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USPshnik [31]
2 years ago
5

Of 6.5 hectoliters of fuel, the private spilled 350 milliliters. how many liters did the private spill?

Mathematics
1 answer:
sergiy2304 [10]2 years ago
7 0

The private spilled 0.35 litres of fuel

<h3>How to calculate the amount of litres spilled ? </h3>

The first step is to convert 6.5 hectolitres to litres

= 6.5 × 100

= 650 litres

Next is to convert millilitres to litres

= 350/1000

= 0.35 litres

Hence the number of litres spilled is 0.35 litres

Read more here

brainly.com/question/18239674?referrer=searchResults

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The answer would be $12.15
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Ve the equation. 3c + 1 = c +1​
yKpoI14uk [10]
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Hope this helped
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What ratio is equivalent to 9 6
elena-s [515]
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2 years ago
What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10
jasenka [17]

<u>Answer:</u>

 The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

<u>Solution: </u>

The given expression is (2 x-3 y)^{10}

As per binomial theorem, we know,

(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now x^{5}\times y^5 will be the 6 term where k =5

Now, \mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}

So, the coefficient of x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right.

10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252

The coefficient of x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552

6 0
3 years ago
what is the polynomial function of lowest degree with leading coefficient of 1 and roots 2 and 1 + square root2
Marta_Voda [28]
If the roots to such a polynomial are 2 and 1+\sqrt2, then we can write it as

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which would be the correct answer, but clearly this option is not listed. Which is silly, because none of the offered solutions are *the* polynomial of lowest degree and leading coefficient 1.

So this makes me think you're expected to increase the multiplicity of one of the given roots, or you're expected to pull another root out of thin air. Judging by the choices, I think it's the latter, and that you're somehow supposed to know to use 1-\sqrt2 as a root. In this case, that would make our polynomial

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Whoever originally wrote this question should reevaluate their word choice...
5 0
2 years ago
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