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horsena [70]
3 years ago
7

Can anyone help me with these 2 please ! I’ll mark you as a brainliest.

Mathematics
1 answer:
Mars2501 [29]3 years ago
8 0

Answer:

i dont see a question?

Step-by-step explanation:

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Find the area of the region bounded by the y-axis, the line y=6, and the line y = 1/2.
Ghella [55]

Solution:

As region bounded by y-axis, the line y=6, and the line y=1/2 is a line segment of definite length on y-axis.

We consider a line , one dimensional  if it's thickness is negligible.

So, Line is  two dimensional if it's thickness is not negligible becomes a quadrilateral.

So, Area (region bounded by y-axis, the line y=6, and the line y=1/2 is a line segment of definite length on y-axis)= Area of line segment between [,y=6 and  y=1/2.]= 6-1/2=11/2 units if we consider thickness of line as negligible.

4 0
3 years ago
An object is moving at a speed of 850 miles per hour. Express this speed in meters per minute.
MAVERICK [17]

Answer:

it would be 1385

Step-by-step explanation: because 850 miles per hour converted into meter per min would be 22799.04 meters per sec and then u would Find the number in the whole number place 4 and look one place to the right for the rounding digit on the right side of the decimal point 7. Round up if this number is greater than or equal to 5 and round down if it is less than 5

So the answer is 1385

8 0
3 years ago
Read 2 more answers
Prove that the diagonals of a parallelogram bisect each other​
Nady [450]

Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof :\begin{array}{ |c| c |  c |  } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In  \: \triangle ^{s}  \:AOB \: and \: COD  } \\  \sf{(i)}&  \sf{ \angle \: OAB =  \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\  \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\  \sf{(iii)} &\sf{ \angle \: OBA=  \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\  \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}.          Proved ✔

♕ And we're done! Hurrayyy! ;)

# STUDY HARD! So, Tomorrow you can answer people like this , " Dude , I just bought this expensive mobile phone but it is not that expensive for me" [ - Unknown ] :P

☄ Hope I helped! ♡

☃ Let me know if you have any questions! ♪

\underbrace{ \overbrace  {\mathfrak{Carry \: On \: Learning}}} ☂

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5 0
3 years ago
A bank features a savings account that has an annual percentage rate of 4.8 % with interest compounded monthly. Umbrosia deposit
Scilla [17]

S(8)=3500(1+(.047/4))^32

S(8)=$5086.40 in the account after 8 years.

a)The relative growth rate is .25, or 25%

b)at t=0, the population is 955e^.25(0)=955

c)at t=5; the population is 955*e^.25(5)=955*3.49=3333.28 bacterium.

5 0
1 year ago
Help because I DONT know how to even start
Ierofanga [76]

Slope is rise/run or y2-y1/x2-x1. So to find the slope, find two points where the point is exact, then do the equation. That's the best I can explain, sorry if this doesn't help at all.

6 0
3 years ago
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