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3 years ago

How would you write this as a single function?

Mathematics

1 answer:

dexar [7]3 years ago

3 0

Written as a single function is sin(20+30) which is sin(50deg)

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(A3- 2a2) - ( 3a2- 4a3)

Tanzania [10]

If you would like to calculate (a^3 - 2 * a^2) - (3 * a^2 - 4 * a^3), you can do this using the following steps:

(a^3 - 2 * a^2) - (3 * a^2 - 4 * a^3) = a^3 - 2 * a^2 - 3 * a^2 + 4 * a^3 = 5 * a^3 - 5 * a^2

The correct result would be 5 * a^3 - 5 * a^2.

8 0

3 years ago

Read 2 more answers

Evaluate the expression when b=2<br> 12-b

exis [7]

Answer:

10 $\huge\checkmark$

Step-by-step explanation:

Hi there, hope you are having a nice day!

All we should do is plug in the value of b:

12-2

10 (Answer)

Hope you find it helpful.

Feel free to ask if you have any questions.

$\bf{-MistySparkles^**^*$

7 0

2 years ago

Read 2 more answers

What is the answer and how do you solve <br><br><br> 21x^3/3x^−1

Sunny_sXe [5.5K]

Answer:

as written: 7x^2

perhaps intended: 7x^4

Step-by-step explanation:

As written, the Order of Operations specifies that the only denominator factor is 3, so the expression is interpreted as ...

$\dfrac{21x^3}{3}x^{-1}=\dfrac{21}{3}x^{3-1}=7x^2$

___

Sometimes grouping of denominator factors is intended, but parentheses are not written as they should be. If the intention is ...

21x^3/(3x^-1)

then the simplification is ...

$\dfrac{21x^3}{3x^{-1}}=\dfrac{21}{3}x^{3-(-1)}=7x^4$

_____

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

a^b/a^c = a^(b-c)

6 0

3 years ago

ASAP WHO EVER ANSWERS THIS WILL BE MARKED BRANLIEST

Kaylis [27]

Answer:

option A is the answer hope you find it helpful

3 0

3 years ago

Using a property of operations, what can you say about the sums of (-13.2) 8.1 and 13.2 (-8.1)

artcher [175]

No matter where the negative sign is, the product will always be negative, so is the number itself.

13.2(-8.1)=-106.92

twice of that would be -213.84.

first do the multiplying, then the adding(the adding of negative numbers, so it would be basically subtracting).

7 0

3 years ago