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3 years ago

Estimate by rounding to the nearest ten. Then, find the sum.441+345=?

Mathematics

1 answer:

cricket20 [7]3 years ago

3 0

441 to the nearest ten is 440
345 to the nearest ten is 350

The sum is;

440 + 350 = 790

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Given:

$\log_34\approx 1.262$

$\log_37\approx 1.771$

To find:

The value of $\log_3\left(\dfrac{4}{49}\right)$ .

Solution:

We have,

$\log_34\approx 1.262$

$\log_37\approx 1.771$

Using properties of log, we get

$\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349$ $\left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]$

$\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2$

$\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37$ $[\log x^n=n\log x]$

Substitute $\log_34\approx 1.262$ and $\log_37\approx 1.771$ .

$\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)$

$\log_3\left(\dfrac{4}{49}\right)=1.262-3.542$

$\log_3\left(\dfrac{4}{49}\right)=-2.28$

Therefore, the value of $\log_3\left(\dfrac{4}{49}\right)$ is $-2.28$ .

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3 years ago

Estimate by rounding to the nearest ten.​ Then, find the sum.441+345=?

Estimate by rounding to the nearest ten. Then, find the sum.441+345=?