All categories

2 years ago

If the exchange rate is €1.20 to £1, how many pounds sterling would you get for €180?

Mathematics

1 answer:

Len [333]2 years ago

4 0

Answer:

150 pounds

Step-by-step explanation:

1.20 euros = 1 pound

180 euros = x pounds

Set up the proportion

1.20 Euros/1 pound = 180 Euros / x pound Cross multiply

1.20*x = 1 * 180 Combine

1.20 x = 180 Divide by 1.2

x = 180 / 1.20

x = 150 pounds.

As a bonus, you might want to try and guess which is the stronger currency.

You might be interested in

Layana’s house is located at (2 and two-thirds, 7 and one-third) on a map. The store where she works is located at (–1 and one-t

NeTakaya

We have been given that Layana’s house is located at $(2\frac{2}{3}, 7\frac{1}{3})$ on a map. The store where she works is located at $(-1\frac{1}{3}, 7\frac{1}{3})$ .

We are asked to find the distance from Layana’s home to the store

We will use distance formula to solve our given problem.

Let us convert our given coordinates in improper fractions.

$2\frac{2}{3}\Rightarrow \frac{8}{3}$

$7\frac{1}{3}\Rightarrow \frac{22}{3}$

$-1\frac{1}{3}\Rightarrow -\frac{4}{3}$

Now we will use distance formula to solve our given problem.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Upon substituting coordinates of our given point in above formula, we will get:

$D=\sqrt{(\frac{22}{3}-\frac{22}{3})^2+(\frac{8}{3}-(-\frac{4}{3}))^2}$

$D=\sqrt{(0)^2+(\frac{8}{3}+\frac{4}{3})^2}$

$D=\sqrt{0+(\frac{8+4}{3})^2}$

$D=\sqrt{(\frac{12}{3})^2}$

$D=\sqrt{(4)^2}$

$D=4$

Therefore, the distance from Layana's home to the store is 4 units and option A is the correct choice.

3 0

3 years ago

Read 2 more answers

Simplify the expression

vekshin1

Answer:

the answer is c

Step-by-step explanation:

6 0

3 years ago

Which statement about the graph of the function f(x)=2x^2-x-6 are true?

Snezhnost [94]

C the vertex of the function

7 0

3 years ago

Read 2 more answers

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

3 years ago

If the hypotenuse of a right triangle is 2 <img src="https://tex.z-dn.net/?f=%5Csqrt%7B3%7D" id="TexFormula1" title="\sqrt{3}" a

Svetradugi [14.3K]

Answer:

Step-by-step explanation:

a = the other leg:

$a^{2} =(2\sqrt{3} )^{2} -(\sqrt{3}) ^{2} =(4)(3)-3=12-3$