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Yuliya22 [10]
3 years ago
9

Whats the is the value

Mathematics
1 answer:
telo118 [61]3 years ago
5 0
Can't see the graph can you make it more clear?
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PLS HELP IM TIMED
shutvik [7]

Answer:

0.56

Step-by-step explanation:

0.56 Because you would have yo subtract 0.77 and 0.21

8 0
2 years ago
"Write two equations: one parallel, one perpendicular to (5,-2) y= 1/5x-3
guajiro [1.7K]

Answer:

For parallel:

gradient is 1/5

y = mx + c

consider (5, -2):

- 2 = ( \frac{1}{5}  \times 5) + c \\  - 2 = 1 + c \\ c =  - 3

{ \boxed{ \bf{equation : y =  \frac{1}{5}x - 3 }}}

For perpendicular:

gradient, m1:

m _{1} \times m_{2} =  - 1 \\ m _{1} \times  \frac{1}{5}  =  - 1 \\  \\ m _{1} =  - 5

gradient = -5

y = mx + c \\  - 2 = (5 \times -  5) + c \\  - 2 =  - 25 + c \\ c =  23

{ \boxed{ \bf{equation :y =  - 5x + 23 }}}

7 0
3 years ago
Read 2 more answers
A.10<br> B.70<br> C.110<br> D.170<br><br> Help pls asap
nirvana33 [79]
Vertical angles are congruent
2x+50=7x
50=5x
x=10

2(10)+50
20+50
Angle AEB = 70
5 0
2 years ago
Simplest form for 10 sec 1 min
Mademuasel [1]

Answer:

70 seconds

I think that's what you are looking for?

3 0
3 years ago
An object is launched from a platform.
tankabanditka [31]

Answer:

  10

Step-by-step explanation:

Ground level is where h = 0, so solve the equation ...

  h(x) = 0

  -5(x -4)^2 +180 = 0 . . . . substitute for h(x)

  (x -4)^2 = 36 . . . . . . . . . . divide by -5, add 36

  x -4 = 6 . . . . . . . . . . . . . . positive square root*

  x = 10 . . . . . . add 4

The object will hit the ground 10 seconds after launch.

_____

* The negative square root also gives an answer that satisfies the equation, but is not in the practical domain. That answer would be x = -2. The equation is only useful for time at and after the launch time: x ≥ 0.

8 0
2 years ago
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