Alaina has $28 in her account. She wants to purchase a pair of shoes that costs $45. If Alaina makes the purchase, which

If £1 = 1.60 euros, what is £4.50 worth in euros

Vedmedyk [2.9K]

I think it’s 7.20 euros..

1 x 4.5 = £4.50
1.60 x 4.5 = 7.20 euros ?

8 0

3 years ago

Which equation represents this problem? The time it takes Tomás to bike to the community center is 2 minutes less than half the

Helen [10]

Answer:

The equation that represents this problem is $\frac{1}{2}$ p - 2 = 18 ⇒ D

Step-by-step explanation:

Let us solve the question

∵ p is the time it takes Tomás to bike to the park

∵ The time it takes Tomás to bike to the community center is 2 minutes

less than half the time it takes him to bike to the city park

→ That means multiply p by $\frac{1}{2}$ , then subtract 2 from it

∴ The time to the community center = $\frac{1}{2}$ (p) - 2

∴ The time to the community center = $\frac{1}{2}$ p - 2

∵ It takes Tomás 18 minutes to bike to the community center

→ Equate The time to the community center by 18

∴ $\frac{1}{2}$ p - 2 = 18

∴ The equation that represents this problem is $\frac{1}{2}$ p - 2 = 18

7 0

3 years ago

Please help me on this problem

Goshia [24]

D: {1,2,3,4}
R: {1,2,3,4}
Function? Yes, the relation is a function.

6 0

4 years ago

The angles of depression of two ships from the top of the light house are 45° and 30° towards east. If the ships are 200 m apart

azamat

Answer:

273 meters

Step-by-step explanation:

See image attached for the diagram I used to represent this scenario.

The distance between the ships, at angles 30 and 45, is 200 meters. The distance between the left ship and the lighthouse is x meters.

We can use trigonometric ratios to solve this problem. We can use the tangent ratio $\big{(} \frac{\text{opposite}}{\text{adjacent}} \big{)}$ to create an equation with the two angles.

$\displaystyle \text{tan(45)} = \frac{h}{x}$
$\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Let's take these two equations and solve for x in both of them.

<h2> $\textbf{Equation I}$ </h2>

$\displaystyle \text{tan(45)} = \frac{h}{x}$

tan(45) = 1, so we can rewrite this equation.

$\displaystyle 1=\frac{h}{x}$

Multiply x to both sides of the equation.

$\displaystyle x = h$

<h2> $\textbf{Equation II}$ </h2>

$\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Multiply x + 200 to both sides and divide h by tan(30).

$\displaystyle \text{x + 200} = \frac{h}{\text{tan (30)}}$

Subtract 200 from both sides of the equation.

$\displaystyle \text{x} = \frac{h}{\text{tan (30)}} - 200$

Simplify h/tan(30).

$x=\sqrt{3}h - 200$

<h2> $\textbf{Equation I = Equation II}$ </h2>

Take Equation I and Equation II and set them equal to each other.

$h=\sqrt{3}h-200$

Subtract √3 h from both sides of the equation.

$h-\sqrt{3}h=-200$

Factor h from the left side of the equation.

$h(1-\sqrt{3}) =-200$

Divide both sides of the equation by 1 - √3.

$\displaystyle h=\frac{-200}{1-\sqrt{3} }$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate.

$\displaystyle h=\frac{-200}{1-\sqrt{3} } \big{(} \frac{1+\sqrt{3} }{1+\sqrt{3}} \big{)}$
$\displaystyle h=\frac{-200+200\sqrt{3} }{1-3}$
$\displaystyle h =\frac{-200+200\sqrt{3} }{-2}$