneous. Seem to work but indeed don't work in the *original* equation
 
6 0
3 years ago
Read 2 more answers
Solve for x <br> A 15 <br> B 16<br> C 2.4<br> D 12
Ne4ueva [31]

Answer: bd=8 dc=2 and ae=4 so value of ab=16

Step-by-step explanation:

7 0
3 years ago
2. Andrew made an error in determining the polynomial equation of smallest degree whose roots are 3, 2+2i and 2-2i.
inna [77]

The polynomial equation for the given roots is x³-7x²+20x-24=0.

Given that, the polynomial equation of the smallest degree whose roots are 3, 2+2i and 2-2i.

Now, we need to write the polynomial equation.

<h3>What is a polynomial equation? </h3>

A polynomial equation is an equation where a polynomial is set equal to zero. i.e., it is an equation formed with variables, non-negative integer exponents, and coefficients together with operations and an equal sign. It has different exponents. The highest one gives the degree of the equation. For an equation to be a polynomial equation, the variable in it should have only non-negative integer exponents. i.e., the exponents of variables should be only non-negative and they should neither be negative nor be fractions.

Now, x=3, x=2+2i and x=2-2i

x-3, x-2-2i and x-2+2i

So, the polynomial equation is (x-3)(x-2-2i)(x-2+2i)=0

⇒(x-3)(x(x-2-2i)-2(x-2-2i)+2i(x-2-2i))=0

⇒(x-3)(x²-2x-2xi-2x+4+4i+2xi-4i-4i²)=0

⇒(x-3)(x²-2x-2x+4-4i²)=0

⇒(x-3)(x²-4x+4+4)=0 (∵i²=-1)

⇒(x-3)(x²-4x+8)=0

⇒x(x²-4x+8)-3(x²-4x+8)=0

⇒x³-4x²+8x-3x²+12x-24=0

⇒x³-7x²+20x-24=0

Therefore, the polynomial equation for the given roots is x³-7x²+20x-24=0.

To learn more about the polynomial equation visit:

brainly.com/question/20630027.

#SPJ1

3 0
2 years ago
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