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spayn [35]
3 years ago
11

HELPPPPP‼️ IM BEGGING FLR HELP

Mathematics
2 answers:
Gwar [14]3 years ago
5 0

Answer:

1/25

Step-by-step explanation:

Leni [432]3 years ago
3 0

Answer:

1/25

Step-by-step explanation:

lol hope i could helped:)

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What numbers are a distance of 3/4 unit from 1/8 on a number line? Select the locations on the number line to plot the points.
ICE Princess25 [194]
3/4 = 6/8. so 6/8-1/8= 5/8. 5/8 is the answer.
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Pls help i need to know if this is right i got 7d+50
vodka [1.7K]

Answer:

l think d=1.4

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Select the correct answer.
Dovator [93]

{\qquad\qquad\huge\underline{{\sf Answer}}}

Here's the explanation ~

Zeros or roots of a function is the value of independent variable (usually x - coordinate or absicca) where the value of dependent variable (y - coordinate or ordinate) is 0.

That is :

The point where it cuts the x - axis, it has an x - coordinate = -6

Therefore, we can conclude that The zero of given function is -6

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2 years ago
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10 + (8 - 4)2 ÷ 2<br> Anyone
Marina CMI [18]
The answer for this question 15 i hope i have helped you today
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2 years ago
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Find the solution to the differential equation dz/dt = 7te^4z that passes through the origin.
andreev551 [17]

Answer:

z=(\frac{-1}{4} )ln|1-14t^{2}|\\

Step-by-step explanation:

from the equation, \frac{dz}{dt}=7te^{4z}\\.

we can give different approach to the equation, but to make it simple and direct, let separate the equation by bringing all like terms to the same side i.e

\frac{dz}{e^{4z}}=7tdt\\e^{-4z} dz=7tdt.

if we integrate both side,

\int\limits^a_b{e^{-4z} } \,dz =\int\limits^a_b {7t} \,dt

-1/4e^{-4z} +c_{1}=7/2t^{2} +c_{2}\\-1/4e^{-4z}= 7/2t^{2} +c_{2}-c_{1}\\let c_{2}-c_{1}=c \\-1/4e^{-4z}= 7/2t^{2} +c

since the equation passes through the origin, and at the origin z=0 and t=0

we substitute this values and solve for the constant c

e^{-4*0}= 7/2*0 +c\\c=1.

If we substitute the value of c into the equation we arrive at

(-1/4)e^{-4z}= (7/2)t^{2}+1\\ e^{-4z}=1-14t^{2} \\

if we the the natural logarithm of both sides, we arrive at

-4z=ln|1-14t^{2}|\\z=(\frac{-1}{4} )ln|1-14t^{2}|\\

3 0
3 years ago
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