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3 years ago

PLEASE HELP

Mathematics

2 answers:

tekilochka [14]3 years ago

7 0

Answer:

PLEASE HELP

For the right triangles below, find the values of the side lengths b and c.

Round your answers to the nearest tenth.

Step-by-step explanation:

PLEASE HELP

For the right triangles below, find the values of the side lengths b and c.

Round your answers to the nearest tenth.

Damm [24]3 years ago

6 0

Answer:

b = 1.7

c = 7.1

Step-by-step explanation:

the triangle with side 'b' is a 30-60-90° Δ where the ratio of sides, respectively, are 1 : $\sqrt{3}$ : 2

to find 'b' we could set up this proportion:

b/3 = 1/ $\sqrt{3}$

cross-multiply:

b $\sqrt{3}$ = 3

b = 3 / $\sqrt{3}$

if we rationalize the denominator, this becomes (3 $\sqrt{3}$ )/3 which is $\sqrt{3}$ or 1.7

the other triangle is a 45-45-90°Δ where the ratio of the sides are 1 : 1 : $\sqrt{2}$

if the legs are 5 then the hypotenuse is 5 $\sqrt{2}$ or 7.1

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In ACE, G is the centroid and BE=9. Find BG and GE

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4 years ago

(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.