All categories

2 years ago

Simplify to create an equivalent expression. −3(2+4k)+7(2k−1)

Mathematics

2 answers:

Genrish500 [490]2 years ago

7 0

Answer:

$\huge\boxed{2k-13}$

Step-by-step explanation:

$\sf -3(2+4k)+7(2k-1)\\Expanding \ brackets\\-6-12k+14k-7\\Combining \ like \ terms\\-12k+14k-6-7\\2k -13$

prisoha [69]2 years ago

6 0

2k-13
First step= -3(2+4K)+7(2k-1)
Second step= -6 - (12k) + 14k- 7
Third step= -12k+14k - 6 -7= 2k-13

You might be interested in

WILL GIVE BRAINLIST AS SOON AS I CAN is this 32, maybe? i have no idea.

Feliz [49]

Answer:

y ≈ 32.0°

Step-by-step explanation:

Since the triangle is right use the sine ratio to find y

sin y = $\frac{opposite}{hypotenuse}$ = $\frac{9}{17}$ , thus

y = $sin^{-1}$ ( $\frac{9}{17}$ ) ≈ 32.0° ( nearest tenth )

8 0

2 years ago

The supercontinent that existed about 200 million years ago was named

mylen [45]

Pangaea is the answer I had the same question.

3 0

2 years ago

Read 2 more answers

Help! Prove the equality<br><br>arccos √(2/3) - arccos (1+√6)/(2*√3) = π/6

stealth61 [152]

Answer:

<em>Proof in the explanation</em>

Step-by-step explanation:

<u>Trigonometric Equalities</u>

Those are expressions involving trigonometric functions which must be proven, generally working on only one side of the equality

For this particular equality, we'll use the following equation

$\displaystyle cos(x-y)=cos\ x\ cos\ y+sin\ x\ sin\ y$

The equality we want to prove is

$\displaystyle arccos\ \sqrt{\frac{2}{3}}-arccos\left(\frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\frac{\pi}{6}$

Let's set the following variables:

$\displaystyle x=arccos\ \sqrt{\frac{2}{3}},\ y=arccos(\frac{1+\sqrt{6}}{2\sqrt{3}})$

And modify the first variable:

$\displaystyle x=arccos\ \frac{\sqrt{6}}{3}}=>\ cos\ x= \frac{\sqrt{6}}{3}}$

Now with the second variable

$\displaystyle y=arccos\ \frac{1+\sqrt{6}}{2\sqrt{3}}=>cos\ y=\frac{1+\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+3\sqrt{2}}{6}$

Knowing that

$sin^2x+cos^2x=1$

We compute the other two trigonometric functions of X and Y

$\displaystyle sin \ x=\sqrt{1-cos^2\ x}=\sqrt{1-(\frac{\sqrt{6}}{3})^2}=\sqrt{1-\frac{6}{9}}=\frac{\sqrt{3}}{3}$

$\displaystyle sin\ y=\sqrt{1-cos^2y}=\sqrt{1-\frac{(\sqrt{3}+3\sqrt{2})^2}{36}}}$

$\displaystyle sin\ y=\sqrt{\frac{36-(3+6\sqrt{6}+18)}{36}}=\sqrt{\frac{15-6\sqrt{6}}{36}}$

Computing

$15-6\sqrt{6}=(3-\sqrt{6})^2$

Then

$\displaystyle sin\ y=\frac{3-\sqrt{6}}{6}$

Now we replace all in the first equality:

$\displaystyle cos(x-y)=\frac{\sqrt{6}}{3}.\frac{\sqrt{3}+3\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{3-\sqrt{6}}{6}$

$\displaystyle cos(x-y)=\frac{3\sqrt{2}+6\sqrt{3}}{18}+\frac{3\sqrt{3}-3\sqrt{2}}{18}$

$\displaystyle cos(x-y)=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}=cos\ \pi/6$

Thus, proven

5 0

3 years ago

What is the area of a parallelogram with a height of 7 feet and a base of 12 feet?

Vesna [10]

Answer:

84ft^2

Step-by-step explanation:

Relying on formula S(area) =a(base) *h(height )

3 0

2 years ago

Read 2 more answers

Find the equation of a line with the given points.(put your answers in the form y=mx+b)(6,2)(5,5)

Nesterboy [21]

The slope-intercept form: y= mx + b

m - slope

b - y-intercept

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (6, 2) and (5, 5). Substitute:

$m=\dfrac{5-2}{5-6}=\dfrac{3}{-1}=-3$

Therefore we have y = -3x + b.

Put the coordinates of the point (5, 5) to the equation:

5 = -3(5) + b

5 = -15 + b <em>add 15 to both sides</em>

20 = b

<h3>Answer: y = -3x + 20</h3>

6 0

2 years ago

Read 2 more answers