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3 years ago

Write an inequality that represents the graph below:

Mathematics

1 answer:

Ray Of Light [21]3 years ago

5 0

Answer:

X is greater than 3

X > 3

Step-by-step explanation:

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Please help me with this! I will give you a thanks!

Grace [21]

-2 + 6
.................

4 0

3 years ago

Read 2 more answers

Please solve this simultaneous equation, i'm not that good at it. 2r-s=10 rs-s^2=12

o-na [289]

Answer:

s = 6, r = 8 or s = 4, r = 7

Step-by-step explanation:

2r - s = 10

2r = 10 + s

r = 5 + s/2 --(1)

rs - s^2 = 12 --(2)

sub (1) into (2):

(5 + s/2)s - s^2 = 12

5s + 0.5s^2 - s^2 = 12

-0.5s^2 + 5s - 12 = 0

s^2 - 10s + 24 = 0

(s - 6)(s - 4) = 0

therefore s = 6 or s = 4

when s = 6, r = 8 and when s = 4, r = 7

if you would like to discover more about simultaneous equations, you can have a look at my Instagram page (learntionary). I'll be posting mathematics related stuff there and some of my notes (simultaneous enq notes are already posted!)

7 0

2 years ago

I need your Help, please! THX:)

USPshnik [31]

450 miles
1 in. Per 60 miles
460/60 = 7.5 inches

7 0

2 years ago

Read 2 more answers

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

Matthew purchased a desk that was on sale for 45% off the orginal price of $480.If the sales tax was 8% (of the sale price), how

Yanka [14]

Answer:

285.12

Step-by-step explanation:

discount amount = 45/100 x 480= 45x48/10

=216.0

price after discount= 480-216=264

sales tax= 8/100 x 264=21.12

total money spent = 264+21.12=285.12

3 0

2 years ago

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