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Alinara [238K]
3 years ago
15

Bruno bought 96 pounds of sugar for his bakery. Every day he used the same amount of sugar to make bakery items. After 10 days,

Bruno was left with 15.2 pounds of sugar. On average, how many pounds of sugar did Bruno use each day? (1 point)
a 5.60 pounds b 8.08 pounds c 11.12 pounds d 18.46 pounds
Mathematics
2 answers:
Kazeer [188]3 years ago
8 0

Answer:

9.6

Step-by-step explanation:

you divide 96 by 10

ale4655 [162]3 years ago
6 0

Answer:

The answer is B

Step-by-step explanation:

The answer is B. Since Bruna used 96 pounds of sugar each day you would have to use the equation 96-x(10 times)=15.2. You would have to plug in each answer to see if it would give you 15.2. I plugged in 5.60 and Bruno was left with too many pounds left over. 11.2 pounds each day would be to low. I then plugged in 8.08 and subtracted by 96 ten times and got the answer 15.2. I hope this helps.

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A rectangle has a length of
Marina CMI [18]

rectangle has a length of

18 feet. The width is x feet

less than the length. If the

area must be less than 216

square feet, what could be

the measurement of the

width?

6 0
3 years ago
At a bargain store.Tanya bought 5 items that each cost the same amount. Tony bought 6 items that each cost the same amount, but
Ray Of Light [21]

(A) For x representing the cost of one of Tanya's items, her total purchase cost 5x. The cost of one of Tony's items is then (x-1.75) and the total of Tony's purchase is 6(x-1.75). The problem statement tells us these are equal values. Your equation is ...

... 5x = 6(x -1.75)

(B) Subtract 5x, simplify and add the opposite of the constant.

... 5x -5x = 6x -6·1.75 -5x

... 0 = x -10.50

... 10.50 = x

(C) 5x = 5·10.50 = 52.50

... 6(x -1.75) = 6·8.75 = 52.50 . . . . . the two purchases are the same value

(D) The individual cost of Tanya's iterms was $10.50. The individual cost of Tony's items was $8.75.

6 0
3 years ago
Please can anyone help me please​
Step2247 [10]
4.5, 4, 3.5, 3, 2.5, 2 and use the pencil and ruler, love a bit of MathsWatch
7 0
3 years ago
Read 2 more answers
What Numbers Go Into The Venn Diagram
antiseptic1488 [7]

Answer:

(a)The total number of outcomes where the sum is 9 or

    greater than 9 is  10

(b)Total number of outcomes where the sum is odd = 18

(c)Total number of outcomes where the  sum greater or equal to 9 and is

     also odd = 6

Step-by-step explanation:

Here, Sample Space = { Sum of the two digits when two dices are thrown together}

or,   S   =   {2,3,4,5,6,7,8,9,10,11,12}

(a)  The number of ordered pairs where sum is  9 or greater than 9

    = { sum is 9  ,  Sum is 10 ,  Sum is 11,   Sum is 12}

    = {(6,3)(3,6),(4,5)(5,4) ,   (5,5), (6,4),(4,6)  , (6,5)(5,6),   (6,6)}

Hence the total number of outcomes where the sum is 9 or

greater than 9 is  10

(b)  The number of ordered pairs where sum is odd.

    = { Sum is 3 ,  Sum is 5,   Sum is 7, Sum is 9, Sum is 11}

    = {(1,2)(2,1),   (4,1)(1,4),(2,3)(3,2) ,   (6,1), (1,6),(5,2),(2,5),(4,3)(3,4)  ,

        (6,3)(3,6), (4,5)(5,4),  (6,5),(5,6)}  = 18

Hence total number of outcomes where the sum is odd = 18

(c) Intersection point refers the outcomes which have sum greater or equal to 9 and is odd

Here, the possible outcomes are  = { Sum is 9 ,  Sum is 11}

                                            ={ (6,3)(3,6), (4,5)(5,4),  (6,5),(5,6)}  = 6

Hence total number of outcomes where the  sum greater or equal to 9 and is also odd = 6

6 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}}
\\\\\\
\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
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