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sukhopar [10]
3 years ago
8

Find the quotient. If possible, rename the quotient as a mixed number or a whole number. Write your answer in simplest form, usi

ng only the blanks needed. 2 3/5 divided by 10
Mathematics
1 answer:
irina [24]3 years ago
3 0

9514 1404 393

Answer:

  13/50

Step-by-step explanation:

The mixed number 2 3/5 is equal to the improper fraction (2·5 +3)/5 = 13/5.

When that is multiplied by 1/10, the result is ...

  (13/5)×(1/10) = 13/50 . . . your quotient

_____

<em>Additional comment</em>

Division is the same as multiplication by the reciprocal. In this case, division by 10 is the same as multiplication by 1/10.

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Norman bought 5/6 pounds of cookies. He ate 3/5 cookies. What was the weight of the cookies he ate
schepotkina [342]

Answer:

Half a pound.

Step-by-step explanation:

To find the number of pounds he ate, multiply the fractions.

\frac{5}{6}*\frac{3}{5} = \frac{15}{30}=\frac{1}{2}

He ate 1/2 a pound.

8 0
3 years ago
A researcher computes a 2 x 3 factorial anova. in this example, how many interactions can be observed?
Ivan

 

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7 0
3 years ago
If 11 is subtracted from twice a number the answer is 7. Find the number?
KonstantinChe [14]
Answer:
9

Solution:

Write it in an algebraic form (x is your number)
2x-11=7
Now solve
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6 0
3 years ago
Read 2 more answers
There are 25boys and 15girls in a class.if a student is picked at random .what is the probability that the student is a girl
emmasim [6.3K]

Answer:

Step-by-step explanation:

Total Students in Class = 25+ 15=40

Probability= (Number of Girls in Class) ÷ (Total students in Class)

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7 0
3 years ago
Someone help me with this
Jet001 [13]

2a. 14,9; 2b. 15,4; 2c. 30,8; 3. 32

For 2a., you have to set it up like this: csc 48° = 20⁄x [OR sin 48° = ˣ⁄20]. Then you would have to isolate the variable by getting rid of the denominator. The cosecant function has an extra step because you will get xcsc 48° = 20. As stated, isolate the variable; this time, divide by csc 48°. This is what 20 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

For 2b., you have to set it up like this: sec 39° = ˣ⁄12 [OR cos 39° = 12⁄x. Then you would have to isolate the variable by getting rid of the denominator. The cosine function has an extra step because you will get xcos 39° = 12. As stated, isolate the variable; this time, divide by cos 39°. This is what 12 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

For 2c., you have to set it up like this: cot 64° = 15⁄x [OR tan 64° = ˣ⁄15]. Then you would have to isolate the variable by getting rid of the denominator. The cotangent function has an extra step because you will get xcot 64° = 15. As stated, isolate the variable; this time, divide by cot 64°. This is what 15 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

Now, for 3., it is unique, but similar concept. In this exercise, we are solving for an angle measure, so we have to use inverse trigonometric ratios. So, we set it up like this: cot⁻¹ 1⅗ = m<x [OR tan⁻¹ ⅝ = m<x]. We simply input this into our calculator and we get 32,00538321°. When rounded to the nearest degree, we get 32°.

WARNING: If you use a graphing calculator, you have to input it uniquely because most graphing calculators do not have the inverse trigonometric ratios programmed in their systems. This is how you would write this: tan⁻¹ 1⅗⁻¹. You set 1⅗ to the negative first power, ALONG WITH the inverse tangent function, because without it, your answer will be thrown off. Since Cotangent and Tangent are multiplicative inverses of each other, that is the reason why the negative first power is applied ALONG WITH the inverse tangent function.

**NOTE: 1⅗ = 8⁄5

Take into consideration:

sin <em>θ</em> = O\H

cos <em>θ</em><em> </em>=<em> </em>A\H

tan <em>θ</em><em> </em>= O\A

sec <em>θ</em><em> </em>= H\A

csc <em>θ</em><em> </em>= H\O

cot <em>θ</em><em> </em>= A\O

I hope this helps you out alot, but if you are still in need of assistance, do not hesitate to let me know and subscribe to my channel [username: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.

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3 years ago
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