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shtirl [24]
3 years ago
8

Solve each system by elimination 5x + y - 2z = -12

Mathematics
1 answer:
kumpel [21]3 years ago
6 0
X = -12 - y + 2z / 5

I hope this helps. :)
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3. The circumference of a circle is 6 pie yards.
LiRa [457]

Answer:

113.1 yards

Step-by-step explanation:

...........

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Max constructed a scale model of the Eiffel tower. The base area of the model is 0.25 square meters, while the base area of the
Vanyuwa [196]

Answer:

1:250

Step-by-step explanation:

\sqrt{15625/0.25}

=\sqrt{62500}

=250

8 0
3 years ago
tom has a large photo he wants to shrink to wallet size its width is 20 centimeters and its length is 30 centimeters if he wants
Dafna11 [192]
The length should be 7.5 centimeters
7 0
3 years ago
Newton’s law of cooling states that dx/dt= −k(x − A) where x is the temperature, t is time, A is the ambient temperature, and k
Vadim26 [7]

Answer:

(a) The solution to the differential equation is x = A_0Coswt + Ce^(-kx)

(b) The initial condition t > 0 will not make much of a difference.

Step-by-step explanation:

Given the differential equation

dx/dt= −k(x − A); t > 0, A = A_0Coswt

(a) To solve the differential equation, first separate the variables.

dx/(x - A) = -kdt

Integrating both sides, we have

ln(x - A) = -kt + c

x - A = Ce^(-kt) (Where C = ce^(-kt))

x = A + Ce^(-kx)

Now, we put A = A_0Coswt

x = A_0Coswt + Ce^(-kx) (Where C is constant.)

And we have the solution.

(b) Since temperature t ≠ 0, the initial condition t > 0 will not make much of a difference because, Cos(wt) = Cos(-wt).

It is not any different from when t < 0.

7 0
2 years ago
HELPP!!what is the answer to this picture above^
nasty-shy [4]

Answer:

\fbox{ \fbox { \sf{Distance  =  \sf{15 \: units}}}}

{ \fbox { \fbox { \sf{Midpoint = { \sf{ \: (12 \: , \: 10.5)}}}}}}

Step-by-step explanation:

\star{ \:  \sf { \: Let \: the \: points \: be \: A \: and \: B}}

\star { \sf{Let \: A(6 \:, 6) \: be \: (x1 ,\: y1) \: and \: B(18 ,\: 15) \: be \: (x2 \:, y2)}}

\underline{ \underline{ \tt{Finding \: the \: distance}}}

\boxed{ \sf{Distance =  \sqrt{ {(x2 - x1)}^{2}  +  {(y2 - y1)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{ {(18 - 6)}^{2}  +  {(15 - 6)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{ {(12)}^{2}  +  {(9)}^{2} } }}

\mapsto{ \sf{Distance =  \sqrt{144 + 81}}}

\mapsto{ \sf{Distance =  \sqrt{225} }}

\mapsto{ \sf{Distance =  \sqrt{ {15}^{2} } }}

\mapsto{  \boxed{\sf{Distance = 15 \: units}}}

\underline{ \underline {\tt{Finding \: the \: Midpoint}}}

\boxed{ \sf{Midpoint = ( \frac{x1 + x2}{2}  \: , \:  \frac{y1 + y2}{2} )}}

\mapsto{ \sf{Midpoint = ( \frac{6 + 18}{2}  \: , \:  \frac{6 + 15}{2} )}}

\mapsto{ \sf{Midpoint = ( \frac{24}{2}  \: , \:  \frac{21}{2} )}}

\mapsto{ \boxed{ \sf{Midpoint = (12 \: , \: 10.5)}}}

Hope I helped!

Best regards! :D

~\sf{TheAnimeGirl}

3 0
2 years ago
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