All categories

3 years ago

4 thousands equal how many hundreds

Mathematics

2 answers:

Zinaida [17]3 years ago

7 0

4 thousands equals 40 hundreds.

KiRa [710]3 years ago

5 0

4 thousands equals 40 hundreds

Hope this helps.

oops i had it wrong.

You might be interested in

Two numbers have a product of 48 and a sum of 19. What are the numbers??

Elis [28]

16 * 3 = 48

and

16 + 3 = 19

5 0

3 years ago

Read 2 more answers

Alex wrote 103.903 in expanded form as 100+3 x 1+9 x 1/100 +3 x 1/1000. Explain his mistake then tell how to correct it.

Damm [24]

Your mistake was in this part 9 x 1/100 . After the decimal point the place value starts with tenths so you should write 9 x 1/10 .

Place values

Before the decimal point

3 ones = 3 x 1

0 tens = 0 x 10

1 hundreds = 1 x 100

After the decimal point

9 tenths = 9 x 1/10

0 hundredths = 0 x 1/100

3 thousandths= 3 x 1/1000

1 x 100 + 0 x 10 + 3 x 1 + 9 x 1/10 + 0 x 1/100 + 3 x 1/1000

100 + 0 + 3 + 0.9 + 0 + 0.003

100 + 3 + 0.9 + 0.003 = 103.903

8 0

3 years ago

Read 2 more answers

Simplify the expression(a4−6a2b2+ b4)−(−2a4+5a2b2+ 3b4) and hence, find its value for a = 2and b = −1.

ELEN [110]

Answer:

(a) $3a^4 - 11a^2b^2 -2b^4$

(b) $3a^4 - 11a^2b^2 -2b^4= 2$

Step-by-step explanation:

Given

$(a^4 - 6a^2b^2+ b^4) - (-2a^4+5a^2b^2+ 3b^4)$

Solving (a): Simplify

$(a^4 - 6a^2b^2+ b^4) - (-2a^4+5a^2b^2+ 3b^4)$

Open brackets

$a^4 - 6a^2b^2+ b^4 +2a^4-5a^2b^2- 3b^4$

Collect Like Terms

$a^4 +2a^4- 6a^2b^2-5a^2b^2+ b^4 - 3b^4$

Simplify Like Terms

$3a^4 - 11a^2b^2 -2b^4$

Solving (b): Simplify when a = 2 and b = -1

$3a^4 - 11a^2b^2 -2b^4$

$3*(2)^4 - 11*(2^2)*(-1)^2 -2*(-1)^4$

$3 * 16 - 11 * 4 * 1 - 2 * 1$

$48 - 44 - 2$

$= 2$

Hence:

$3a^4 - 11a^2b^2 -2b^4= 2$

4 0

2 years ago

Round 53,579 to the nearest thousand

ololo11 [35]

54,000 if the column before is 5 or more, you round up

4 0

2 years ago

Jan is x years old, her sister is twice as old as her, and her brother is three times older than her. Their ages add up to 29. H

victus00 [196]

Jan's age = $4\frac{5}{6}$ years

Sister's age = $9\frac{2}{3} \: years$

Step-by-step explanation:

Jan's age = x years

Sister's age = 2x years

Brother's age = 3x years

According to the given condition:

x + 2x + 3x = 29

6x = 29

x = 29/6

x = $4\frac{5}{6}$ years

2x = $2\times \frac{29}{6}= \frac{29}{3}=9\frac{2}{3} \: years$

Therefore,

Jan's age = $4\frac{5}{6}$ years

Sister's age = $9\frac{2}{3} \: years$

3 0

3 years ago