All categories

3 years ago

Is this the correct solution? 6x ≥ 50 --> x ≥ 8 1/3 Yes or No

Mathematics

1 answer:

nika2105 [10]3 years ago

6 0

Answer:

Yes this would be correct!!!!

Step-by-step explanation:

You might be interested in

one rectangle has a length 10 cm and width 5 cm the second rectangle has length 12 cm and width 4 cm are the two rectangles simi

tatuchka [14]

Answer:

No, they aren't similar.

Step-by-step explanation:

<u><em>Let's take the proportionality of their sides to check whether they are similar or not:</em></u>

$\frac{10}{5} = \frac{12}{4}$

=> 2 ≠ 3

So, they are not similar.

8 0

3 years ago

If m=2 and b=-5, what is the equation of the line in slope intercept form

nordsb [41]

Answer:

y=2x-5

explanation:

y=mx+b

6 0

2 years ago

If 400 bricks measuring 81" by 31" are required to build a wall 42 ft. high, how long is the wall

musickatia [10]

Idek lol but yeah have a good day

7 0

2 years ago

Which lines are parallel?

kenny6666 [7]

AC and DF are parallel

8 0

2 years ago

Given sin x = -4/5 and x is in quadrant 3, what is the value of tan x/2

love history [14]

bearing in mind that, on the III Quadrant, sine as well as cosine are both negative, and that hypotenuse is never negative, so, if the sine is -4/5, the negative number must be the numerator, so sin(x) = (-4)/5.

$\bf sin(x)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-(-4)^2}=a\implies \pm\sqrt{9}=a\implies \pm 3=a \\\\\\ \stackrel{III~Quadrant}{-3=a}~\hfill cos(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill$

$\bf tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)}\qquad \leftarrow \textit{let's use this one} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf tan\left( \cfrac{x}{2} \right)=\cfrac{~~\frac{-4}{5}~~}{1-\frac{3}{5}}\implies tan\left( \cfrac{x}{2} \right)=\cfrac{~~\frac{-4}{5}~~}{\frac{2}{5}}\implies tan\left( \cfrac{x}{2} \right)=\cfrac{-4}{5}\cdot \cfrac{5}{2} \\\\\\ tan\left( \cfrac{x}{2} \right)=\cfrac{-4}{2}\cdot \cfrac{5}{5}\implies tan\left( \cfrac{x}{2} \right)=-2$

4 0

3 years ago

Read 2 more answers