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iragen [17]
3 years ago
6

PLEASE HELP!

Mathematics
1 answer:
Vladimir79 [104]3 years ago
3 0

Answer:

1) 4

2)I think 2

3)4

4)3

5)2

6)1

7)3

Step-by-step explanation:

1) since the x is to squared it is a quadratic equation

3) A and D have same slope=0.5

others were solved using Excel

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Pls help pls pls pls
Sindrei [870]
\bf \begin{array}{llccll}
y=&1x^2&-6x&+9\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad discriminant\implies b^2-4ac

so... what does that give you for the discriminant?

if it's a negative value, no solutions
if it's a positive value, two solutions
if it's 0, one solution
8 0
3 years ago
MATHS | Cousin needs help❤️ question bellow
Licemer1 [7]
Equation: y = 6 - 2x

Fill in the x numbers in the x which is next to -2 —> y = 6 - 2(0), y = 6 -2(1), y = 6 -2(3), and etc.

A) 0 —> 6
1 —> 4
2 —> 2
3 —> 0
4 —> -2
5 —> -4

B) Put a dot on y axis on number 6, then go down by 2 because your subtracting 2

C) 6 - 2x = 3
-6 -6
-2x = -3
-2/-2 -3/-2

X = 1.5
3 0
3 years ago
For A Free 50 Points... Tell me how to give an answer the "Brainliest Answer" Thingy... Also the person who answers with how to
KatRina [158]

Answer:

ok so when a person answers a question and then theres two people who answered it a little crown will appear at the top and you can decide if you want to mark them brainliest for the answer, Brainliest means when you think or decide that it is the best answer theoretically.

Step-by-step explanation:

               

                                :c

3 0
3 years ago
Read 2 more answers
A car is on a driveway that is inclined 10 degrees to the horizontal. A force of 490 lb is required to keep the car from rolling
lara31 [8.8K]

Answer:

a) weight  of the car = 2816,1 lbs

b) 2773 lbs

Step-by-step explanation:

The equilibrium force is 490 lbs. That force keep the car at rest, then

∑ Fy  =  0     and  ∑Fx  =  0

Forces acting on the car:

The external force   490 lbs

weight of the car   uknown

Normal force

sin∠10°  =  0,174

cos∠10° = 0.985

∑Fx  =  0         mg*sin10°- 490 = 0      ∑Fy =  0      mg*cos10° - N  =  0

mg*0,174= 490

mg  =  490 / 0,174

mg = 2816,1 lbs

weight  of the car = 2816,1 lbs

The Normal force

mg*cos10° - N  =  0        2816,1 * 0,985 = N

N = 2773 lbs

Then equal force in magnitude and in opposite direction will car exets on the driveway

3 0
3 years ago
Which of the equations should anthony pick ?
Komok [63]

Answer:

the third answer

Step-by-step explanation:

hope it helps

7 0
2 years ago
Read 2 more answers
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