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worty [1.4K]
2 years ago
12

HELP ASAP PLS!! An explanation too pls

Mathematics
1 answer:
NISA [10]2 years ago
4 0

Answer:

Step-by-step explanation:

Polygon is regular when all angles are congruent and all sides are congruent.

So, the given polygon is square.

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Jack uses a probability simulator to roll a six-sided number cube 100 times and to flip a coin 100 times. The results of the exp
Mandarinka [93]

The experimental probability is the number of specific outcomes divided by the sample size...


P(6)=27/100  (27%)


P(H)=41/100  (41%)


Not sure, but if you meant rolling a 6 AND getting a head then:


P(6 AND H)=(27/100)(41/100)=1107/10000  (11.07%)




5 0
3 years ago
Read 2 more answers
Is x = -7, the solution to -11 + 2x = -26? Justify your answer in complete sentences.
Natali5045456 [20]
First of all, you can move the +2x to the other side:
-11 = -26 - 2x

Then you can move the -26 to the other side and then simplify:
-11+26 = -2x
 15 = -2x

Then you can divide both sides by 2:
7.5 = -x

Then you can flip the sides and the signs around:
x = -7.5

No, The answer is not -7, it is -7.5
Hope this helps! :)
4 0
3 years ago
use the information provided to write the standard form equation of each hyperbola vertices:(4,14), (4,-10) foci: (4,15), (4,-11
Evgen [1.6K]
So... if you notice the picture below, based on the given vertices, is a hyperbola with a vertical traverse axis

meaning for the equation, the fraction with the "y" variable is the positive fraction, thus \bf \cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1

so... what' the point h,k for the center?  well, just take a peek at the graph, the center is half-way between both vertices, that's h,k

what's the "a" component or the traverse axis... well, is the distance from one vertex to the center, notice the picture, notice how many units from either vertex to the center, that's "a"

now.. what's "b"?  well, "b" comes from the conjugate axis, or the other runnning over the x-axis   hmm the smaller one in this case.... well, we dunno what "b" is

however, we know the distance from either focus, to the center of the hyperbola, and that distance "c", is  \bf c=\sqrt{a^2+b^2}

notice the picture, notice the distance from either focus to the center, that's the distance "c", thus  \bf c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b

that'd be "b"

bearing in mind that   \bf \cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center= ({{ h}},{{ k}})\\\\ b=\sqrt{c^2-a^2}
\end{cases}

8 0
3 years ago
Does a/9=25 i dont know
Mamont248 [21]

Answer:

225

Step-by-step explanation:

225 ÷ 9 =25

8 0
2 years ago
Read 2 more answers
Estimate the student's walking pace, in steps per minute, at 3:20 p.m. by averaging the slopes of two secant lines from part (a)
ehidna [41]

This question is incomplete, the complete question is;

A student bought a smart-watch that tracks the number of steps she walks throughout the day. The table shows the number of steps recorded (t) minutes after 3:00 pm on the first day she wore the watch.

t (min)       0          10          20         30         40

Steps   3,288    4,659    5,522    6,686    7,128

a) Find the slopes of the secant lines corresponding to the given intervals of t.

1) [ 0, 40 ]

11) [ 10, 20 ]

111) [ 20, 30 ]

b) Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part (a). (Round your answer to the nearest integer.)

Answer:

a)

1) for [ 0, 40 ], slope is 96

11) for [ 10, 20 ],  slope is 86.3

111) for  [ 20, 30 ], slope is 116.4

b) the student's walking pace is 101 per min

Step-by-step explanation:

Given the data in the question;

t (min)       0          10          20         30         40

Steps   3,288    4,659    5,522    6,686    7,128

SLOPE OF SECANT LINES

1) [ 0, 40 ]

slope =  ( 7,128 - 3,288 ) / ( 40 - 0

= 3840 / 40 = 96

Hence slope is 96

11)  [ 10, 20 ]

slope = ( 5,522 - 4,659 ) / ( 20 - 10 )

= 863 / 10 = 86.3

Hence slope is 86.3

111)  [ 20, 30 ]

slope = ( 6,686 - 5,522 ) / ( 30 - 20 )

= 1164 / 10 = 116.4

Hence slope is 116.4

b)

Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part .

Since this is recorded after 3:00 pm

{ 3:20 - 3:00 = 20 }

so t = 20 min

so by average;

we have ( [ 10, 20 ] + [ 20, 30 ] ) /2

⇒ ( 86.3 + 116.4 ) / 2

= 202.7 /2

= 101.35 ≈ 101

Therefore, the student's walking pace is 101 per minutes

3 0
3 years ago
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