All categories

4 years ago

3(3 - 8x) - 1 = -6x - 28

Mathematics

2 answers:

Wewaii [24]4 years ago

6 0

Answer:

x=2

Step-by-step explanation:

3(3 - 8x) - 1 = -6x - 28

Distribute

9 -24x -1 = -6x -28

Combine like terms

8 -24x = -6x -28

Add 24x to each side

8-24x+24x = -6x-28+24x

8 = 18x-28

Add 28 to each side

8+28 = 18x-28+28

36 = 18x

Divide each side by 18

36/18 = 18x/18

2 =x

Llana [10]4 years ago

5 0

Answer:

Step-by-step explanation:

3(3-8x)-1=-6x - 28

(3×3)-(3×8x)-1=-6x -28

6-3×8x-1=-6x -28

6-3-1×8x=-6x -28

2×8x=-6x -28

+6x +6x

2×8x+6x=-28

2×14x=-28

-2 -2

14x=-28 -2

14x=-30

÷14 ÷14

x=-30÷14

x=-2.15

You might be interested in

Please this will determine if I pass

PSYCHO15rus [73]

The answer is B through ASA (Angle-Side-Angle) similarity criterion.

5 0

3 years ago

Plz help Fast!!!!! Which is faster 15 feet per second or 12 miles per hour

Misha Larkins [42]

I think 15 feet per second

3 0

3 years ago

Read 2 more answers

Write a word sentence for this equation 25 equals K divided by 5

STALIN [3.7K]

Answer: (K) marbles divided into five sections equal 25 marbles into each section.

4 0

3 years ago

Define the opposite and the absolute value of 6

Yakvenalex [24]

The opposite is a negative so the answer would be -6

5 0

3 years ago

Read 2 more answers

It is known that the straight line l is tangent to the circle x^2+y^2=4 at a point on the x-axis, and intersects with the straig

Talja [164]

Step-by-step explanation:

The general equation of a circle is

$(x \ - \ h)^2 \ + \ (y \ - \ k)^2 \ = \ r^{2}$ ,

where h and k forms the coordinates of the centre of the circle.

When the circle has a centre at the origin, the equation reduces into

. $x^2 \ + \ y^2 \ = r^2$ .

Now, we are interested in solving for the x-intercepts (the x-coordinates when the circle intersects the x-axis), of the circle

$x^2 \ + \ y^2 \ = \ 4$ .

Thus,

$x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2$ .

Geometrically speaking, the tangent to the circle at the point defined by one of the x-intercepts of the circle is actually a vertical line, more specifically the lines $x \ = \ \pm \ 2$ .

First and foremost, for the vertical line $x \ = \ 2$ , it intersects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ , giving the y-coordinate for point P,

$y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1$ .

Hence, the coordinates of point P are $(1, \ 1)$ .

However, since there are no boundaries given in the question and a circle is symmetrical about its centre, thus, point P also exists when the vertical line $x \ = \ -2$ and interdects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ .

$y \ = \ \displaystyle\frac{1}{2}(-2) \\ \\ \\ y \ = \ -1$ .

Therefore, the coordinates of point P are also $(1, \ -1)$ .

8 0

2 years ago