Can someone help with these 3 questions, they are in the pictures!

Mathematics

1 answer:

Eduardwww [97]3 years ago

4 0

Question One
y = v - 1/2 g t^2 Subtract v from both sides.
y - v = - 1/2 gt^2 Multiply both sides by - 1
-(y - v) = 1/2gt^2 Remove the brackets.
v - y = 1/2 gt^2 Divide by g
(v - y)/g = 1/2 t^2 Multiply by 2 on both sides
2(v - y)/g = t^2 Move the 2 to the denominator on the left.
(v - y) / 0.5g = t^2 Take the square root of both sides.
sqrt [ (v - y) / 0.5g ] = t

Answer: B or second one down <<<<<<<

Question Two
There are no minuses in the answer. That eliminates 2 and 3
km^2 - c = n Add c to both sides.
km^2 = c + n divide both sides by k
m^2 = [c + n]/k Take the square root of m^2
m = sqrt [(c + n)/k]

Answer A or the first one.

Problem 3
This is an estimation question. One value of x is <0 the other > 0
The two graphs cross when x = a bit over 2 and x < about - 5
The answer is C 2.1 and - 5.1 about.

Answer: 2.1 and -5.1
Answer: C <<<<<<<<<

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2 years ago

Solve for x<br> (1/x) -2/3 = (4x)

Vadim26 [7]

The value of x is $x=\frac{-1+\sqrt{37}}{12}$ and $x=\frac{-1-\sqrt{37}}{12}$

Step-by-step explanation:

The equation is $\frac{1}{x}-\frac{2}{3}=4 x$

Subtracting by $4x$ on both sides,

$\frac{1}{x}-\frac{2}{3}-4 x=0$

Taking LCM,

$\frac{3-12 x^{2}-2 x}{3 x}=0$

Multiplying by 3x on both sides,

$-12 x^{2}-2 x+3=0$

Dividing by (-) on both sides,

$12 x^{2}+2 x-3=0$

Using quadratic formula, we can solve for x.

$\begin{aligned}x &=\frac{-2 \pm \sqrt{2^{2}-4 \cdot 12 \cdot(-3)}}{2 \cdot 12} \\&=\frac{-2 \pm \sqrt{4+144}}{2 \cdot 144} \\&=\frac{-2 \pm \sqrt{148}}{24} \\&=\frac{-2 \pm 2 \sqrt{37}}{24}\end{aligned}$