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3 years ago

3+ 6x - x = 33 Can you find the answer

Mathematics

2 answers:

larisa [96]3 years ago

4 0

Answer:

x = 6

Step-by-step explanation:

Subtract x from 6 x .

3 + 5 x = 33

Move all terms not containing x to the right side of the equation.

Subtract 3 from both sides of the equation.

5 x = 33 − 3

Subtract 3 from 33 . 5 x = 30

Divide each term by 5 and simplify.

Divide each term in 5 x = 30 by 5 .

Reduce the expression by cancelling the common factors.

x = 30

$\frac{5x}5} = \frac{30}5}$ 5

Divide 30 by 5 .

x = 6

SVEN [57.7K]3 years ago

3 0

Answer:

x=6

Step-by-step explanation:

6x-x=5x

33-3=30

30/5=6

x=6

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What number divided by 5 equals 6

djyliett [7]

The answer your looking for is 30
30 ÷ 5 = 6

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3 years ago

Read 2 more answers

(1 point) (a) Find the point Q that is a distance 0.1 from the point P=(6,6) in the direction of v=⟨−1,1⟩. Give five decimal pla

natima [27]

Answer:

following are the solution to the given points:

Step-by-step explanation:

In point a:

$\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}$

$=\sqrt{1+1}\\\\=\sqrt{2}$

calculating unit vector:

$\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}$

the point Q is at a distance h from P(6,6) Here, h=0.1

$a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071$

the value of Q= (5.92928 ,6.07071 )

In point b:

Calculating the directional derivative of $f (x, y) = \sqrt{x+3y}$ at P in the direction of $\vec{v}$

$f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\$

$=\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557$

$\vec{v} = 1.97651557$

In point C:

Computing the directional derivative using the partial derivatives of f.

$f_x(x,y)= \frac{1}{2 \sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{2 \sqrt{22}}\\\\f_x(x,y)= \frac{1}{\sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{\sqrt{22}}\\\\f_{(PQ)}(P)= (f_x \vec{i} + f_y \vec{j}) \cdot \frac{\vec{v}}{|\vec{v}|}\\\\= (\frac{1}{2 \sqrt{22}}\vec{i} + \frac{1}{\sqrt{22}} \vec{j}) \cdot \frac{-1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}} \vec{j}$