A straight line has equation 3y-3x=4. Write down the equation of another straight line parallel to it

Can someone please explain why x= -2! ASAP!!

bearhunter [10]

Because you add both negative x’s which is -2 and add the -6 in the right side so then it’s -2x=4
Divide and you get x=-2

6 0

3 years ago

Read 2 more answers

G(x) = 6x^2-7<br> Find g(x+2)

Ahat [919]

Answer:

= 6x² + 24x + 17

Step-by-step explanation:

g(x+2) = 6(x+2)² -7

= 6(x² + 4x + 4) - 7

= 6x² + 24x + 24 - 7

= 6x² + 24x + 17

7 0

3 years ago

TrianglesABC and DEF are <br> Similar<br> Not similar

puteri [66]

Answer:

Similar

Step-by-step explanation:

We can say that the triangles are similar by looking at the angle measurements

Angle B = 35

Angle F = 55

Both angles A and D = 90 because these are right triangles

We can say the measure of angle E = 35 because the sum of angles in a triangle equal to 180

Same rule applies to angle C

4 0

2 years ago

1. (i) Express 2704 as a product of its prime factors.

lidiya [134]

Answer:

all answers below

Step-by-step explanation:

(i) 2704 = 2^4 x 13^2

(ii) as both indices (the powers) are divisible by 2, it is a perfect square.

(iii) as both indices are not divisible by 3, it is not a perfect square.

8 0

3 years ago

Find the roots of the equation<br> x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits

scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

$x^{2} + 3x - 8^{- 14} = 0$

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

The above solution is for the quadratic equation of the form:

$ax^{2} + bx + c = 0$

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

From the given eqn

a = 1

b = 3

c = $- 8^{- 14}$

Now, using the above values in the formula mentioned above:

$x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)$

Now, Rationalizing the above eqn:

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

$x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

Solving the above eqn:

$x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}$

Solving with the help of caculator:

$x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}$

The precise value upto three decimal places comes out to be:

$x_{1, 1'} = 0.758\times 10^{- 14}$

5 0

3 years ago