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3 years ago

What is the unit value of 6 in 216

Mathematics

2 answers:

mylen [45]3 years ago

6 0

The 6 in 216 is in the ones place.

fomenos3 years ago

6 0

The correct answer is:

The ones place.

Explanation:

From the end of a whole number left, we start with the ones place. The next place to the right is the tens place; then the hundredths place. Beyond this would be the thousands place.

Since 6 is the first number at the end of the whole number, 6 is in the ones place. Its value is 6(1) = 6.

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Jim is showing his work in simplifying (−8.5 + 6.1) − 1.3.

dexar [7]

9514 1404 393

Answer:

In step 4, Jim's answer is incorrect.

Step-by-step explanation:

In step 1, Jim swaps the order of addends using the commutative property of addition.

In step 2, Jim uses the distributive property to factor -1 from the final two terms. (The associative property lets Jim move parentheses.)

6.1 +(-8.5 -1.3) . . . associative property

6.1 +(-1)(8.5 +1.3) . . . distributive property

In step 3, Jim has used the properties of real numbers to form the sum of two of them.

In step 4, Jim wrote an answer of 1.1, when the answer should have been -3.7. Jim's answer is incorrect.

The descriptive statements about steps 2 and 4 are both true.

8 0

2 years ago

Given: 5x + 3 > 4x + 7. Choose the graph of the solution set.

Natali5045456 [20]

Answer:

5x+3>4x+7

-4x -4x

x+3>7

-3 -3

x>4

so the answer will be the second

Step-by-step explanation:

7 0

3 years ago

Simplify: cos2x-cos4 all over sin2x + sin 4x

GrogVix [38]

Answer:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

Step-by-step explanation:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}$

Apply formula:

$\cos\left(A\right)-\cos\left(B\right)=-2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$ and

$\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$

We get:

$=\frac{-2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\sin\left(\frac{2x-4x}{2}\right)}{2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\cos\left(\frac{2x-4x}{2}\right)}$