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slavikrds [6]
3 years ago
12

LOOK AT PICTURE!! PLEASSEEEE HELP! (the one about the quadrilateral)

Mathematics
2 answers:
liraira [26]3 years ago
6 0
The answer too this question is D
pashok25 [27]3 years ago
5 0

Answer:

D

Step-by-step explanation:

They say it's a Parallelogram, meaning that you would make a square to make it fit in.

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Which graph represents an exponential growth function?
mezya [45]

Answer:

The fourth graph shows and exponential growth function.

Step-by-step explanation:

The last graph shows the value increasing by a large amount.

6 0
3 years ago
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First-order linear differential equations
kkurt [141]

Answer:

(1)\ logy\ =\ -sint\ +\ c

(2)\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c

Step-by-step explanation:

1. Given differential equation is

  \dfrac{dy}{dt}+ycost = 0

=>\ \dfrac{dy}{dt}\ =\ -ycost

=>\ \dfrac{dy}{y}\ =\ -cost dt

On integrating both sides, we will have

  \int{\dfrac{dy}{y}}\ =\ \int{-cost\ dt}

=>\ logy\ =\ -sint\ +\ c

Hence, the solution of given differential equation can be given by

logy\ =\ -sint\ +\ c.

2. Given differential equation,

    \dfrac{dy}{dt}\ -\ 2ty\ =\ t

=>\ \dfrac{dy}{dt}\ =\ t\ +\ 2ty

=>\ \dfrac{dy}{dt}\ =\ 2t(y+\dfrac{1}{2})

=>\ \dfrac{dy}{(y+\dfrac{1}{2})}\ =\ 2t dt

On integrating both sides, we will have

   \int{\dfrac{dy}{(y+\dfrac{1}{2})}}\ =\ \int{2t dt}

=>\ log(y+\dfrac{1}{2})\ =\ 2.\dfrac{t^2}{2}\ + c

=>\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c

Hence, the solution of given differential equation is

log(y+\dfrac{1}{2})\ =\ t^2\ +\ c

8 0
4 years ago
Find the quotient. Write the answer in the simplest form (reduce).
Anit [1.1K]
6 x 1 = 6 /2 =3. For 6*1/2

3/5 x 1/4 = 3*1/5*4 = 3/20
5 0
3 years ago
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What percent of the area underneath<br> this normal curve is shaded?<br> the answer is not 36%
klemol [59]

Answer:

95%

Step-by-step explanation:

8 0
3 years ago
Help answer these two pretty please
4vir4ik [10]

Answer:

none of these triangles are similar

Step-by-step explanation:

for a triangle to be similar, they have to have the same angle measures

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3 years ago
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