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3 years ago

X+2y=10 X+y=6 Using eliminations to solve each system

Mathematics

2 answers:

Svetllana [295]3 years ago

7 0

Answer:

x= 2

y=4

Just subtract by like numbers.

shusha [124]3 years ago

4 0

x + 2y = 10

x + y = 6

Both x's in the equation are equal. We want the x of the last equation to be negative so we can eliminate.

x + 2y = 10

-x - y = -6

Subtract.

y = 4

Now that we figured out y, we have to figure out x. Substitute y into one on the equations.

x + 4 = 6

x = 2

Now we figured out x. The answer is (2,4). I hope this helps! Let me know if I got anything wrong or if you need any more help with this problem.

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The interquartile range (IQR) for the data: 32,7,3,-5,1,8 , 10,36 is :

GenaCL600 [577]

Answer:

Your Interquartile range (IQR) would be 19.

Step-by-step explanation:

-5, 1, 3, 7, 8, 10, 32, 36

Median: 7.5

Lower quartile: 2

Upper quartile: 21

Interquartile range: 21 - 2 = 19

7 0

3 years ago

When Anna was born 16 years ago, her small town of Lewisville had a population of 15,000. The population has increased 11% each

ZanzabumX [31]

16 years ago = - 16
Present population -16 = 15,000 (population)

11% increase = every year =0.11

present year = 15,000 x 0.11^16
= 15,000 x 4.594
approx = 68,925

7 0

3 years ago

Read 2 more answers

Bring the fraction a/a−4 to a denominator of 16−a^2 really do appreciate this thx

RSB [31]

Answer:

-a(a+4)/(16 - a²)

Step-by-step explanation:

a/(a - 4) Multiply by (a + 4)/(a + 4)

= a(a + 4)/[(a – 4)(a + 4)] Multiply the denominatorator terms

= a(a + 4)/(a² - 16) Multiply by -1/(-1)

= -a(a+4)/(-a² + 16) Reorder terms in denominator

= -a(a+4)/(16 - a²)

5 0

3 years ago

Round 12.5478 to the nearest hundredths

solong [7]

Answer:

12.04

Step-by-step explanation:

Find the number in the hundredth place 4 and look one place to the right for the rounding digit 1 . Round up if this number is greater than or equal to 5 and round down if it is less than 5 .

6 0

3 years ago

Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

3 years ago