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3 years ago

Help please

Mathematics

1 answer:

Ivahew [28]3 years ago

7 0

QUESTION:

The amount you spend at the fair can be represented by the function: y=2x+10

Which statement describes what's happening correctly?

ANSWER:

[a] For each additional game played, the cost increases by $10.

QUESTION:

What will the function f(x) = x² look like translated 3 units right and 8 units down?

ANSWER:

To move up or down just add or subtract the constant equal to the units moved.

So, just to move it 8 units down, would give us

g(x) = x² - 8

Now, To move horizontally replace each x with (x ± c).

If you move it to the right, you need x - c
If you move it to the left , you need x + c

So , since we are moving 3 units to the right, we replace the x with x - 3.

g(x) = (x - 3)² - 8.

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What is the slope of the line that passes through the points (-2, -3) and (5,4) ? -3 3 -1 1

Andrews [41]

Answer:

$\boxed{\textsf{ The slope of the line is \textbf{1}.}}$

Step-by-step explanation:

Two points are given to us and we need to find the slope of the line that passes through these two points . The two points are (-2,-3) and (5,4) .As we know that the slope of the line is given by $\tan\theta$

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On using this formula we can find the slope of the line .

Putting the respective values :-

$\sf\implies Slope = \dfrac{y_2-y_1}{x_2-x_1}\\\\\sf\implies Slope = \dfrac{ -3-4}{-2-5} \\\\\sf\implies Slope =\dfrac{-7}{-7}\\\\\sf\implies\boxed{\pink{\sf Slope = 1 }}$

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3 years ago

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Step-by-step explanation:

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3 years ago

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(10 pts) A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstruc

dimulka [17.4K]

Answer:

309

Step-by-step explanation:

To determine the estimated free-flow speed

$FFS = 75.4 -f_{LW}-f_{LC} - 3.22 TRD^{0.84}$

From the table "Adjustment for lane width," which corresponds to a lane width of 12 feet. As a result, $f_{LW}$ equals 0 mi/h.

Take the meaning from the table "Adjustment for right-shoulder lateral clearance," which corresponds to 6 feet of right shoulder lateral clearance and two lanes in one direction.

The $f_{LC} = 0 mi/h$

$FFS = 75.4 -0-0-3.22 ( \dfrac{5}{6})^{0.84}$

$= 72.64 \ mi/h$

$\simeq 73 \ mi/h$

Determine the peak-hour factor

$PHF = \dfrac{V}{V_{15}\times 4} \\ \\ =\dfrac{1800}{700\times 4}\\ \\ = 0.6429$

Now, Find the heavy-vehicle adjustment factor.

$v_P = \dfrac{V}{PHF\times N \times f_{HV}\times f_P} --- (1)$

Take the value for the 15-minute passenger car equivalent flow rate from the table "LOS requirements for freeway segments" for the FFS and LOS C conditions. The free-flow speed is estimated to be 73 miles per hour

$v_p = 1735+ \dfrac{73-70}{75-70}(1775-1735)$

$v_p = 1759 \ pc/h/In$

Think about for familiar users the value of f_p = 1.00

Replace all of the values obtained in (1)

$v_p = \dfrac{V}{PHF \times N\times f_{HV}\times f_p}$

$1759 = \dfrac{1800}{0.6429\times 2 \times f_{HV}\times 1}$

$f_{HV} = \dfrac{1800}{0.6429\times 2 }\times \dfrac{1}{1759}$

= 0.795

Calculate the percentage of trucks and buses in the flow of traffic stream.

$f_{HV} = \dfrac{1}{1+P_{\tau}(E_{\tau}-1) + P_R(E_R-1)}$

Take the values from the table "Passenger car equivalent for extended highway segments" referring to trucks and buses and rolling terrains for passenger car equivalent for trucks and buses and recreational vehicles. As a result, $E_T$ has a value of 2.5 and $E_R$ has a value of 2.0.