9514 1404 393
Answer:
10s place
Step-by-step explanation:
When you line up the three digits of the divisor with the three most significant digits of the dividend, you see that the least significant digit of the divisor lines up with the dividend digit that is in the 10s place.
The first (most-significant) quotient digit will be in the 10s place.
_____
In the attached, the left-most non-zero quotient digit is 4, in the tens place.
I think the one you chose is correct but I am not sure tho
Since we are using 3 sides of measurement and are looking for a volume, or a 3rd dimensional amount, the correct unit of measure is going to cm3, which is B.).
I hope this helps!
Answer:
y= 1/12(x-0)^2+0
this answer works as an upward parabola
Step-by-step explanation:
- The formula for a veritcal parabola is y=1/4p(x-h)^2+k
- (h,k)= coordinates of the vertex of the parabola
- p= absolute value of the distance from the vertex to the focus/directrix
- In this problem, it is given that the vertex is at the origin (0,0) and the focus (the bulb), is 3 centimeters away from the vertex.
- Now, you know the values of the variables. Fill in the values
- FROM THE FORMULA: 1/4p turns into 1/12 since p is 3.
- (x-h)^2+k turns into (x-0)^2+0, since h and k where the values of the vertex which was 0,0
- once all the variables are given values (except x and y) you have made your equation!
- The answer is y=1/12(x-0)^2+0
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Answer:
<u>a) x = 3</u>
<u>b) z = 10</u>
<u>c) p = 2</u>
<u>d) x = 7</u>
<u>e) u = 1</u>
Step-by-step explanation:
a) 2x = 6
Despejamos x dividiendo por 2 a amabos lados de la eacuacion.
(2/2)x = 6/2
<u>x = 3</u>
Si remplazamos x en la ecuación original:
2(3)=6
6 = 6
Queda demostrado.
b) 10 + z = 20
Despejamos z restando 10 en amabos lados de la eacuacion.
10-10+z = 20-10
<u>z = 10</u>
Si remplazamos z en la ecuación original:
10 + 10=20
20 = 20
Queda demostrado.
c) p + 9 = 11
Despejamos p restando 9 en amabos lados de la eacuacion.
p + 9 - 9 = 11-9
<u>p = 2</u>
Si remplazamos p en la ecuación original:
2 + 9 = 11
11 = 11
Queda demostrado.
d) 3x + 8 = 29
Despejamos x restando 8 en amabos lados de la eacuacion y luego divideindo por 3 en ambos lados de la ecuación.
3x+8-8 = 29-8
3x = 21
(3/3)x = 21/3
<u>x = 7</u>
Si remplazamos x en la ecuación original:
3(7) + 8 = 29
21 + 8 = 29
29 = 29
Queda demostrado
e) 2u + 8 = 10
Despejamos u restando 8 en amabos lados de la eacuacion y luego divideindo por 2 en ambos lados de la ecuación.
2u+8-8 = 10-8
2x = 2
(2/2)x = 2/2
<u>x = 1</u>
Si remplazamos x en la ecuación original:
2(1) + 8 = 10
2 + 8 = 10
10 = 10
Queda demostrado
Espero te haya sido de ayuda!
