Find R if p = 1.5 and R = (1400

Rewrite the equations in the form ax^2+bx+c =0:<br> -5x (x +6)=4(x-3) -10<br> 4x^2-2x(3x+1)=5

Rainbow [258]

Step-by-step explanation:

this is my answer...sorry if I made any mistake

4 0

3 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.

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

Simplify this equation.

$\displaystyle h=100+100\sqrt{3}$
$h=273.20508075$

The height of the lighthouse is about 273 meters.

5 0

3 years ago

Read 2 more answers

Frank says that 50% of a number will always be greater than 20% of any other number. Complete one inequality to support Frank's

Archy [21]

Hi there! I saw your question and decided to help you out.

To support his claim:

50 of 100 and 20 of 100

To debunk his claim:

50 of 10 and 20 of 100

I hope this answers your question !

5 0

3 years ago

Read 2 more answers

1/3 - 1/2 x 1/2 ?? Help please

Liula [17]

Answer:

1/2

Step-by-step explanation:

1

3

−

1

2

(1)

2

=

1

3

−

1

2

=

1

3

−

1

4

=

1/12

5 0

3 years ago

7 Point Question Tyler was trying to prove that all rhombuses are similar. He thought, “If I draw 2 rhombuses, I can find dilati

Viefleur [7K]

A rhombus can have the same shape and dimension as a square or a

parallelogram, therefore, all rhombuses are not similar.

A step in the proof that have flaws is <u>step 3</u>.

Reasons:

By reading through the given steps, we have, the first step that is incorrect

is step 3.

Step 3: The requirement for the similarity of two figures based on the

proportionality of the corresponding sides of the figures is related with

triangles as given by the triangle proportionality theorem.

This is so because, the lengths of the sides of a triangle depends on the

opposite interior angle, while the side of a rhombus can have a

combination of large length and and acute angle.

Two rhombuses that have proportional sides may be dissimilar,

because they can have different interior angles.

Therefore;

<u>The step in Tyler's proof that have a flaw is Step 3</u>

Learn more about similarity between geometric figures here:

brainly.com/question/10270676

7 0

2 years ago

Find R if p = 1.5 and R = (1400 – 500p)p R=?