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DochEvi [55]
3 years ago
15

2. Point P is shown on the number line below. The distance between point Q and point P is 6 & 1/2 units. Which number could

represent the location of point Q? *
Mathematics
1 answer:
leva [86]3 years ago
8 0

Complete question :

Point P is shown on the number line below. The distance between point Q and point P is 6 1/2 units. Which number could represent point Q?

A) -9 1/2

B) 1 1/2

C) 2 1/2

D) 10 1/2

Answer:

1 1/2

Step-by-step explanation:

To obtain the possible position of point Q:

Add or subtract the distance between the points to the position of point P on the number line ;

Point P is located at - 5

Therefore, point Q will be located at :

Either :

Point P + 6 1/2 = - 5 + 6.5 = 1 1/2

Point P - 6 1/2 = - 5 - 6.5 = - 11 1/2

From the option ;

Q = 1 1/2

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\qquad\qquad\huge\underline{{\sf Answer}}

Both the given lines in graph intersect at point (5 , -2), the point must satisfy both the equations.

and if we check the options, we will observe that D. option has both the equations that satisfy the given conditions.

that is ~

\qquad \sf  \dashrightarrow \: x + y = 3

put y = - 2,we will get :

\qquad \sf  \dashrightarrow \: x - 2 = 3

\qquad \sf  \dashrightarrow \: x = 5

so, 1st equation satifys the condition. let's check out 2nd one.

\qquad \sf  \dashrightarrow \: x + 2y = 1

put y = -2

\qquad \sf  \dashrightarrow \: x + (2 \times  - 2) = 1

\qquad \sf  \dashrightarrow \: x - 4 = 1

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5 0
2 years ago
The initial size of the population is 300. After 1 day the population has grown to 800. Estimate the population after 6 days. (R
Cloud [144]

Solution :

Given initial population = 300

Final population after 1 day = 800

Number of days = 6

∴ $\frac{dP}{dt} =kt^{1/2} $

P(0) = 300    P(1) = 300

We need to find P(8).

$dP = kt^{1/2} dt$

$ \int 1 dP = \int kt^{1/2} dt$

$P(t) = k \left(\frac{t^{3/2}}{3/2}\right)+c$

$P(t)= \frac{2k}{3}t^{3/2} + c$

When P(0) = 300

$300 = \frac{2k}{3} (0)^{3/2} + c$

∴ c = 300

∴ $P(t)= \frac{2k}{3}t^{3/2} + 300$

When P(1) = 800

$800 = \frac{2k}{3} (1)^{3/2} + 300$

$500 = \frac{2k}{3}$

∴ k = 750

$P(t)= 500t^{3/2} + 300$

So, P(8) is

$P(t)= 500(8)^{3/2} + 300$

        = 11,614

So the population becomes 11,614 after 8 days.

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