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kakasveta [241]
3 years ago
15

Mr.russo exchanged 200 euros for 270 dollars how many dollar he get foe 300 euros

Mathematics
2 answers:
dimulka [17.4K]3 years ago
6 0
We must fine the conversion for this, so we have to divide $270 by 200€ to get 1.35 Dollars for every Euro..
Now we multiply 300 by 1.35 to get 405 dollars
alukav5142 [94]3 years ago
6 0

Answer:

$405

Step-by-step explanation:

This is a proportion problem!

Build fractions that have euros on top and dollars on the bottom (or the other way around, you decide, but make it the same in both fractions).

Let's do euros over dollars.

Setup:

\frac{200}{270}=\frac{300}{d}

"Cross multiply."

200d=300\times 270\\\\200d=81000\\\\d=\frac{81000}{200}=405

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two positive integers are 3 units apart on a number line. their product is 180. wich equation can be used to solve for m, the gr
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Answer:

m^2-3m-180=0

m=15

Step-by-step explanation:

greater integer = m

small integer = n

we know that m = n+3 or n = m-3

we also know that n*m = 180

replace n for m-3

(m-3)*m=180

or

m^2-3m-180 = 0

(m-15)(m+12)=0

so m = 15 or -12, but since problem states positive, m=15

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3 years ago
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BartSMP [9]
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3 years ago
9/25 ddivided by 729/1000
Nat2105 [25]
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5 0
3 years ago
Read 2 more answers
If tan e = 1.3, what is the value of cot (90° - 9)?
ella [17]

Answer:

A

Step-by-step explanation:

We are given that:

\tan(\theta)=1.3

And we want to find:

\cot(90^\circ -\theta)

Remember that tangent and cotangent are co-functions. In other words, they follow the cofunction identities:

\tan(\theta)=\cot(90^\circ-\theta)\text{ and } \cot(90^\circ-\theta)=\tan(\theta)

Therefore, since tan(θ) = 1.3 and cot(90° - θ) = tan(θ), then cot(90° - θ) must also be 1.3.

Our answer is A.

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%20lnx%20%7D%20" id="TexFormula1" title=" \sqrt[4]{ lnx } " alt=" \sqrt[4]
Kamila [148]

Use the power rule for differentiation:

\dfrac{\text{d}}{\text{d}x} (f(x))^k = k(f(x))^{k-1}f'(x)

You can use this formula if you remember that a root is just a rational exponential:

\sqrt[4]\ln(x) = (\ln(x))^{\frac{1}{4}}

So, remembering that the derivative of the logarithm is 1/x, you have

\dfrac{\text{d}}{\text{d}x} (\ln(x))^{\frac{1}{4}} = \frac{1}{4}(\ln(x))^{\frac{1}{4}-1}\dfrac{1}{x}

Which you can rewrite as

\dfrac{1}{4}(\ln(x))^{\frac{1}{4}-1}\dfrac{1}{x} =\dfrac{1}{4}(\ln(x))^{\frac{-3}{4}}\dfrac{1}{x} =\dfrac{1}{4}\dfrac{1}{\sqrt[4]{\ln(x))^3}}\dfrac{1}{x} = \dfrac{1}{4x\sqrt[4]{\ln(x))^3}}

3 0
3 years ago
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