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3 years ago

14

Estimate 462809 + 256738

1 answer:

mafiozo [28]3 years ago

5 0

462809 round off to 500000.
256738 round off to 300000.
500000+300000=800000

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Kyle's car average 380 miles on a 16-gallon tank of gas. How far can he drive on 6 gallons of gas?

mr Goodwill [35]

380 / 16 = x / 6
16x = 2,280
x = 142.5

Answer is C!

6 0

4 years ago

Water has a density of 1.0 g/cm3. What is the mass of 10.0 cm3 of water?

fgiga [73]

1.0g/cm3 means that the mass of one cm3 is 1.0g
The easiest method to use is the rule of three, and let x be the mass of 10.0 cm3 of water
1g -- > 1.0 cm3
x --> 10.0 cm3

x= (10*1)/1
x=10.0 g

So the mass of 10.0 cm3 of water is 10.0g

Hope this Helps! :)

8 0

4 years ago

Read 2 more answers

Rewrite these expressions using words: 3/4x(2 2/5-5/6)

myrzilka [38]

Answer:

Step by atep explaination :

$\frac{3}{4}$ means three₋fourth

$2\frac{2}{5}$ means two and two₋fifths

$\frac{5}{6}$ means five₋sixth

An algebraic expression in words can be written as

A numbers three₋fourth times the subtractions of five₋sixths from two and two₋fifths

5 0

4 years ago

Arrange the following in ascending order<br><br><br><br> 3/4 , -1/3 , 1/-2 , 3/10

Over [174]

Answer:

Ans: 1/-2 , -1/3 , 3/10 , 3/4

Step-by-step explanation:

The LCM of 2, 3, 4 and 10 is 60.

Let's make denominator equal,

→ 3/4 , -1/3 , 1/-2 , 3/10

→ 3×15/4×15 , -1×20/3×20 , -1×30/2×30 , 3×6/10×6

→ 45/60 , -20/60 , -30/60 , 18/60

Now we can order in ascending,

→ -30/60 , -20/60 , 18/60 , 45/60

→ 1/-2 , -1/3 , 3/10 , 3/4 {final answer}

7 0

3 years ago

Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)

Drupady [299]

Answer:

The equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

Let us solve the expression step by step to get the equivalent

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

as

$\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=x^{2\cdot \frac{1}{7}}$

$=x^{\frac{2}{7}}$

also

$\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=y^{3\cdot \frac{1}{5}}$

$=y^{\frac{3}{5}}$

so the expression becomes

$\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

$=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$ ∵ $\:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}$

Thus, the equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

5 0

3 years ago