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3 years ago

6 over 2/3 I need help once again

1 answer:

6 0

$\bf \cfrac{6}{~~\frac{2}{3}~~}\implies \cfrac{\frac{\stackrel{}{6}}{1}}{~~\frac{2}{3}~~}\implies \cfrac{\stackrel{3}{~~\begin{matrix} 6 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{1}\cdot \cfrac{3}{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies 9$

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Simplify the expression completely. 7b+3(4b−2)+6÷3

Karo-lina-s [1.5K]

Answer:

19b-4

Step-by-step explanation:

7b+3(4b-2)+6÷3

Bracket comes first so we distribute 3 over the terms in the bracket

= 7b + 12b -6 +6 ÷ 3

First we solve division

= 7b + 12b -6 +6 ÷ 3

= 7b + 12b -6 +2

= (7b+12b)+(-6+2)

= 19b -4

This can't be simplified further as it can't be solved anymore

4 0

3 years ago

Read 2 more answers

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slavikrds [6]

Answer:Y=3x(x-4)

Step-by-step explanation:

4 0

3 years ago

around a square with diagonal d = 4.6 cm a circle is described. Calculate the perimeter of the circle

Nadya [2.5K]

Answer:

14.45cm2

Step-by-step explanation:

The diagonal is the diameter of the circle.

4.6cm

The perimeter of the circle is

Pi x4.6= 3.142 * 4.6= 14.45cm2

3 0

3 years ago

Please I need help on this question, A line passes through (6,3), (8,4), and (n,-2). Find the value of n.

Vika [28.1K]

Answer:

n= -4

Step-by-step explanation:

First we have to find the equation of line which passes through (6,3) and (8,4).

Then we have to pass this line through the point (n , -2).

we know, equation of line passing through (x1,y1) and (x2,y2) is

$\frac{x-x1}{x1-x2}=\frac{y-y1}{y1-y2}$

So, the equation of line passing through (6,3) and (8,4) is

$\frac{x-6}{6-8}=\frac{y-3}{3-4}$

or, $\frac{x-6}{-2} =\frac{y-3}{-1}$

multiplying both sides by -2

x-6=2(y-3)

or, x-6=2y-6

or, x-2y=6-6

or, x-2y=0

Now, we have to pass this line through (n,-2)

So, using the point in the line, that is using x=n , y=-2

we get

n-2(-2)=0

or, n+4=0

or, n=-4

6 0

3 years ago

Read 2 more answers

Solve the equation. 2x squared-3x+1=0

natali 33 [55]

$2x^2-3x+1=0\\\\2x^2-2x-x+1=0\\\\2x(x-1)-1(x-1)=0\\\\(x-1)(2x-1)=0\iff x-1=0\ \vee\ 2x-1=0\ \ \ \ |+1\\\\x=1\ \vee\ 2x=1\ \ \ |:2\\\\\boxed{x=1\ \vee\ x=0.5}$

4 0

3 years ago