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AysviL [449]
3 years ago
14

Needed now PLEASE HELP ME FOR BRAINIEST on a graph what does the solution of a system of equations look like.

Mathematics
1 answer:
siniylev [52]3 years ago
7 0
On a graph with two or more different lines representing the two or more different equations in a system of equations, the solution to the system of equations is the point at which the different lines intersect.

Hope this helps!
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Jacob has some cookies. 3/4 of his cookies are chocolate chip. Out of the chocolate chip cookies 1/8 of them have nuts as well.
ryzh [129]
Well, we can find the answer by multiplying the two fractions. 3/4*1/8 equals 3/32.

7 0
3 years ago
Read 2 more answers
Anyone know how to do this?
Lera25 [3.4K]

Answer:

Part A) 25

Part B) 225

Step-by-step explanation:

Part A

The slope of the graph gives us the mile per gallon of the car.

Recall that the slope of the graph is rise over run.

m =  \frac{200}{8}  = 25

Therefore car gets 25 miles per gallon

Part B) Since we know the slope of the direct variation line, we substitute into y=kx to get:

y = 25x

To find the number of miles the car travels using 9 gallons, we substitute x=9 to get:

y = 25 \times 9

This gives us 225 miles

8 0
3 years ago
How do you solve #18?
kompoz [17]
Budget ≤ 1250 - 439
 m = $811
3 0
3 years ago
Read 2 more answers
Mrs Richmond bought some apples and pears in the ratio 2:5. An apple costs 30 cents. A pear costs 15 cents mire than an apple. H
Lera25 [3.4K]

Answer:

$2.85

Step-by-step explanation:

30×2=60

30+15=45

60+(45×5)=?

45×5=225

60+225=285

$2.85

Cuz every 100cents is one dollar

5 0
3 years ago
Solve only if you know the solution and show work.
SashulF [63]
\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx

For the remaining integral, let t=\tan\dfrac x2. Then

\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}
\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}

and

\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt

Now the integral is

\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt

The first integral is trivial, so we'll focus on the latter one. You have

\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt

Decompose the integrand into partial fractions:

\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}

so you have

\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt

which are all standard integrals. You end up with

\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt
=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C
=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C
=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C

To try to get the terms to match up with the available answers, let's add and subtract \ln\left|1+\tan\dfrac x2\right| to get

2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C
2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C

which suggests A may be the answer. To make sure this is the case, show that

\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1

You have

\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}
\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}
\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}

So in the corresponding term of the antiderivative, you get

\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|
=\ln2-\ln|\cos x+\sin x+1|

The \ln2 term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C
5 0
3 years ago
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