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frutty [35]
3 years ago
5

If you take a number, times by 3 then add 6. You get the same as if you took the number, times by 6 then subtract 3. What is the

number?
Mathematics
2 answers:
PtichkaEL [24]3 years ago
8 0

Answer:

3

Step-by-step explanation:

write the question in an algebraic way:

3x+6 = 6x - 3

Solve:

6x-3 = 3x+6

6x - 3 - 3x = 6

3x = 9

x  = 3

Hope this helps.

Good Luck

Romashka-Z-Leto [24]3 years ago
4 0

Answer:

3

Step-by-step explanation:

Let X be the unknown number

3X+6=6X-3

3X-6X=-3-6

-3X=-9

X=-9÷-3

X=3

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A jar has 570 marbles of different colors.The ratio of red to blue marbles are 2:3 and red to green is 4:5.how many marbles of e
iVinArrow [24]

Answer:

The number of red marbles = 152

The number of blue marbles = 228 and

the number of green marbles = 190

Step-by-step explanation:

Let x be the number of red marbles, y be the number of blue marbles and z be the number of green marbles.

It is given that the total marbles = 570.

Therefore, x + y + z = 570 --- (1)

It is also given that the ratio of red to blue marbles is 2 : 3.

Therefore, x : y = 2 : 3

or 3x = 2y --- (2)

It is also given that the ratio of red to green marbles is 4 : 5.

Therefore, x : z = 4 : 5

or 5x = 4z --- (3)

From (2), y=\frac{3x}{2}

From (3), z=\frac{5x}{4}

Substituting the above in (1), we get,

x+\frac{3x}{2} +\frac{5x}{4} =570

Multiply both sides by 4.

4x + 6x + 5x = 2280

15x = 2280

Divide both sides by 15.

x = 152

Hence,

y = \frac{3(152)}{2}

y = 228

z = \frac{5(152)}{4}

z = 190

Therefore,

The number of red marbles = 152

The number of blue marbles = 228 and

the number of green marbles = 190

<u>Check:</u>

Total number of marbles = 152 + 228 + 190 = 570

Ratio of red to blue marbles = 152 : 228 = 2 : 3

Ratio of red t green marbles = 152 : 190 = 4 : 5


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3 years ago
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The expression -16t ^2+ 48t can be written as the product of -16t with which of the following?
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Answer:

4) t -3

Step-by-step explanation:

-16t² = -16 * t * t

48t = (-16t) * (-3)

-16t² + 48t = (-16t) * t + (-16t) *(-3)

                = (-16t) [ t - 3]

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3 years ago
Find an equation of the tangent plane to the given parametric surface at the specified point.
Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

Step-by-step explanation:

Given equation

      r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}---(1)

Normal vector  tangent to plane is:

\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}

\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}

Normal vector  tangent to plane is given by:

r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]

Expanding with first row

\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\

at u=5, v =π/3

                  =\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k} ---(2)

at u=5, v =π/3 (1) becomes,

                 r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

From above eq coordinates of r₀ can be found as:

            r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})

From (2) coordinates of normal vector can be found as

            n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)  

Equation of tangent line can be found as:

  (\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

5 0
3 years ago
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