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2 years ago

Help please help help

Mathematics

1 answer:

AnnZ [28]2 years ago

7 0

Make image more close pls.Then I will answer

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It is sometimes possible to write two inequalities as a inequality.

goblinko [34]

Hi!

Yes, it is. When we have one inequality composed of two inequalities, we call this a double inequality.

A double inequality would look something like -5 < 4x < 10.

Hopefully, this helps! =)

7 0

3 years ago

Are the leap years divisible by 2

Helen [10]

Any year that is evenly divisible by 4 is a leap year.

8 0

3 years ago

Use rounding to estimate the product

sdas [7]

Answer:

The estimate of the product is about 35 answer C

Step-by-step explanation:

- We use rounding to estimate the product of two fractions

- The mixed number has whole number and fraction

- To round the mixed number we look to the fraction

- If the numerator is half or greater than half the denominator, then we

add the whole number by 1 and cancel the fraction

- If the numerator is less than half the denominator, then we keep the

whole number as it and cancel the fraction

- Ex: If the mixed number is $2\frac{3}{4}$ , 3 is more than

half 4 then we will add 2 by 1 and cancel the fraction part, the

mixed number $2\frac{3}{4}$ after rounding is 3

- Ex: If the mixed number is $5\frac{1}{4}$ , 1 is less than

half 4 then we will keep 5 as it and cancel the fraction part, the

mixed number $5\frac{1}{4}$ after rounding is 5

* Lets solve the problem

∵ In the part $4\frac{3}{4}$ 3 is more than half 4

∴ The rounding value of $4\frac{3}{4}$ is 5

∵ In the part $7\frac{2}{17}$ 2 is less than half 17

∴ The rounding value of $7\frac{2}{17}$ is 7

∴ $4\frac{3}{4}*7\frac{2}{17}=5*7=35$

* The estimate of the product is about 35

6 0

3 years ago

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/mi

Margaret [11]

Answer:

a) $y(t)=0.65\frac{Kg}{min}(tmin)$

b) $y(40)=26Kg$

Step-by-step explanation:

Data

Brine a (Ba)

$V_{Ba}=5\frac{Lt}{min}\\ Concentration(Bca)=0.05\frac{Kg}{Lt}$

Brine b (Bb)

$V_{Bb}=10\frac{Lt}{min}\\ Concentration(Bcb)=0.04\frac{Kg}{Lt}$

we have that per every minute the amount of solution that enters the tank is the same as the one that leaves the tank (15 Lt / min)

, then the amount of salt (y) left in the tank after (t) minutes: $y=V_{Ba}*B_{ca}+V_{Bb}*B_{cb}=5\frac{Lt}{min}*0.05\frac{Kg}{Lt}+10\frac{Lt}{min}*0.04\frac{Kg}{Lt}=\\0.25\frac{Kg}{min}+0.4\frac{Kg}{min}=0.65\frac{Kg}{min}$

Finally:

a) $y(t)=0.65\frac{Kg}{min}(tmin)$

b) $y(40)=0.65\frac{Kg}{min}(40min)=26Kg$

being y(t) the amount of salt (y) per unit of time (t)

5 0

2 years ago

PLEASE HELP QUICK!!

OleMash [197]

The answer should be D

5 0

3 years ago