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FinnZ [79.3K]
3 years ago
10

What type of triangle is shown in the figure?

Mathematics
2 answers:
Katen [24]3 years ago
8 0
The answer is D. Isosceles because the shape is Triangle Isosceles.
HOPE THIS HELP!
Juliette [100K]3 years ago
4 0
Its an isosceles triangle because there are two sides that are the same length
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(-3+7i)(1-2i) can you help me with the answer to this
lozanna [386]

Answer:

11 + 13i

Step-by-step explanation:

A must know is that

\sqrt{i}  =  - 1

Complex numbers

1st: foil the equations

- 3(1 - 2i) + 7i(1 - 2i)

- 3 + 6i + 7i- 14 {i}^{2}

2nd Combine like terms and substitute

i ^{2} for \:  - 1

- 3 + 14  + 13i

11  +  13i

4 0
1 year ago
Please help!!! 20 POINTS!!
Sedbober [7]

Answer:

it would be 9 by 6

Step-by-steit would be 9 by 6p explanation:

it would be 9 by 6

8 0
2 years ago
A rocket is launched straight up from the ground with an initial velocity of 192 feet per second. The equation for the height of
Ivenika [448]

The equation for the height of the rocket at time t given

h= -16t^2+192t

We have to find the time t, when the rocket reaches 560 feet.

That means we have to find t when h = 560 ft. we will place 560 in the place of h to find t now.

h= -16t^2+192t

560 = -16t^2+192t

In the right side, we can check -16 is the common factor. So we will take out -16 from the rigbht side.

560 = -16(t^2 - 12t)

To get rid of -16 from the right side and move it to left side, we will divide both sides by -16.

560/-16 = -16(t^2-12t)/-16

-35 = t^2 -12t

Now we will move -35 to the righ side by adding 35 to both sides.

-35+35 = t^2-12t+35

0 = t^2 -12t+35

t^2-12t+35 = 0

We will factorize thee left side to find the values of t now. We need to find a pair of factors of 35 that by adding them we will get -12.

The pair of factors of 35 are -5 and -7 and by adding -5-7 we will get -12.

t^2-12t+35 =0

(t-5)(t-7) =0

So by using zero product property we will get

t-5 =0

t-5+5 = 0+5

t=5

Also t-7 =0

t-7+7 = 0+7

t=7

So we have got the rocket reaches at 560ft when t = 5 seconds and also when t = 7 seconds.

Now part b.

When the rocket completes its trajectory and hits the ground then the height or h = 0. So we will place h = 0 there in the equation.

h= -16t^2+192t

0= -16t^2 + 192 t

0 = -16(t^2-12t)

-16(t^2-12t) = 0

We will move -16 to the other side by dividing it to both sides.

-16(t^2-12t)/-16 = 0/-16

t^2-12t = 0

We will take out the common factor t from the left side. By taking out t we will get,

t(t-12) = 0

We will use zero product property now. By using that we will get,

t = 0

ans also t-12 = 0

t-12+12 = 0+12

t = 12

When the rocket completes its trajectory and hits the ground the time t can not be 0. When t =0, the rocket starts the trajectory.

So when the rocket completes its trajectory and hits the ground ,

then t = 12seconds.

So we have got the required answers.

6 0
2 years ago
Calculate the value of P and of Q that satisfy the following simultaneous linear equations.
luda_lava [24]

  1. 0.5p -3q =11
  2. 5p +6q =2

1) ×-2

--> 3. -p +6q =-22

2) - 3)

6p =24

p =4

sub into 2)

5(4) +6q =2

20 +6q =2

6q =2 -20 =-18

q=-3

5 0
3 years ago
Someone please help...​
Rom4ik [11]

Answer:

y=mc

Step-by-step explanation:

4 0
2 years ago
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