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2 years ago

PLEASE HELP! Will give the BRAINLIEST!!

Mathematics

1 answer:

Anton [14]2 years ago

6 0

Answer:

Step-by-step explanation:

448 divided by 32 is 14 so then do 14 times 1/2 to get 7 centimeters

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if 1/5 and -2 are respectively product and sum of the zeroes of a quadratic polynomial. Find the polynomial.

Nikitich [7]

Answer:

this might be the equation

Step-by-step explanation:

To find: Quadratic Polynomial.

where , ( α + β ) is sum of zeroes and αβ is product of zeroes.

4 0

3 years ago

Read 2 more answers

For each function shown:

mestny [16]

#1: y= 0.5+3 x
which means that x= y/3 - 0.5

#2: y= 3(0.5) x
which means that y= 1.5 x
that means x = y/1.5

Hope it can help

5 0

3 years ago

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Find the distance between -6 - 8

Viefleur [7K]

Answer:

Number line.

We start at -6 or -8.

If we start from -6, we will be going __ spaces to the left until we get to -8.

Or start from -8 and go __ spaces to the right until you get to -6.

Then count the spaces.

You'll get 2.

4 0

2 years ago

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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

Eight less than a number divided by seven equals 2

madreJ [45]

Answer:

x=12

Step-by-step explanation:

（X-8)/7=2

X-8=14

X=22

7 0

2 years ago