Find the slope of the line that passing through the points (5,-2) and (5,8)

Pleaseee someone help me :( Ive been stuck on this for an hour

Whitepunk [10]

Answer:

The number of times organism B's population is larger than organism A's population after 8 days is 32 times

Step-by-step explanation:

The population of organism A doubles every day, geometrically as follows

a, a·r, a·r²

Where;

r = 2

The population after 5 days, is therefore;

Pₐ₅ = = 32·a

The virus cuts the population in half for three days as follows;

The first of ta·2⁵ he three days = 32/2 = 16·a

The second of the three days = 16/2 = 8·a

After the third day, Pₐ = 8/2 = 8·a

The population growth of organism B is the same as the initial growth of organism A, therefore, the population, P₈ of organism B after 8 days is given as follows;

P₈ = a·2⁸ = 256·a

Therefore, the number of times organism B's population is larger than organism A's population after 8 days is P₈/Pₐ = 256·a/8·a = 32 times

Which gives, the number of times organism B's population is larger than organism A's population after 8 days is 32 times.

Hope this helps you out! :)

Also, it's OK! This was actually pretty hard to figure out!

6 0

3 years ago

Some body can help me with a geometric mean maze

Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

Theorem 1: The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

$x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.$

2. South-East point from the Start: By the 2nd theorem,

$x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.$

3. West point from the previous: By the 2nd theorem,

$x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.$

4. West point from the previous: By the 1st theorem,

$9^2=x\cdot 15\Rightarrow x=\dfrac{81}{15}=\dfrac{27}{5}=5.4.$

5. West point from the previous: By the 2nd theorem,

$10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.$

6. North point from the previous: By the 1st theorem,

$x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.$

7. East point from the previous: By the 2nd theorem,

$x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.$

8. North point from the previous: By the 1st theorem,

$x^2=7.5\cdot 36=270\Rightarrow x=3\sqrt{30}.$

8. West point from the previous: By the 2nd theorem,

$x^2=12.5\cdot (12.5+13.5)=12.5\cdot 26=25\cdot 13\Rightarrow x=5\sqrt{13}.$

9. North point from the previous: By the 1st theorem,

$12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.$

101. East point from the previous: By the 1st theorem,

$6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.$

11. East point from the previous: By the 2nd theorem,

$20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.$

12. South-east point from the previous: By the 2nd theorem,

$18^2=x\cdot 21.6\Rightarrow x=15.$

13. North point=The end.

6 0

2 years ago

A carpenter constructed a rectangular sandbox with a capacity of 10 cubic feet. If the carpenter were to make a similar sandbox

jeka57 [31]

Answer:

(D)80 ft³

Step-by-step explanation:

The volume volume of the sandbox is = L × W × H

Where L = Length, W= width, H= Height.

A sandbox has the same shape as a cuboid.

A cuboid is a solid shape with rectangular base and sides. It has six rectangular faces if all sides are closed.

From the question,

LWH = 10ft³ ........................ (1).

If a similar sandbox twice as long, twice as wide and twice as high.

∴ L₍n₎= 2L, W₍n) =2W, H₍n₎ = 2H

Where L₍n₎, W₍n), H₍n₎ is the new Length, new Width and new Height of the similar sandbox.

∴L₍n₎ ×W₍n)× H₍n₎ = 2L × 2W × 2H

and Volume of the new sandbox = L₍n₎ ×W₍n)× H₍n₎

⇒Volume of the new sandbox = 2 × 2 ×2 ×L×W×H

Volume of the new sandbox = 8 × LWH....................(2)

In equation(1) LWH = 10ft³.

Substituting the value in equation(2)

∴Volume of the new(similar) sandbox= 8 × 10

=80ft³.

Therefore the capacity of the second sandbox in cubic feet is 80.

The answer is option (D).

7 0

3 years ago

Dhfnfnsngsdbnchfbmfnbmbmhfnc

finlep [7]

4/32 = x/100
answer: x = 12.5

3 0

2 years ago

Maria's first four test scores were 60, 90, 90, and 70. What what does she need to make on the fifth test to maintain a 70 avera

snow_lady [41]

A 40 is the lowest she can make and that would give her exactly a 70 average.

3 0

3 years ago

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