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denpristay [2]
3 years ago
9

32 assignments have been turned in. 14 didn't have names. The rest were put into 3 equal piles to grade over 3 days. How many as

signments are graded each day?
Mathematics
1 answer:
denpristay [2]3 years ago
6 0
Answer: is 6 because if you take away 32 and 14 you will get 18 and if you divide 18 into 3 equal piles for three days you will get 6.......graded assignments

Your Welcome:)
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Mini staw<br> D:<br> R:<br> what is the domain and range
IgorC [24]

The entire range of independent variable values is the domain of a function.

After substituting the domain, the range of just a function is the entire set of all possible values for the dependent variable (often y).

What is domain and range?

  • The collection of all x-values that can cause the function to "work" and produce actual y-values is known as the domain.
  • The range is the set of y-values that are produced when all the conceivable x-values are substituted.

The entire range of independent variable values is the domain of a function.

Keep these things in mind when locating the domain:

  1. A fraction's denominator (bottom) cannot be 0.
  2. In this section, the integer following a square root symbol must be positive.

After substituting the domain, the range of just a function is the entire set of all possible values for the dependent variable (often y).

The variety of potential y-values makes up a function's range (minimum y-value to maximum y-value)

  1. To observe what happens, substitute several x-values into the expression for y.
  2. Be sure to search for the least and highest y values.

Learn more about Domain and Range here:

brainly.com/question/10197594

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3 0
2 years ago
80% of the 55 coffee mugs at Kari's Pancake House are dirty. How many dirty coffee mugs are there at the pancake house?
Pachacha [2.7K]

Answer:

44

Step-by-step explanation:

80% of 55=44

4 0
3 years ago
Read 2 more answers
Simplify using order of operations.Show all your work. 4^3 + (`10-8) divided by 2
Igoryamba

Answer:

<h2>65</h2>

Step-by-step explanation:

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

------------------------------------------------------------------------

4^3+(10-8)\div2=64+2\div2=64+1=65

3 0
3 years ago
How to solve linear and nonlinear equations graphically?
Ede4ka [16]
Plot the equation. If you wish to solve a polynomial, let y= polynomial and plot the graph. Best set up a table of values first.
Where the graph crosses the x axis there is a solution for x. There are also solutions for other horizontal lines (y values) by looking at intersections of the graph with these lines. This technique works for linear and non linear equations. You can also use graphs to solve 2-variable systems of equations by examining where the graphs intersect one another. The disadvantage is that you may not be able to have sufficient detail for high degrees of accuracy because of the scale of the graph and drawing inaccuracies.
8 0
3 years ago
Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
VikaD [51]
Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}&#10;\\\\\\&#10;\textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{adjacent}{opposite}&#10;\\\\\\&#10;% cosecant&#10;csc(\theta)=\cfrac{hypotenuse}{opposite}&#10;\qquad \qquad &#10;% secant&#10;sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{\sqrt{35}}{6}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{-1}{\sqrt{35}}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{\sqrt{35}}{1}&#10;\\\\\\&#10;sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}&#10;\\\\\\&#10;sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
3 0
3 years ago
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