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Snowcat [4.5K]
3 years ago
10

PLEASE HELP!!! 10 POINTS!!

Mathematics
1 answer:
Damm [24]3 years ago
4 0
Well I'm sorry if this isn't right but (a)
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Celia filled the glasses shown below completely with water. The total amount of water that Celia poured into the glasses is 80 c
bagirrra123 [75]
The height of glass 1 is 7.6
so, you find the volume of glass 2,= 26.376
then you subtract the answer from 80 cubic cm to get the volume of glass 1
                           80 - 26.376 = 53.694
then you work backwards to get the height, which is 7.6
7 0
4 years ago
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The annual Gross Domestic Product (GDP) of a country is the value of all of the goods and services produced in the country durin
Vladimir [108]
F=5.77X10^9(1.032^(y-1985)), we want to know when this is:

2.2X10^12=5.77X10^9(1.032^(y-1985))

381.2825=1.032^(y-1985)

ln(3,812825)=(y-1985)ln(1.032)

ln(3.812825)/ln(1.032)=y-1985

y=1985+ln(3.812825)/ln(1.032)

y≈2027.5  

So GDP will reach 2.2 trillion during the year 2027.
5 0
3 years ago
In a G.P the difference between the 1st and 5th term is 150, and the difference between the
liubo4ka [24]

Answer:

Either \displaystyle \frac{-1522}{\sqrt{41}} (approximately -238) or \displaystyle \frac{1522}{\sqrt{41}} (approximately 238.)

Step-by-step explanation:

Let a denote the first term of this geometric series, and let r denote the common ratio of this geometric series.

The first five terms of this series would be:

  • a,
  • a\cdot r,
  • a \cdot r^2,
  • a \cdot r^3,
  • a \cdot r^4.

First equation:

a\, r^4 - a = 150.

Second equation:

a\, r^3 - a\, r = 48.

Rewrite and simplify the first equation.

\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}.

Therefore, the first equation becomes:

a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150..

Similarly, rewrite and simplify the second equation:

\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}.

Therefore, the second equation becomes:

a\, r\, \left(r^2 - 1\right) = 48.

Take the quotient between these two equations:

\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}.

Simplify and solve for r:

\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}.

8\, r^2 - 25\, r + 8 = 0.

Either \displaystyle r = \frac{25 - 3\, \sqrt{41}}{16} or \displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}.

Assume that \displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}. Substitute back to either of the two original equations to show that \displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75.

Calculate the sum of the first five terms:

\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}.

Similarly, assume that \displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}. Substitute back to either of the two original equations to show that \displaystyle a = \frac{497\, \sqrt{41}}{41} - 75.

Calculate the sum of the first five terms:

\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}.

4 0
3 years ago
Rewrite with positive exponents a^-3 b^-5
amm1812

Answer:

1/a^3 and 1/b^5

6 0
3 years ago
What has to be true for corresponding angles to be congruent ​
faust18 [17]

Answer:

The lines intersected by a transversal must be parallel.

4 0
3 years ago
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