All categories

3 years ago

Solve the System of Equations

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

6 0

Answer:

Equations: $x+y=80\,,\,2x+4y=216\,\,$

number of chickens = 52

number of goats = 28

Step-by-step explanation:

Given: A small ranch has chickens and goats

Total number of heads = 80

Total number of legs = 216

To find: number of chickens and goats

To draw: the graph

Solution:

Let x denotes number of chickens and y denotes number of goats.

According to the question,

$x+y=80\,\,(i)\\2x+4y=216\,\,(ii)$

From (i), put $x=80-y$ in (ii)

$2(80-y)+4y=216\\160-2y+4y=216\\2y=216-180\\2y=56\\y=\frac{56}{2}=28$

Put $y=28$ in equation (i)

$x+28=80\\x=80-28=52$

So,

number of chickens = 52

number of goats = 28

Download pptx

You might be interested in

Plz help me out with this and do show work I will be waiting.... btw u will be marked brainliest!!!!!! Fast plz

pychu [463]

Answer:

x=16°

Step-by-step explanation:

5*x-30=2*(x+18)

For the second, I'd say it is:

5*x-30=2*x+18

Let's solve both.

5*x-30=2*(x+18)

5x-30=2x+36

3x=66

x=22

5*x-30=2*x+18

3x+48

x=16°

8 0

3 years ago

josefina surveyed 298 students and found that 52% like scary movies . estimate the number of students that like scary movies .

Inessa [10]

It would be: 298*52 / 100 = 154.96

Approx. 155 students like scary movies

8 0

3 years ago

Keith has $500 in a savings account at the beginning of the summer. He withdraws $25 each week for food, clothes, and movie tick

alexdok [17]

Answer:

You can put this solution on YOUR website!

the inequality is 500 - 25x >= 200

this insures that he will have at least 200 at the end of the summer.

subtract 200 from both sides of that inequality and add 25x to both sides of that inequality to get 500 - 200 >= 25x

simplify to get 300 >= 25x

divide both sides of that equation by 25 to get 300 / 25 >= x

simplify to get 12 >= x

12 >= x means x <= 12.

when x is smaller than or equal to 12, he will be guaranteed to have at least 200 in the account at the end of the summer.

when x = 12, what is left in the account is 500 - 25 * 12 = 200.

when x = 11, what is left in the account is 500 - 25 * 11 = 225.

when x = 13, what is left in the account is 500 - 25 * 13 = 175.

the maximum number of weeks he can withdraw money from his account is 12.

Step-by-step explanation:

6 0

3 years ago

Evaluate 3/7(14)+5/8(8) ?

Misha Larkins [42]

Did you get this answer???!

4 0

3 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