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2 years ago

13. After Valentine's Day, boxes of chocolates go on sale and are marked down 10%. The

Mathematics

1 answer:

Anna007 [38]2 years ago

5 0

Answer:

$27.00

Step-by-step explanation:

First, 10% of 30 = 3.

Second, 30 - 3 = 27

Hope this helps!

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Answer this plssssssssssss

Inessa05 [86]

Answer:

0.84 is the answers for the question

Step-by-step explanation:

please give me brainlest

7 0

2 years ago

Read 2 more answers

Given cos ⁡ ( θ ) = 5/6 , then sin ⁡ ( θ ) = a/b .<br><br> Determine the values of a and b.

slega [8]

Answer:

a = √11 and b = 6

Step-by-step explanation:

Refer to attached picture for reference

for an right triangle with angle θ

we are given

cos θ = 5/6 = length of adjacent side / length of hypotenuse

hence

adjacent length = 5 units

hypotenuse length = 6 units

the missing side is the "opposite" length which we can find with the Pythagorean equation. in our case:

hypotenuse ² = adjacent ² + opposite² (rearrange)

opposite ² = hypotenuse ² - adjacent ²

opposite ² = 6² - 5²

opposite = √ (6²-5²) = √11

sin θ = opposite length / hypotenuse (substitute values above)

sin θ = √11 / 6

hence a = √11 and b = 6

3 0

2 years ago

If a point P(–1, –1) is reflected across the line y = –2, what are the coordinates of its reflection image?

Ksivusya [100]

Since the line of refrection is a horizontal line, the x-coordinate of the image will be the same as the x-coordinate of the point P(-1, -1). P(-1, -1) is one unit above the line y = -2, hence the image will be one unit below the line y = -2. Therefore the coordinate of the image is (-1, -3)

8 0

3 years ago

If a gallon of milk costs $5.25 now and the price is

adelina 88 [10]

Answer: It will take approximately 21 years.

Step-by-step explanation:

6 percent of 5.25 is 0.315 (5.25 x 0.06 = 0.315)

Then you subtract 5.25 from 12 to calculate the number of years. (12 - 5.25 = 6.75)

Divide 0.315 from 6.75 (6.75 / 0.315 ≈ 21.4)

5 0

3 years ago

For the function given state the period f(t) =6sin(3t-pi/6)-1

lapo4ka [179]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now, with that template in mind, let's take a peek at yours

$\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}$

8 0

3 years ago