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seraphim [82]
3 years ago
9

A frog is hopping along in a straight line. Each hop is twice as far as its previous hop. Its first hop covers 1 inch. After 8 h

ops, the frog will have traveled _____ inches.
Mathematics
1 answer:
torisob [31]3 years ago
8 0
The frog's jump represent a geometric series.
The sum of finite geometric series can be solved using the equation:
Sn = ao ( (1-r^n)/(1-r) )

Plugging in the values,
S8 = 1 ( (1-2^8)/(1-2) )
S8 = 255

Therefore, after 8 hops, the frog will have traveled 255 inches
You might be interested in
the width of a rectangle is y feet long. the rectangle is 4 feet longer than it is wide l. what is arena of the rectangle?​
ahrayia [7]
Area= i^2 + 4i
Area= length x width
Length (y)= i + 4
Width (i)=i
8 0
3 years ago
A ball is kicked upward with an initial velocity of 32 feet per second. The ball's height, h (in feet), from the ground is model
stellarik [79]

Answer:

1.        t = 0.995 s

2.       h = 15.92  ft

Step-by-step explanation:

First we have to look at the following formula

Vf = Vo + gt

then we work it to clear what we want

Vo + gt = Vf

gt = Vf - Vo

t = (Vf-Vo)/g

Now we have to complete the formula with the real data

Vo = 32 ft/s      as the statement says

Vf = 0     because when it reaches its maximum point it will stop before starting to lower

g = -32,16 ft/s²        it is a known constant, that we use it with the negative sign because it is in the opposite direction to ours

t = (0 ft/s - 32 ft/s) / -32,16 ft/s²

we solve and ...

t = 0.995 s

Now we will implement this result in the following formula to get the height at that time

h = (Vo - Vf) *t /2

h = (32 ft/s - 0 ft/s) * 0.995 s / 2

h = 32 ft/s * 0.995 s/2

h = 31.84 ft / 2

h = 15.92  ft

4 0
2 years ago
Don't get this
kap26 [50]
I'll talk you through it so you can see why it's true, and then
you can set up the 2-column proof on your own:

Look at the two pointy triangles, hanging down like moth-wings
on each side of 'OC'.

-- Their long sides are equal,  OA = OB, because both of those lines
are radii of the big circle.

-- Their short sides are equal, OC = OC, because they're both the same line.

-- The angle between their long side and short side ... the two angles up at 'O',
are equal, because OC is the bisector of the whole angle there.

-- So now you have what I think you call 'SAS' ... two sides and the included angle of one triangle equal to two sides and the included angle of another triangle.
(When I was in high school geometry, this was not called 'SAS' ... the alphabet
did not extend as far as 'S' yet, and we had to call this congruence theorem
"broken arrow".)

These triangles are not congruent the way they are now, because one is
the mirror image of the other one.  But if you folded the paper along 'OC',
or if you cut one triangle out and turn it over, it would exactly lie on top of
the other one, and they would be congruent.

So their angles at 'A' and at 'B' are also equal ... those are the angles that
you need to prove equal.
5 0
3 years ago
Maria y juan deben realizar el mapamundi en medio pliego de cartulina para la tarea de sociales pero cada uno tiene un cuarto de
kondor19780726 [428]

Answer:

Cuando María afirma que si unen sus dos cuartos de cartulina obtendrán el medio pliego que necesitan, esto es:

  • <u>Verdadero</u>.

Step-by-step explanation:

Para entender mejor el ejercicio vamos a utilizar números cada vez que se habla de cantidades de cartulina, por lo tanto, María y Juan tienen 1/4 de cartulina cada uno, es decir, 1/4 * 2, y necesitan 1/2 pliego para poder realizar su tarea, por lo tanto, con la afirmación de María sobre unir los dos cuartos de cartulina, en caso de que sea verdadero, ocurrirá que la suma de los dos cuartos dará el medio pliego, como se muestra a continuación:

  • Total de cartulina de María y Juan = \frac{1}{4}+\frac{1}{4}
  • Total de cartulina de María y Juan = \frac{4+4}{16}
  • Total de cartulina de María y Juan = \frac{8}{16}

Procedemos a simplificar el fraccionario obtenido, sacando mitad tanto en el numerador como en el denominador:

  • Total de cartulina de María y Juan = \frac{4}{8}
  • Total de cartulina de María y Juan = \frac{2}{4}
  • <u><em>Total de cartulina de María y Juan = </em></u>\frac{1}{2}<u><em /></u>

Como puedes ver al final, <u>la cantidad de cartulina de ambos, al ser sumada, da como resultado el 1/2 (medio) pliego que necesitan para su tarea de sociales, por lo cual la afirmación de María es correcta</u>.

3 0
3 years ago
To solve m/-2+11=14 what steps would you use
Butoxors [25]
M÷-2+11=4
m÷9=4
m=4*9
m=36
5 0
2 years ago
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