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2 years ago

How many times does 150 go into 3000

Mathematics

2 answers:

Tcecarenko [31]2 years ago

7 0

3000/150=20

20 times does 150 go in 3000

Airida [17]2 years ago

6 0

Answer:

Step-by-step explanation:

x × 150 = 3000

x = 3000 ÷ 150
x = 20

many times does 150 go into 3000 is 20

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Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work

valina [46]

Answer:

Step-by-step explanation:

Question

Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work.

perimeter of a triangle = AB+AC+BC

Using the distance formula

AB = sqrt(-2-5)²+(2-2)²

AB = sqrt(-7)²

AB =sqrt(49)

AB =7

BC = sqrt(-2+2)²+(2-5)²

BC = sqrt(0+3²)

BC =sqrt(9)

BC =3

AC= sqrt(-2-5)²+(2-5)²

AC= sqrt(-7)²+3²

AC =sqrt(49+9)

AC =sqrt58

Perimeter = 10+sqrt58

5 0

3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

an octagon has sides that all have a length of y. Write a simplified algebraic expression for the perimeter of the octagon in te

padilas [110]

Answer = 8y

Step-by-step explanation:

if an octagon has 8 sides and each side is = to y then the total side length in the primiter is = to 8sides times y or 8y

5 0

3 years ago

Please help<br><br><br><br> ..........................

Alenkinab [10]

Answer:

9/2

Step-by-step explanation:

perpendicular lines have slopes that are opposite and reciprocal

the reciprocal of -2/9 is 9/2

7 0

3 years ago

What is 50 x 79 x 27 + 69

vova2212 [387]

106,719

Answer:

$50 \times 79 \times 27 + 69 \\ \\ = 106,650 + 69 \\ \\ = 106,719$

8 0

3 years ago

Read 2 more answers