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2 years ago

PLEQSE ANSWER ASAP!!!!!!!!!!!!!!!!

Mathematics

1 answer:

alukav5142 [94]2 years ago

4 0

Answer: (a) identity

Step-by-step explanation:

Anything multiplied by 1 will be considered an equation that is identity property.

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3 100° B. 48 270 o ZA= degrees b= (=

solong [7]

Answer:

Angle A = 53

Step-by-step explanation:

All triangles add up to 180 so...

Lets use x to describe angle A

100+27+x=180

127+x=180

x=53

I'm confused what its asking about b and c but hope this helps a bit :)

6 0

3 years ago

Please help!! ill give BRAINLIEST!!!

kenny6666 [7]

1. Isosceles, right.

2. Scalene, right.

3. Equilateral, acute.

4. Scalene, acute.

5. Isosceles, acute.

6. Scalene, obtuse.

7. Scalene, obtuse.

8. Scalene, obtuse.

9. Equilateral, acute.

10. Isosceles, obtuse.

3 0

3 years ago

Read 2 more answers

Oh please i need help lol

drek231 [11]

Answer:

Step-by-step explanation:

Download pdf

6 0

3 years ago

PLEASE HELP!! this is due soon and i rlly need help

egoroff_w [7]

Answer:

Solution given:

r=3km

area of sector bounded by a 90° arc=90°/360°*πr²

=¼*π*3²

=9/4 π or 7.068km²

is a required answer

6 0

3 years ago

Read 2 more answers

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

6 0

3 years ago