All categories

3 years ago

Melanie draws with great fervor, using up pencils at an incredible rate: 14 pencils every

1 answer:

5 0

4 pencils . 4/9 of an hour = 4/9 of 60 minutes = 4/9 * 60 = 80/3 minutes
x pencils = ? . one hour = 60 minutes

14 * 60 = 80/3 * x /*3
14 * 60 * 3 = 80 * x
80 * x = 2520 /80
x = 2520 / 80
x = 31.5 pencils

The result would be 32 pencils in one hour.

You might be interested in

Pablo put 1260 into a savings account that earns 3% simple interest per year he does not make any deposits ir withdrawls how muc

pav-90 [236]

Si = prt
Si = 1260 (0.03) (2)
Si= 1260 (0.06)
Si = 75.60.
So it’s 1260 + 75.60 = $ 1335.60

5 0

3 years ago

What is 9minus w = to

skad [1K]

9 - w = 2

9 - 7 = 2

If this is what you mean.

7 0

3 years ago

Read 2 more answers

Write the quadratic equation y=x^2-6+7

sveticcg [70]

Do you have answer choices?

8 0

3 years ago

Find the midpoint of the line segment whose endpoints are (-2, 5) and (4, -9).Find the midpoint of the line segment whose endpoi

jek_recluse [69]

Answer:

The midpoint is $(1,-2)$

Step-by-step explanation:

we know that

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

we have

$(x1,y1)=(-2,5)$

$(x2,y2)=(4,-9)$

substitute

$M(\frac{-2+4}{2},\frac{5-9}{2})$

$M(1,-2)$

8 0

2 years ago

4, Find a number x such that x = 1 mod 4, x 2 mod 7, and x 5 mod 9.

olchik [2.2K]

4, 7 and 9 are mutually coprime, so you can use the Chinese remainder theorem.

Start with

$x=7\cdot9+4\cdot2\cdot9+4\cdot7\cdot5$

Taken mod 4, the last two terms vanish and we're left with

$x\equiv63\equiv64-1\equiv-1\equiv3\pmod4$

We have $3^2\equiv9\equiv1\pmod4$ , so we can multiply the first term by 3 to guarantee that we end up with 1 mod 4.

$x=7\cdot9\cdot3+4\cdot2\cdot9+4\cdot7\cdot5$

Taken mod 7, the first and last terms vanish and we're left with

$x\equiv72\equiv2\pmod7$

which is what we want, so no adjustments needed here.

$x=7\cdot9\cdot3+4\cdot2\cdot9+4\cdot7\cdot5$

Taken mod 9, the first two terms vanish and we're left with

$x\equiv140\equiv5\pmod9$

so we don't need to make any adjustments here, and we end up with $x=401$ .

By the Chinese remainder theorem, we find that any $x$ such that

$x\equiv401\pmod{4\cdot7\cdot9}\implies x\equiv149\pmod{252}$

is a solution to this system, i.e. $x=149+252n$ for any integer $n$ , the smallest and positive of which is 149.

3 0

3 years ago