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Anna [14]
3 years ago
13

The equation of a line is 3x−5y=8. Select three lines that are parallel to the given line.

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
5 0

Answer:

The Given Equation is

3x - 5y = 8

Slopes of paralell lines are equal

So option 3, 4, 5 are correct

--------------

Hope it helps..

Have a great day!!!

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Find the 15th term of the geometric sequence 2, 6, 18,...
kenny6666 [7]

Answer:

<h3>           a₁₅ = 9 565 938</h3>

Step-by-step explanation:

6÷2 = 3  and   18÷6 = 3  so:  q = 3

a_n=a_1\cdot q^{n-1}\\\\a_1=2\\q=3\\n=15\\\\a_{15}=2\cdot3^{15-1}\\\\a_{15}=2\cdot3^{14}\\\\a_{15}=9\,565\,938

7 0
3 years ago
1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is
sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by d(t)=\frac{t^{2}}{6}+4t where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}

so let's take the first interval:

(a) [1.95, 2.05]

V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}

we get that:

d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125

d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167

so:

V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s

(b) [1.995, 2.005]

V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}

we get that:

d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831

d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835

so:

V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s

(c) [1.9995, 2.0005]

V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}

we get that:

d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358

d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358

so:

V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s

(d) [2, 2.00001]

V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}

we get that:

d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333

d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333

so:

V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

8 0
3 years ago
The three steps below were used to find the value of the expression [(-10+2)-1]+(2+3)​
brilliants [131]

Answer:

value of expression = 12

Step-by-step explanation:

[(-10+2)-1]+(2+3)​

=  [(8)-1]+(5)​              ∵according to PEMDAS, we first solve the Parenthesis

= (7)+(5)

= 12

8 0
3 years ago
use the coordinate plane above. write the ordered pairs that names each point. then identify the quadrant where each point is lo
m_a_m_a [10]

Answer:

there is no picture

Step-by-step explanation:

i wish i could help tho lol

7 0
2 years ago
six scholars want to share 9 bags of popcorn.how much popcorn will each scholar get?.draw a model to prove your answer
gregori [183]

each person will get 1 and a half bag of popcorn. to draw a model you can either do a table or you can draw graph

6 0
3 years ago
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