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valentinak56 [21]

3 years ago

8

Points A and B are midpoints of the sides of triangle QRS. Triangle Q R S is cut by line segment A B. Point A is the midpoint of

side Q S and point B is the midpoint of side R S. The length of Q R is 6 meters, the length of Q A is 4 meters, the length of A B is 3 meters, and the length of R B is 3 meters. What is SA? 2 m 3 m 4 m 6 m

2 answers:

Murrr4er [49]3 years ago

6 0

Answer:

its A. 2 units

Step-by-step explanation:

Scilla [17]3 years ago

3 0

Answer: 4m

Step-by-step explanation:

Note : midpoints divide a line into two equal parts.

To calculate the length of SA;

We can consider the large triangle QRS with

Two midpoints :

A on the line QS, QA = 4m ;

and B on the line RS with RB = 3m

Fron the definition of midpoint :

QS = QA + AS

where QA = AS

QS = 2QA = 2AS

Therefore,

QA = QS

4m = 4m

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stich3 [128]

Given:

The expression is

$\dfrac{2}{a-2}-\dfrac{8}{a^2-4}$

To find:

The simplified form of the given expression.

Solution:

We have,

$\dfrac{2}{a-2}-\dfrac{8}{a^2-4}$

It can be written as

$=\dfrac{2}{a-2}-\dfrac{8}{a^2-2^2}$

$=\dfrac{2}{a-2}-\dfrac{8}{(a-2)(a+2)}$ $[\because a^2-b^2=(a-b)(a+b)]$

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$=\dfrac{2(a+2)-8}{(a-2)(a+2)}$

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$=\dfrac{2a-4}{(a-2)(a+2)}$

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A local club is arranging a charter flight to Hawaii. The cost of the trip is $586 each for 80 passengers, with a refund of $

professor190 [17]

Answer:

a) The number of passengers that will maximize the revenue received from the flight is 99.

b) The maximum revenue is $48,609.

Step-by-step explanation:

We have to analyse two cases to build a piecewise function.

If there are 80 or less passengers, we have that:

The cost of the trip is $586 for each passenger. So

$R(n) = 586n$

If there are more than 80 passengers.

There is a refund of $5 per passenger for each passenger in excess of 80. So the cost for each passenger is

$R(n) = (586 - 5(n-80))n = -5n^{2} +400n + 586n = -5n^{2} + 986n$ .

So we have the following piecewise function:

$R(n) = \left \{ {{586n}, n\leq 80 \atop {-5n^{2} + 986n}, n > 80} \right$

The maxium value of a quadratic function in the format of $y(n) = an^{2} + bn + c$ happens at:

$n_{v} = -\frac{b}{2a}$

The maximum value is:

$y(n_{v})$

So:

(a) Find the number of passengers that will maximize the revenue received from the flight.

We have to see if $n_{v}$ is higher than 80.

We have that, for $n > 80$ , $R(n) = -5n^{2} + 986n$ , so $a = -5, b = 986$

The number of passengers that will maximize the revenue received from the flight is:

$n_{v} = -\frac{b}{2a} = -\frac{986}{2(-5)} = 98.6$

Rounding up, the number of passengers that will maximize the revenue received from the flight is 99.

(b) Find the maximum revenue.

This is $R(99)$ .

$R(n) = -5n^{2} + 986n$

$R(99) = -5*(99)^{2} + 986*(99) = 48609$

The maximum revenue is $48,609.

8 0

3 years ago

The ratio of 15 days to a week is

olga2289 [7]

<h2>Answer :</h2>

The ratio of 15 days to a week is 15 : 7.

$\\$

<h2>Explanation :</h2>

Ratio of 15 days to a week :

$\sf : \implies 15 \: days : 1 \: week = \dfrac{15 \: days}{1 \: week}$

$\\$

Now, we know that 1 week has 7 days.

$\sf : \implies 15 \: days : 1 \: week = \dfrac{15 \: \cancel{days}}{7 \: \cancel{days}}$

$\sf : \implies 15 \: days : 1 \: week = \dfrac{15}{7}$

$\sf : \implies 15 \: days : 1 \: week = 15 : 7$

$\\$

Hence, The ratio of 15 days to a week is 15 : 7.

7 0

3 years ago