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jok3333 [9.3K]
3 years ago
15

Write a word problem that goes with expression

Mathematics
1 answer:
pshichka [43]3 years ago
6 0
Thirty-Six Divided By Four Equals Nine
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Can someone help me
sergij07 [2.7K]

Answer:

1)  (c)∠ G = 34°

2) (a)  x = 33°

3) (d) The perpendicular segments are  JM, KL, PQ AND NR.

Step-by-step explanation:

Here in the given questions:

1)  p II r

⇒ ∠ G  = 34°  (ALTERNATE EXTERIOR ANGLES)

Hence, the measure of  ∠ G = 34°

2) In the given triangle:

The measure of exterior angle is always the sum of interior opposite angles.

⇒ Here, x + 72°  = 105°

or, x = 105° - 72° =   33°   , or x = 33°

3) The given plane in the cube is JKPN

The segments which have J , K , P and N as their one vertex are perpendicular to the given plane.

Hence, the segments are  JM, KL, PQ AND NR.

7 0
3 years ago
A gasoline pump delivers 4 2/5 gallons of gas per minute. How many minutes will it take to fill a gas tank that holds 17 1/4 gal
Liono4ka [1.6K]
Well if it pumps 4 2/5 per minute that is equal to 4 4/10 in a minute or 4.4 because you multiply the top and bottom of the fraction by two. it’s still the same fraction but it is easy to convert to a decimal now. you could use this fraction but it’s much easier to use decimals. Because the first number is a decimal you should make the second one a decimal too. To do this you need to know that 1/4 of 100 is 25. so now we have these numbers and all you do now is divide 17.25 by 4.4 which could be rounded to 4 because it’s a long decimal
8 0
3 years ago
How can both parents be tall, even though their genes for height are not exactly the same?
SSSSS [86.1K]
Both of the Billy’s parents (calling the parent billy) being tall is a recessive trait therefore both Billy’s parents has to have at least one recessive trait to make a tall child. Billy’s genotype would be way tt (aka recessive, recessive).
5 0
3 years ago
SUPER EASY
goblinko [34]
<span>3(x + 5) = 3 · x + 3 · 5</span>
6 0
3 years ago
Read 2 more answers
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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