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Serjik [45]
3 years ago
13

frank is 5 years younger than twice the age of his cousin Alonso. Frank is 17 years old. Which equation could be used to determi

ne a, alonzos age?
Mathematics
1 answer:
monitta3 years ago
5 0

Answer:

Hi there!

Your answer is:

17= 2A-5

Alonzo is 11

Step-by-step explanation:

Frank=2A -5

Frank = 17

17= 2A-5

A = Alonzo's age

Alonzo's age is:

17= 2A-5

+5

22 = 2A

/2

11= A

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Planes Q and R are parallel. Lines a and b are shown on planes Q and R, respectively. Planes Q and R are parallel. Plane Q conta
natita [175]

Answer:

They are skew lines.

Step-by-step explanation:

Which statement is true about lines a and b?

They are parallel lines.

They are perpendicular lines.

They are skew lines.

They will intersect.

As they both are in different directions they are skew lines .

Skew lines are not parallel neither they .They are also not co planar i.e they lie in different planes.

We have two plane Q and R . We have two line a and b on the different planes Q and R. Both planes are parallel but the lines a and b are in different directions. Therefore they are skew lines . They do not intersect and are also not parallel neither co planar.

7 0
3 years ago
Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species
kotegsom [21]

Answer:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

Step-by-step explanation:

The logistic equation is the following one:

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that P(0) = 80, K = 2000.

The number of fish tripled in the first year. This means that P(1) = 3P(0) = 3(80) = 240.

Using the equation for P(1), that is, P(t) when t = 1, we find the value of r.

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}

280*(2000 + 80(e^{r} - 1)) = 160000e^{r}

280*(2000 + 80e^{r} - 80) = 160000e^{r}

280*(1920 + 80e^{r}) = 160000e^{r}

537600 + 22400e^{r} = 160000e^{r}

137600e^{r} = 537600

e^{r} = \frac{537600}{137600}

e^{r} = 3.91

Applying ln to both sides.

\ln{e^{r}} = \ln{3.91}

r = 1.36

This means that the expression for the size of the population after t years is:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

4 0
3 years ago
Para embalar una caja se emplea 4,2 m de cinta adhesiva. ?Cuántas cajas se podrán embalar con tres rollos que tienen 3 hm, 7 dam
kompoz [17]

For this case, we perform the conversions:

First roll:

1 \ Hectometer --------> 100 \ meters\\3 \ Hectometer --------> x

x = \frac {3 * 100} {1}\\x = 300 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 300 meters of adhesive tape.

1 -----------> 4.2

c -----------> 300

c = \frac {300 * 1} {4.2}\\c = 71.42857143\\c = 71

You can pack 71 boxes.

Second roll:

1 \ Decametro --------> 10 \ meters\\7 \ Decameter --------> x\\x = \frac {7 * 10} {1}\\x = 70 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 70 meters of adhesive tape.

1 -----------> 4.2

c -----------> 70

c = \frac {70 * 1} {4.2}\\c = 16.66666667\\c = 16

You can pack 16 boxes.

Third roll:

1 -----------> 4.2

c -----------> 50

c = \frac {50 * 1} {4.2}\\c = 11.9047619\\c = 11

You can pack 11 boxes.

Thus, in total you can pack11 + 16 + 71 = 98 \ boxes

Answer:

98 boxes

4 0
3 years ago
I need help i dot get it.
gayaneshka [121]

Answer:

the answer is about 8 cuz the answer shows 7.12 if you divide mentally

6 0
3 years ago
Read 2 more answers
Fractions from 6 to 7, with an interval of between each pair of fractions
EastWind [94]

Answer:

6.7

Step-by-step explanation:

3 0
3 years ago
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