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Ganezh [65]
3 years ago
5

Find the eighth term of an arithmetic sequence whose fourth term is 19 and whose fifteenth is 52.

Mathematics
2 answers:
Ganezh [65]3 years ago
7 0

Answer:

The correct answer is actually B, 31

Step-by-step explanation:

Here is a link to a video giving an in depth explanation, just delete all of the (space)s

https://( )www.numerade( ).com/questions/find-the-( )indicated-term-( )of-each-sequence-the( )-eighth-term-of-( )the-arithmetic-( )sequence-whose-( )fourth-ter/

Elina [12.6K]3 years ago
3 0

Answer:

its C can u give me brainly

Step-by-step explanation:

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If 3 workers can paint a room in 2 hours, then approximately how long does it take 4 workers to paint the same room? Assume the
Olin [163]

It takes 1.5 hours for 4 workers to paint the same room

<em><u>Solution:</u></em>

Given that 3 workers can paint a room in 2 hours

To find: Time taken for 4 workers to paint the same room

Assume the time needed to paint the room is inversely proportional to the number of worker

time $ \propto \frac{1}{\text { number of workers }}\\\\time =k \times \frac{1}{\text { number of workers }}

Where, "k" is the constant of proportionality

<em><u>3 workers can paint a room in 2 hours</u></em>

Substitute number of workers = 3 and time = 2 hours

time =k \times \frac{1}{\text { number of workers }}\\\\2 = k \times \frac{1}{3}\\\\k = 6

Therefore,

\text {time}=6 \times \frac{1}{\text { number of workers }}

To find time taken for 4 workers to paint the same room, substitute number of workers = 4 in above expression

time = 6 \times \frac{1}{4} = 1.5

Thus it takes 1.5 hours for 4 workers to paint the same room

6 0
3 years ago
Read 2 more answers
You are designing a ramp where the horizontal distance is twice the vertical rise. what will be the ramp angle to the nearest te
ankoles [38]
Use the definition of tangent

or inv tangent (arctan or atan)

tan(theta) = opposite side/ adjacent side

the vertical rise would be the opposite side, the horozontal distance would be the adjacent side

let rise = y
let distance= x
x=2y

so tan(theta) = y/2y
tan(theta) = 1/2

theta = atan(1/2)

grab a handy dandy calculator, unless you know this conversion off the top of your head
7 0
3 years ago
What is the domain restrictions for x=-6 and x=1
slavikrds [6]

Answer:

the domain in general is negative Infinity, infinity.

Step-by-step explanation:

Infinity means it is a straight line that goes on forever it never stops. graph the -6 and 1. I am not totally sure what domain restrictions are if you have an equation for it go a head and put it there and I can help you more.

8 0
3 years ago
How do I find the measurement of this ARC m&lt;IEF
andrey2020 [161]
The measure of a central angle is equal to measure of a minor arc. That makes m<GEH=17x+12. By the Vertical Angles Theorem, m<GEH and m<IEF are equal to each other (m<GEH=17x+12=m<IEF). By the same theorem, m<FEG and m<IEH are also equal (m<FEG=8x-7=m<IEH). The angles in a circle must all add up to 360 degrees, 2(17x+12)+2(8x-7)=360. Solve for x, then plug x into the equation for m<IEF.
Hope this helps!
4 0
3 years ago
Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de
Ganezh [65]

Answer:

a. \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

b. \mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

Step-by-step explanation:

The initial value problem is given as:

y' +y = 7+\delta (t-3) \\ \\ y(0)=0

Applying  laplace transformation on the expression y' +y = 7+\delta (t-3)

to get  L[{y+y'} ]= L[{7 + \delta (t-3)}]

l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

Taking inverse of Laplace transformation

y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

Recall that:

y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

Then

y(t) = 7 -7e^{-t}  +e^{-(t-3)} u (t-3)

y(t) = 7 -7e^{-t}  +e^{-t} e^{-3} u (t-3)

\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

3 0
3 years ago
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