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4 years ago

What do you suppose a decimal value for a day in the month means

Mathematics

2 answers:

Tasya [4]4 years ago

4 0

It would be 0 till the end of the month because that’s where it started from and when it’s going to end

Juliette [100K]4 years ago

3 0

Say you’re on the 14th day of a month, and it’s noon on the 14th. I’d assume a decimal place after the day would mean how far you are through that day

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determine the kinetic energy of a 1000-kg roller coaster car that is moving with a soeed if 20.0 m/s.

Alona [7]

200000J/200KJ

Step-by-step explanation:

using k.e=1/2(mv2)

k.e=1×1000×20²/2

k.e=400000/2

k.e=200000J/200KJ

8 0

3 years ago

Passes through (3, 1) and is perpendicular to the line y=2x+1

butalik [34]

Answer:

y = -1/2x + 5/2 or y = -0.5x + 2.5

Step-by-step explanation:

perpendicular lines have reciprocated slopes (opposite signs and values)

y = 2x + 1 → y = (-1/2x) + b

plug in (3,1) to find b

1 = -1/2(3) + b

1 = -1.5 + b

2.5 = b

y = -1/2x + 5/2 or y = -0.5x + 2.5

5 0

4 years ago

Read 2 more answers

If −2x + 3 = 13, then 2x =____?

marysya [2.9K]

Step by step: -2x + 3= 13
-3 = -3
-2x = 10 divide -2 on each side and get -5, so x=-5, plug that in 2x would be 2( -5) would be -10
ANSWER: -10

3 0

3 years ago

Complete the number sentence.<br><br> 0.30 ÷ 0.06

ahrayia [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What is the period and midline?

OverLord2011 [107]

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

\end{array}\qquad$

$\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}\\
\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\


\end{array}$

$\bf \begin{array}{llll}
\bullet \textit{function period}\\
\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}
$

so if you notice yours $\bf \begin{array}{llll}
3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\
&\ \uparrow&\uparrow \\
&B&D 
\end{array}$

now.. normally the function $\bf 3.2cos&\left( \frac{5}{3}\theta \right)$
has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis

now, with D = 6.1, that moves the midline up vertically that much

now.. the period, well, B = 5/3, normal period of cosine is $2\pi$
so, the new period will be $\bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}$

notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)

6 0

3 years ago