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Nady [450]
3 years ago
9

This assignment is already late cause I’m terrible with time management but please help

Mathematics
2 answers:
madreJ [45]3 years ago
7 0

45°

B

the sum of the angles of triangles is 180°

60°+75°+x=180°

135°+x=180°

135°-135°+x=180°-135°

x°=45

B

make me brainiest

k0ka [10]3 years ago
3 0

To solve this question, we have to know some rules

-All tringles' angles must add up to 180 digress

-If the triangles angles do not add up to 180 then the triangles are not a triangle (Just a little reminder if you get asked on it)

Now the steps are easy.

Since you know that all triangles' angles must add up to 180 the formula to find one missing angle is

(Angle number one) + (Angles number two) - 180

If we apply it to the triangle here, it would be

180-(60+75)

Let's solve it slowly

P.E.M.D.A.S

60+75 is 135

180-135=45

Therefore, the answer is 45 or B

I hope this helped if it's wrong pls blame me

ps.Brianlist would really help

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Place the indicated product in the proper location on the grid. 3a2 · 2a3 next question ask for help turn it in
marishachu [46]
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4 years ago
Find a formula for Y(t) with Y(0)=1 and draw its graph. What is Y\infty?
tino4ka555 [31]

Answer:

(a)\ y(t)\ =\ -2e^{-2t}+3

(b)\ y(t)\ =\ 4e^{-2t}-3

Step-by-step explanation:

(a) Given differential equation is

   Y'+2Y=6

=>(D+2)y = 6

To find the complementary function, we will write

D+2=0

=> D = -2

So, the complementary function can be given by

y_c(t)\ =\ C.e^{-2t}

To find the particular integral, we will write

y_p(t)\ =\ \dfrac{6}{D+2}

          =\ \dfrac{6.e^{0.t}}{D+2}

           =\ \dfrac{6}{0+2}

           = 3

so, the total solution can be given by

y_(t)\ =\ C.F+P.I

         =\ C.e^{-2t}\ +\ 3

y_(0)=C.e^{-2.0}\ +\ 3

but according to question

1 = C +3

=> C = -2

So, the complete solution can be given by

y_(t)\ =\ -2.e^{-2.t}\ +\ 3

(b) Given differential equation is

   Y'+2Y=-6

=>(D+2)y = -6

To find the complementary function, we will write

D+2=0

=> D = -2

So, the complementary function can be given by

y_c(t)\ =\ C.e^{-2t}

To find the particular integral, we will write

y_p(t)\ =\ \dfrac{-6}{D+2}

           =\ \dfrac{-6.e^{0.t}}{D+2}

           =\ \dfrac{-6}{0+2}

           = -3

so, the total solution can be given by

y_(t)\ =\ C.F+P.I

         =\ C.e^{-2t}\ -\ 3

y_(0)\ =C.e^{-2.0}\ -\ 3

but according to question

1 = C -3

=> C = 4

So, the complete solution can be given by

y_(t)\ =\ 4.e^{-2.t}\ -3

4 0
3 years ago
Simplified form of -3a5-(-2a5)
Andreas93 [3]
6a^2x-10 beccause u have to multiply

6 0
3 years ago
Read 2 more answers
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