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Svetlanka [38]
3 years ago
14

The number doing the dividing in a division problem

Mathematics
2 answers:
pickupchik [31]3 years ago
8 0

Answer:

Divisor

Step-by-step explanation:

It's called the "divisor"

Pavlova-9 [17]3 years ago
6 0

Answer:

divisor

Step-by-step explanation:

I just looked it up bro lol

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Select "Growth" or "Decay" to classify each function. Select "Growth" or "Decay" to classify each function. Function Growth Deca
Oksanka [162]
Equations with a number in brackets that is less then one is a decaying function.
Y=200(0.5)2t Decay
Y=12(2.5)t6 Growth
Y=(0.65)t4 decay
Hope this helps!
6 0
3 years ago
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The radius of a sphere is 6 inches. Find the length of a chord connecting two perpendicular radii. x = inches
Lilit [14]
<h3>Answer:</h3>

6√2 ≈ 8.485 inches

<h3>Step-by-step explanation:</h3>

The radii and the chord together make an isosceles right triangle with legs 6 inches long. The hypotenuse of such a triangle is √2 times the leg length. So, the chord will be 6√2 in long.

_____

<em>Comment on isosceles right triangle</em>

It is worth remembering that the hypotenuse of an isosceles right triangle is √2 times the leg length. This is easily found using the Pythagorean theorem:

... c² = a² + b²

... c² = 1² + 1² = 2 . . . . for legs of length 1

... c = √2 . . . . . . . . . . take the square root.

Scale this result as needed for any particular problem. Here, the scale factor is 6 inches.

3 0
3 years ago
What is the side length of a square with a perimeter of 96 meters
zloy xaker [14]

Answer:

24

Step-by-step explanation:

96 divided by 4 = 24

3 0
3 years ago
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Need Help Please!!!!
denis-greek [22]

Answer:

4ab and 3ab

Step-by-step explanation:

4<em>ab</em> and 3<em>ab</em> because they have the same variables <em>a</em> and <em>b</em>

4 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}&#10;\\ \quad \\&#10;% function transformations for trigonometric functions&#10;\begin{array}{rllll}&#10;% left side templates&#10;f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}&#10;\\\\&#10;f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\\\&#10;f(x)=&{{  A}}tan({{  B}}x+{{  C}})+{{  D}}&#10;&#10;\end{array}

\bf \begin{array}{llll}&#10;% right side info&#10;\bullet \textit{ stretches or shrinks}\\&#10;\quad \textit{horizontally by amplitude } |{{  A}}|\\\\&#10;\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\&#10;\end{array}

\bf \begin{array}{llll}&#10;&#10;&#10;\bullet \textit{vertical shift by }{{  D}}\\&#10;\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\&#10;\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\&#10;\bullet \textit{function period or frequency}\\&#10;\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\&#10;\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)&#10;\end{array}


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

\bf \begin{array}{lllcclll}&#10;f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\&#10;&\uparrow &\uparrow &\uparrow &\uparrow \\&#10;&A&B&C&D&#10;\end{array}\\\\&#10;-----------------------------\\\\&#10;period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}
8 0
4 years ago
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