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2 years ago

I’m wondering if you could give me answers

Mathematics

1 answer:

leonid [27]2 years ago

3 0

Answer:

the answer is b bro

Step-by-step explanation:

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Mort bought his house in 2004 for $279,000. it was recently appraised at $186,000. What is the approximate percent of decrease i

Natalka [10]

Answer:

33.3%

Step-by-step explanation:

(279,000 - 186,000) / 279,000 = 0.33333333333

0.33333333333*100= 33.3333333333

Formula used:

$V_{2} -V_{1}$ / $V_{2}$ × 100

V2= 279,000

V1= 186,000

7 0

3 years ago

(9, 18)and<br> (3, n) .<br> Find<br> n

Maru [420]

I don't seem to comprehend what you are asking?

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2 years ago

Solve 2x-8=-2-8. How do I do this

faust18 [17]

Is that negative 2 minus 8? if so put that into the calculator and add 8 to both sides and divide by 2 after that. simple algebra

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2 years ago

Read 2 more answers

If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the v

pochemuha

Answer:

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.

Step-by-step explanation:

Given:

V = $5000(1 - \frac{1}{40}t )^2$ , where 0≤t≤40.

Here we have to find the derivative with respect to "t"

We have to use the chain rule to find the derivative.

V'(t) = $2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )$

V'(t) = $2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )$

When we simplify the above, we get

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.

4 0

3 years ago

Find the smallest positive $n$ such that \begin{align*} n &\equiv 3 \pmod{4}, \\ n &\equiv 2 \pmod{5}, \\ n &\equiv

Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number $x$ such that taking it mod 4, 5, and 7 leaves the desired remainders:

$x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6$

Taken mod 4, the last two terms vanish and we have

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4$

so we multiply the first term by 3.

Taken mod 5, the first and last terms vanish and we have

$x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5$

so we multiply the second term by 2.

Taken mod 7, the first two terms vanish and we have

$x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7$

so we multiply the last term by 7.

Now,

$x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147$

By the CRT, the system of congruences has a general solution

$n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}$

or all integers $27+140k$ , $k\in\mathbb Z$ , the least (and positive) of which is 27.

3 0

2 years ago