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3 years ago

Evaluate the expression: 2/5 x 0.6

Mathematics

2 answers:

zysi [14]3 years ago

8 0

You can do 2/5 =0.4 than multiply 0.4 * 0.6 to get 0.24

astra-53 [7]3 years ago

5 0

I believe the answer would be 6/25 simplified. 12/50 would be the original answer

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photoshop1234 [79]

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3 years ago

How do I work this type of problem

stich3 [128]

Answer:

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3 years ago

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Question 3

Crank

Step-by-step explanation:

Given that,

The mass of a patient = 70 kg

A 70kg patient has approximately 8 pints of blood.

The patient donates 470mL of blood.

We know that,

1 pint = 568 mL

8 pints = 4544 mL

Required fraction,

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So, the required fraction is approximately $\dfrac{1}{10}$ .

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3 years ago

joseph says that there is 5/8 of a ham sandwich left and 1/2 of a turkey sandwich left . what is not a pair of equivalent fracti

tatyana61 [14]

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7 0

3 years ago

Simplify the expression below and write it as a single logarithm:

OLEGan [10]

The simplification of 3log(x + 4) – 2log(x – 7) + 5log(x - 2) - log(x^2) is $\log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

<u>Solution:</u>

Given, expression is $3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

We have to write in as single logarithm by simplifying it.

Now, take the given expression.

$\rightarrow 3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

Rearranging the terms we get,

$\left.\rightarrow 3 \log (x+4)+5 \log (x-2)-2 \log (x-7)+\log \left(x^{2}\right)\right)$

$\text { since a } \times \log b=\log \left(b^{a}\right)$

$\rightarrow \log (x+4)^{3}+\log (x-2)^{5}-\left(\log (x-7)^{2}+\log \left(x^{2}\right)\right)$

$\text { We know that } \log a \times \log b=\log a b$

$\rightarrow \log \left((x+4)^{3} \times(x-2)^{5}\right)-\left(\log \left((x-7)^{2} \times\left(x^{2}\right)\right)\right.$

$\text { We know that } \log a-\log b=\log \frac{a}{b}$

$\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

Hence, the simplified form $\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

4 0

3 years ago

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