Question 6 of 10<br> sin 30º = √3/2 and cos 30º = 1/2<br><br> O A. True<br> OB. False

Will give brainliest for correct answer.

lukranit [14]

The numerator and denominator must be polynomials so

the answer to your question is - 7x / x^2 ( the last one on the right)

5 0

3 years ago

Read 2 more answers

Graph the image of the figure after a dilation with a scale factor of 2 centered at (-7, - 2).

tigry1 [53]

Answer:

Sorry this is late!

Step-by-step explanation:

8 0

3 years ago

a box contains 4 yellow, 5 green and 3 blue balls. find the probability of obtaining a yellow ball in a single random draw.

Alex

Notice that the box has a total os 4 + 5 + 3 = 12 balls, since there are 4 that are yellow, then, the probability to randomly obtaining a yellow ball in a single draw is:

$P(Yellow)=\frac{\text{ number of yellow balls}}{\text{total number of balls}}=\frac{4}{12}=0.33$

therefore, the probability is 0.33 = 33%

7 0

1 year ago

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$