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Elodia [21]
3 years ago
7

Rodney bought a 50 pound bag of dog food. HIS dog ate 2/5 of the food in the first mon th and 2/10 of the food in the second mon

th. How much dog food, in pounds was remaining in the bag at the end of the two months? Show your work
Mathematics
1 answer:
NeTakaya3 years ago
4 0

Answer:

20 lbs

Step-by-step explanation:

2/5ths of the food would be 20 pounds because:

2/5 * 50 = 20

2/10ths of the food would be 10 pounds because:

2/10 * 50 = 10

If we add those two together, we would get the total amount that the dog ate, which would be 30 pounds. SInce the dog ate 30 pounds of food out of the 50 pounds, there would be 20 pounds left in the bag at the end of two months:

50- ( 20+ 10) = 20

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In the figure,Line x is parallel to line y. identify the angles congruent to angle 3
san4es73 [151]

Answer:

For example, slide ∠ 1 down the transversal and it will coincide with ∠2. are equal in measure. If two parallel lines are cut by a transversal, the corresponding angles are congruent. If two lines are cut by a transversal and the corresponding angles are congruent, the lines are parallel.

Step-by-step explanation:

8 0
2 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}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\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}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}

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

\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
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}
\\\\\\
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
What is the range of the function graphed below?
Nadya [2.5K]

Answer:

D

Step-by-step explanation:

Range is the 'y' values a graph can have

  you can see that the 'y' values can be anythig from - inf to + 2

3 0
1 year ago
What’s the correct answer for this question?
Ad libitum [116K]

Answer:

C:

Step-by-step explanation:

Both angles add up to 180°

<BCG + <BFG = 180°

2x+146+4x+238=180

6x+384 = 180°

6x = 180-384

6x = -204

Dividing both sides by 6

x = -34

8 0
3 years ago
Work out the volume of this sphere?
Elina [12.6K]

Answer:

434.9

Step-by-step explanation:

\frac{4}{3} × π (4.7)²

434.893

4 0
2 years ago
Read 2 more answers
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