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4 years ago

217x + 99 = 25 what is the answer

Mathematics

1 answer:

EleoNora [17]4 years ago

4 0

217x+99=25 217x=25-99 217x= -74 X=-74/217

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Jessica bought two cans of peas three cans of tomato sauce five cans of corn and 12 rolls what is the total cost of her purchase

Paraphin [41]

She has 22 items in total

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4 years ago

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What is the mode? 22, 7, 22, 1, 7, 18, 18, 16, 6, 6, 7

kolbaska11 [484]

Answer:

Step-by-step explanation:

Put them in L to G

1 6 6 7 7 7 16 18 18 22 22

and u see how many times 7 repeated den all numbers.

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3 years ago

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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

Turn 20/12 into a mixed number

luda_lava [24]

Answer:

1 2/3

Step-by-step explanation:

well first reduce it which would be 5/3 since the greatest common denominator would be 4. Then put 3 in the 5 which goes only once and 2 is left over. Keep the threeas the denominator instead of the numerator.

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3 years ago

Law of cosines: a2 = b2 + c2 – 2bccos(A) What is the measure of N to the nearest whole degree? 35° 45° 55° 65°

oksano4ka [1.4K]

Answer:

The measure of angle N is 55°

Step-by-step explanation:

we know that

the law of cosines established

a² = b² + c² – 2*b*c*cos(A)

In this problem we have

a=ML=5.74 units

b=LN=4 units

c=MN=7 units

∠A=∠N=?

see the attached figure to better understand the problem

a² = b² + c² – 2*b*c*cos(A)------> cos (A)=[b²+c²-a²]/[2*b*c]

cos (A)=[4²+7²-5.74²]/[2*4*7]------> 0.57236

A=arc cos (0.57236)--------> A= 55°

therefore

The measure of angle N is 55°

3 0

3 years ago

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