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steposvetlana [31]
1 year ago
14

5th grade math! Can someone help me out? Offering 20 points for both of these!

Mathematics
2 answers:
Kamila [148]1 year ago
7 0

Answer:

8\frac{2}{6}\\\\2\frac{1}{5}

Fractions are sometimes the same as dividing, for this case, that rule applies.

\frac{50}{6}\\\\\frac{11}{5}

↑ These improper fractions translate to...

\frac{50}{6} = 8\frac{1}{3} -- > 8\frac{2}{6}  \\\\ \frac{11}{5} = 2\frac{1}{5}

Kipish [7]1 year ago
4 0

Answer:

1)  D. 8 2/6 hotdogs

2)  B. 2 1/5 cakes

Step-by-step explanation:

1) If she's dividing 50 hotdogs <u>equally</u> among 6 people, then you would divide 50 by 6 to find how many each person gets.
note: 50 ÷ 6 is the same as \frac{50}{6}.

\frac{50}{6} =  8\frac{2}{6}

6 can go into fifty 8 times, so there is a whole number of 8. There is 2/6 left over, so the final result is 8\frac{2}{6}.

2) If Sako makes 11 cakes and wants to share it <u>equally</u> among 5 people, we can divide 11 by 5 to find how many each person gets.

\frac{11}{5} =2\frac{1}{5}

5 can go into eleven 2 times, so there is a whole number of 2. There is 1/5 leftover, so the final result is 2\frac{1}{5}.

Hence, the answers are D, and B.

hope this helps!

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Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter
Levart [38]

Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

.............x/sqrt(64+x^2)

.............sqrt(64-x^2)/8

..............x/sqrt(64-x^2)

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A. ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and  \sqrt{8^{2} - x^{2} }or  \sqrt{64 - x^{2} }

thus tan(arcsin(x/8))  = \frac{x}{\sqrt{64 - x^{2} }}  =

Therefore  ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B

Here we have cos = adjacent/hypotenuse where adjacent side is \sqrt{64 - x^{2} } and hypothenuse = 8 we have \sqrt{64 - x^{2} }/8

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

4 0
3 years ago
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Arturiano [62]

Answer:

See attached picture.

Step-by-step explanation:

To graph linear inequalities, use the y=mx + b form to graph using the slope and y-intercept.

y ≤ -4x + 40 has slope -4 and y-intercept (0,40).

Start at (0,40) and mark it. Then move down 4 units and to the right 1. Mark this point at (1,36). Connect the points with a solid line since the inequality has equal to. Substitute a point like (0,0) to test where the solution set is.

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This is true so shade to the left of the line.

To graph y ≤ 10 mark a point on the y-axis at (0,10). Draw a horizontal solid line through the point. Then shade below the line.

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