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Svetllana [295]

2 years ago

11

a charter bus holding 80 passengers costs $59.50 an hour. If a round trip took 5 hours, what did it cost each passenger (round t

o the nearest cen

1 answer:

nika2105 [10]2 years ago

7 0

Answer:

$23,800

Step-by-step explanation:

59.5 x 5 x 80 = $23,800

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Write the equation....

BaLLatris [955]

1-0/-1-0 = -1
Y- 0 = -(x-0)
Y=-x
Y - 1 = -(x+1)
Y-1= -x-1
Y = -x

5 0

3 years ago

To rationalize the denominator of 2/square 13+ squared 11 , you should multiply the expression by which fraction?

eimsori [14]

<u><em>Answer:</em></u>

You should multiply the expression by $\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}$

<u><em>Explanation:</em></u>

To rationalize any expression, you must multiply it by its conjugate. A conjugate is defined as a similar expression to the original one but with an opposite sign

<u>This means that:</u>

The conjugate of a + b would be a - b

Now, the given expression is $\frac{2}{\sqrt{13}+\sqrt{11}}$

<u>Consider the denominator:</u>

From the above, we can conclude that the conjugate of $\sqrt{13}+\sqrt{11}$ is $\sqrt{13}-\sqrt{11}$

<u>And, remember that</u> we need to keep the value of the expression unchanged. This means that we must multiply both the numerator and the denominator by the same value

<u>Therefore:</u>

You should multiply the expression by $\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}$ in order to rationalize the denominator

Hope this helps :)

5 0

3 years ago

Read 2 more answers

1. Clare has a piece of paper that is 8 inches by 10 inches. How many pieces of paper

pishuonlain [190]

4 papers that are 2 inches by 2 1/2 inches

7 0

2 years ago

Read 2 more answers

NO LINKS!!! What is the equation of these two graphs?

S_A_V [24]

Answer:

$\textsf{1.} \quad y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}\:\:\textsf{ where}\:\:x \geq \dfrac{15-\sqrt{769}}{2}\\\\\quad \textsf{or} \quad y=2\sqrt{x+5}$

$\textsf{2.} \quad y=-|x+1|+5$

Step-by-step explanation:

<h3><u>Question 1</u></h3>

<u>Method 1 - modelling as a quadratic with restricted domain</u>

Assuming that the points given on the graph are points that the <u>curve passes through</u>, the curve can be modeled as a quadratic with a limited domain. Please note that as the x-intercept has not been defined on the graph, I am not including this in this first method.

Standard form of a quadratic equation:

$y=ax^2+bx+c$

Given points:

(-4, 2)
(-1, 4)
(4, 6)

Substitute the given points into the equation to create 3 equations:

<u>Equation 1 (-4, 2)</u>

$\implies a(-4)^2+b(-4)+c=2$

$\implies 16a-4b+c=2$

<u>Equation 2 (-1, 4)</u>

$\implies a(-1)^2+b(-1)+c=4$

$\implies a-b+c=4$

<u>Equation 3 (4, 6)</u>

$\implies a(4)^2+b(4)+c=6$

$\implies 16a+4b+c=6$

Subtract Equation 1 from Equation 3 to eliminate variables a and c:

$\implies (16a+4b+c)-(16a-4b+c)=6-2$

$\implies 8b=4$

$\implies b=\dfrac{4}{8}$

$\implies b=\dfrac{1}{2}$

Subtract Equation 2 from Equation 3 to eliminate variable c:

$\implies (16a+4b+c)-(a-b+c)=6-4$

$\implies 15a+5b=2$

$\implies 15a=2-5b$

$\implies a=\dfrac{2-5b}{15}$

Substitute found value of b into the expression for a and solve for a:

$\implies a=\dfrac{2-5(\frac{1}{2})}{15}$

$\implies a=-\dfrac{1}{30}$

Substitute found values of a and b into Equation 2 and solve for c:

$\implies a-b+c=4$

$\implies -\dfrac{1}{30}-\dfrac{1}{2}+c=4$

$\implies c=\dfrac{68}{15}$

Therefore, the equation of the graph is:

$y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}$

$\textsf{with the restricted domain}: \quad x \geq \dfrac{15-\sqrt{769}}{2}$

<u>Method 2 - modelling as a square root function</u>

Assuming that the points given on the graph are points that the <u>curve passes through</u>, and the x-intercept should be included, we can model this curve as a <u>square root function</u>.

Given points:

(-4, 2)
(-1, 4)
(4, 6)
(0, -5)

The parent function is:

$y=\sqrt{x}$

Translated 5 units left so that the x-intercept is (0, -5):

$\implies y=\sqrt{x+5}$

The curve is stretched vertically, so:

$\implies y=a\sqrt{x+5} \quad \textsf{(where a is some constant)}$

To find a, substitute the coordinates of the given points:

$\implies a\sqrt{-4+5}=2$

$\implies a=2$

$\implies a\sqrt{-1+5}=4$

$\implies 2a=4$

$\implies a=2$

$\implies a\sqrt{4+5}=6$

$\implies 3a=6$

$\implies a=2$

As the value of a is the same for all points, the equation of the line is:

$y=2\sqrt{x+5}$

<h3><u>Question 2</u></h3>

<u>Vertex form of an absolute value function</u>

$f(x)=a|x-h|+k$

where:

(h, k) is the vertex
a is some constant

From inspection of the given graph:

vertex = (-1, 5)
point on graph = (0, 4)

Substitute the given values into the function and solve for a:

$\implies a|0-(-1)|+5=4$

$\implies a+5=4$

$\implies a=-1$

Substituting the given vertex and the found value of a into the function, the equation of the graph is:

$y=-|x+1|+5$

7 0

1 year ago

Solve the following equation.<br> 5t - 3 = 14

sweet [91]

Answer:

Step-by-step explanation:

5t - 3 = 14

5t=14+3

5t=17

t=17/5

t=3.4

6 0

2 years ago

Read 2 more answers