All categories

2 years ago

Please help i attached it

Mathematics

1 answer:

lapo4ka [179]2 years ago

4 0

Answer:

C. plumber A has a greater initial value

You might be interested in

Prove:

Levart [38]

Given the statement, "If $n^2$ is odd, then $n$ is odd," its contrapositive claims that, "If $n$ is not odd, then $n^2$ is not odd."

So assume $n$ is not odd, i.e. $n$ is even. This means there is an integer $k$ for which $n=2k$ . Squaring this gives $n^2=(2k)^2=4k^2$ .

Well, we can write $4k^2=2(2k^2)$ , and $2k^2$ is just another integer, which means $4k^2=(2k)^2=n^2$ must be even.

3 0

4 years ago

What is dangerous about living for 7 days on just one can of sardines

pshichka [43]

Sardines aren't very healthy; not to mention, living off of just one food for a week is horrible for your body, because you are lacking several kinds of nutrition.

8 0

3 years ago

Suppose that 21 inches of wire costs 63 cents.

defon

Answer:

162 cents

Step-by-step explanation:

to find the rate you want take the inches which is 21 and divide by the cost which is 63 and find the rate is 3 which means each inch costs 3 cents and so with that rate you can find the cost of 54 inchest of wire so you take your 54 inches and multiply it by 3 which is the cents per inch and find that 54 inches of wire would cost you 162 cents

6 0

2 years ago

Simplify: (3–a)·2+a =

IrinaK [193]

Answer:

The simplified form of the given expression is 6-a.

Step-by-step explanation:

The given expression is

$(3-a)\cdot 2+a$

According to distributive property.

$a\cdot (b+c)=ab+ac$

Use distributive property.

$3(2)-a(2)+a$

$6-2a+a$

Combine like terms.

$6+(-2a+a)$

$6-a$

Therefore the simplified form of the given expression is 6-a.

8 0

3 years ago

given the relationship 2x^2 + y^3 =10, with y > 0 and dy/dt = 3 units/min., find the value of dx/dt at the instant x = 1 unit

dedylja [7]

$\dfrac{\mathrm d}{\mathrm dt}\left[2x^2+y^3\right]=\dfrac{\mathrm d}{\mathrm dt}[10]\implies 4x\dfrac{\mathrm dx}{\mathrm dt}+3y^2\dfrac{\mathrm dy}{\mathrm dt}=0$

When $x=1$ , the original equation tells you that you have

$2\times1^2+y^3=10\implies y^3=8\implies y=2$

You also know that when $x=1$ , you have $\dfrac{\mathrm dy}{\mathrm dt}=3$ . Substituting everything you know into the differentiated equation, you get

$4\times1\times\dfrac{\mathrm dx}{\mathrm dt}+3\times2^2\times3=0\implies\dfrac{\mathrm dx}{\mathrm dt}=-9$

So $x$ is changing at a rate of $-9\text{ units/min}$ .

6 0

4 years ago