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3 years ago

Please helppppppp meeee

Mathematics

2 answers:

Svetlanka [38]3 years ago

7 0

First question is D ( x multiplied by x is x^2 so you can easily answer that ) then the 2nd one is D

erica [24]3 years ago

5 0

The first question is c and then b . It easy if u remember from first grade what we learned

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Evaluate please like pretty please

yaroslaw [1]

It would be zero because you can’t raise 0 to any positive power.

8 0

3 years ago

What is the slope of the line that passes through the points (-7, -7) and (-7, 1)? Write your answer in simplest form.

Ronch [10]

Answer:

undefined, stays on x so straight vertical - vertical line is undef slope

3 0

4 years ago

Read 2 more answers

Will had 64 baseball cards in his collection. He then got 8 more cards. He keeps the cards in a binder. Each page in the binder

WITCHER [35]

Answer: 64+8=72...72/9=8

Step-by-step explanation:

6 0

3 years ago

Sla square field has an area of 12,382m², calculate it's side length. show your work

shepuryov [24]

Answer:

side = 111.274........

Explanation:

area of square = side × side

means we're finding area by finding square of side.

now, we're finding side. so, we'll do the opposite of square, which means square root.

$side = \sqrt{area}$

so,

$side = \sqrt{12382}$

side = 111.274........

please mark BRAINLIEST!

3 0

2 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

3 years ago