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3 years ago

What is the slope of a line parallel to the line 2x + 3y = 5?

Mathematics

2 answers:

babunello [35]3 years ago

8 0

Answer:

Step-by-step explanation:

Parallel lines have same slope

Write the equation in slope - intercept form

2x + 3y =5

3y = -2x + 5

Divide the equation by 3

$y = \frac{-2}{3}x +\frac{5}{3}$

So, slope = (-2/3)

SOVA2 [1]3 years ago

6 0

i think its 2/3 but im not sure .

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3 years ago

How to do question 22?

Aleks04 [339]

Answer:

A = 20sinθ(6 + 5 cosθ) cm²

Step-by-step explanation:

Drop perpendiculars DE and CF to AB.

Then, we have congruent triangles ADE and BCF, plus the rectangle CDEF.

The formula for the area of the trapezium is

A = ½(a + b)h

DE = 10sinθ

AE = 10cosθ

BF = 10cosθ

EF = CD = 12 cm

AB = AE + EF + BF = 10cosθ + 12 + 10 cosθ = 12 + 20cosθ

A = ½(a + b)h

= ½(12 +12 + 20 cosθ) × 10 sinθ

=(24 + 20 cosθ) × 5 sinθ

= 4(6 + 5cosθ) × 5sinθ

= 20sinθ(6 + 5 cosθ) cm²

6 0

3 years ago

Which equation is the slope-intercept form of the line that passes through (-3,5) and is parallel to the of y=−2\3x+5

irinina [24]

Answer:

y = -2/3x + 3

Step-by-step explanation:

y = -2/3x + b

5 = -2/3(-3) + b

5 = 2 + b

3 = b

8 0

3 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

3 years ago

Given the volume is 105 inches³ for an 8 pound bowling ball, find the radius of the bowling ball,

mariarad [96]

Answer:

2.92

Step-by-step explanation:

$V=\frac{4}{3}\pi r^3 \\ \\ 105=\frac{4}{3}\pi r^3 \\ \\ r^3=\frac{105}{\frac{4}{3}\pi} \\ \\ r=\sqrt[3]{\frac{105}{\frac{4}{3}\pi}} \\ \\ r \approx 2.92$

7 0

1 year ago