P(3,4), y = x - y 1

Mathematics

1 answer:

Ivanshal [37]3 years ago

7 0

Answer:

i don knwo

Step-by-step explanation:

but thanks for the points lol

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24 boys started a cycle racc and 18 of them

soldier1979 [14.2K]

75% ended the race
25-18=6
6 goes into 24 4 times so it is 25% that did not finish
100%-25%=75%
So 75% finished the race

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3 years ago

A gerbil cage,priced at $18, is discounted 20% for one week, and then this price is marked up 20%. is the price back to 20% expl

stiks02 [169]

No, it isn't!

first, the discount is 20 percent of 18 dollars, that is 18/5=3.6

this means that during the first week the prize is 18-3.6=14.4

now, the prize is marked up 20 % of 14.4! this is 2.88

so the prize later will be 14.4+2.88, that is 17.28, not 18

what is relevant here is what a number is a percent OF

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3 years ago

ANOTHER QUESTION:

iris [78.8K]

If there are 3 red blocks for every 5 blocks
if we multiple both numbers by 12 to make every 60 blocks we get:
36 red blocks every 60 blocks
since you have 36 and you only need 32 then yes you will have enough

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2 years ago

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

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3 years ago

10.1 and 10.5 witch one is greater

OLga [1]

Answer:

10.5

Step-by-step explanation:

Because the digit at tenths place in 10.5 is greater than that of in 10.1

$5 > 1$