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3 years ago

Plz help!!!!!!!!!!!!!

Mathematics

2 answers:

Anvisha [2.4K]3 years ago

8 0

Answer: $\bold{c)\quad x=\dfrac{6\pm \sqrt{(-6)^2-4(3)(8)}}{2(3)}}$

<u>Step-by-step explanation:</u>

$3x^2-6x+8=0\quad \rightarrow \quad a=3,\ b=-6,\ c=8\\\\\text{Quadratic formula is: }x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(-6)\pm \sqrt{(-6)^2-4(3)(8)}}{2(3)}}\\\\\\x=\dfrac{6\pm \sqrt{(-6)^2-4(3)(8)}}{2(3)}$

pantera1 [17]3 years ago

5 0

Answer:

the third one

Step-by-step explanation:

-b to start is -(-6) which is 6

needs the plus and minus

the first expression is the whole numerator

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2 years ago

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6 0

3 years ago

If it takes 6 hours for a plane to travel 720km with a tail wind and 8 hours to make the return trip with a head wind. Find the

denis-greek [22]

The speed of wind and plane are 105 kmph and 15 kmph respectively.

<u>Solution:</u>

Given, it takes 6 hours for a plane to travel 720 km with a tail wind and 8 hours to make the return trip with a head wind.

We have to find the air speed of the plane and speed of the wind.

Now, let the speed wind be "a" and speed of aeroplane be "b"

And, we know that, distance = speed x time.

$\text { Now, at tail wind } \rightarrow 720=(a+b) \times 6 \rightarrow a+b=\frac{720}{6} \rightarrow a+b=120 \rightarrow(1)$

Now at head wind → $720=(a-b) \times 8 \rightarrow a-b=\frac{720}{8} \rightarrow a-b=90 \rightarrow(2)$

So, solve (1) and (2) by addition

2a = 210

a = 105

substitute a value in (1) ⇒ 105 + b = 120

⇒ b = 120 – 105 ⇒ b = 15.

Here, relative speed of plane during tail wind is 120 kmph and during head wind is 90 kmph.

Hence, speed of wind and plane are 105 kmph and 15 kmph respectively.

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3 years ago

A square field has an area of 479 ft what is the approximate length of a size of the field

MissTica

Answer:

21.9 ft

Step-by-step explanation:

The area of a square is the square of the side length, so the side length is the square root of the area:

... s = √(479 ft²) ≈ 21.886 ft

This is approximately 21.9 ft.

4 0

3 years ago