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2 years ago

Find the value of x of the triangle

Mathematics

2 answers:

marysya [2.9K]2 years ago

6 0

Answer:

I think The answer would be 7

irina [24]2 years ago

3 0

Answer:7

Step-by-step explanation:

It’s 7 their the same size as the other side of it

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Fittoniya [83]

Answer:

Step-by-step explanation:Before you can tackle complex word problems, you need to know how to translate simple problems into algebra. Generally, you'll do this by replacing the words with operators that mean the same thing. So for instance, four divided by x could be 4 ÷ x. ÷ and / are just two ways of writing divided. If you were to translate this word for word, you'd change the word times to a multiplication symbol (⋅) and put a plus sign (+) between the four and the two. Your expression would look like this:

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2 years ago

Use the following statements to write a compound statement for -q and ~r. Then find its

gayaneshka [121]

Answer:

pqr

Step-by-step explanation:

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3 years ago

Simplify (5x-7) (2x+3) (7x-8)

Natali5045456 [20]

Answer:

70x³ - 73x² - 155x + 168

Step-by-step explanation:

Given

(5x - 7)(2x + 3)(7x - 8) ← expand the first pair of factors using FOIL

= (10x² + x - 21)(7x - 8)

Each term in the second factor is multiplied by each term in the first factor, that is

10x² (7x - 8) + x(7x - 8) - 21(7x - 8) ← distribute the 3 parenthesis

= 70x³ - 80x² + 7x² - 8x - 147x + 168 ← collect like terms

= 70x³ - 73x² - 155x + 168

3 0

2 years ago

(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.

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