PLEASE HELP

1 answer:

5 0

$(x,\ y)\\\text{Put the values of x and y to the inequality}\ 3y-4x\geq-6\\\\A.\ (-3,\ 8)\\3(8)-4(-3)\geq-6\\24+12\geq-6\\36\geq-6\qquad TRUE\\\\B.\ (5,\ 10)\\3(10)-4(5)\geq-6\\30-20\geq-6\\10\geq-6\qquad TRUE\\\\C.\ (1,\ 2)\\(3)(2)-(4)(1)\geq-6\\6-4\geq-6\\2\geq-6\qquad\ TRUE\\\\D.\ (-5,\ 2)\\(3)(2)-(4)(-5)\geq-6\\6+20\geq-6\\26\geq-6\qquad TRUE\\\\Answer:\ a,\ b,\ c\ and\ d\ are\ solutions\ of\ this\ inequality.$

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Kindly help me!!!!!!

Oxana [17]

Answer:

Step-by-step explanation:

AS we can see the lines are parallel so

2 ( 4x - 3) + 7(x + 3) = 180° ( being so - interior angles)

8x - 6° + 7x + 21° = 180°

15x + 15° = 180°

15x = 180° - 15°

15x = 165°

x = 165° / 15

Therefore x = 11°

Now

2 ( 4x - 3) = 2 ( 4 * 11° - 3°) = 2 ( 44 - 3)° = 2* 41 = 82°

7(x + 3 ) = 7 ( 11° + 3°) = 7 * 14 = 98°

4 0

2 years ago

What's the x-intercept for number 26?

Lunna [17]

It gives function: h = -16t^2 + 40
water is when height is 0 so solve for 0= -16t^2 + 40
subtract 40 from both sides: -40 = -16t^2
divide by -16: -40/-16 = t^2
simplify: 5/2 = t^2
square root both sides: 1.58 = t

the orange hits the river in 1.58 seconds.

6 0

2 years ago

Solve for x. Round to the nearest tenth.

kondor19780726 [428]

Answer:

22.8

Step-by-step explanation:

$a^{2} +b^{2} =c^{2} \\20^{2} +11^{2} =x^{2} \\400+121=x^{2}\\\sqrt{521} =x\\\\or\\\\22.8 = x$