All categories

3 years ago

Sela has 6 times as many coins as she had 4 months ago. If sela has 240 coins now ,How many coins did she have 4 months ago.

Mathematics

2 answers:

saw5 [17]3 years ago

5 0

She has 40 coins since 40 times 6 is 240.

Lesechka [4]3 years ago

4 0

Sela had 40 coins

240/6=40

If you want to double check multiply 40x6

Your answer will give you 240 which means she had 40 coins 4 months ago

You might be interested in

MARKING BRAINLIEST

Schach [20]

Answer

2.25

Step-by-step explanation:

4 0

3 years ago

Work out m and c for the line: y = 10 x

alekssr [168]

Answer:

The slope is 10 and the y intercept 0

Step-by-step explanation:

m is 10 and c is 0. The reason why is because m is the slope and c is the y intercept.

8 0

3 years ago

Read 2 more answers

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

17. Admission prices to Cinema I to see a movie are $9.50 for an adult and $6.50 for a child. The admission charge at Cinema II

maksim [4K]

Answer:

a. 9.5x + 6.5(x+c) < 8 when c>0

b. Must be one child more than the no. of adults.

Step-by-step explanation:

For Cinema 1:

for adult = $9.50

for child = $6.50

For Cinema 2:

Per person regardless of age = $8.00

First of all, we will find out the condition when per person rates in both cinema are equal.

Assume x = no. of adults

y = no. of children

Rate per person in Cinema I = Rate per person in Cinema II

(9.5x + 6.5y)/(x+y) = 8

9.5x + 6.5y = 8(x+y)

9.5x + 6.5y = 8x + 8y

9.5x-8x = 8y-6.5y

=> x = y

So rates are equal when no. of adults equals no. of children

For Cinema I to have better rates, no. of children should be atleast 1 more than the no. of adult. In this way the rate per person of Cinema I will be less than 8

Hence we form an inequality when y = x+c and c > 0

9.5x + 6.5(x+c) < 8 when c>0

Hence there must be 1 more children than the no. of adults attending Cinema I for it to be a better deal.

4 0

3 years ago

Read 2 more answers

Can somebody please help me and tell me what the error is!! PLEASE I NEED HELP ELSE I WILL FAIL MY CLASS AND GO TO SUMMER SCHOOL

VikaD [51]

Answer:

I think that is enlargement

Step-by-step explanation:

i like helping out people oh by the way i'm new will you be my friend

3 0

3 years ago