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2 years ago

What quadratic equation could be solved to find any solutions to the system of equations?

Mathematics

1 answer:

Kipish [7]2 years ago

7 0

Answer:

Step-by-step explanation:

x²+7x-11=-5x+6

x²+7x+5x-11-6=0

x²+12x-17=0

$x=\frac{-12\pm \sqrt{12^2-4*1*(-17)} }{2*1} \\=\frac{-12 \pm \sqrt{144+68} }{2} \\=\frac{-12 \pm \sqrt{212}}{2} \\=\frac{-12+2\sqrt{53}}{2}\\=-6 \pm \sqrt{53}$

You might be interested in

Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay p

garik1379 [7]

Answer:

m = $\frac{1}{20}$
μ = 20
σ = 20

The probability that a person is willing to commute more than 25 miles is 0.2865.

Step-by-step explanation:

Exponential probability distribution is used to define the probability distribution of the amount of time until some specific event takes place.

A random variable X follows an exponential distribution with parameter m.

The decay parameter is, m.

The probability distribution function of an Exponential distribution is:

$f(x)=me^{-mx}\ ;\ m>0, x>0$

Given: The decay parameter is, $\frac{1}{20}$

X is defined as the distance people are willing to commute in miles.

The decay parameter is m = $\frac{1}{20}$ .
The mean of the distribution is: $\mu=\frac{1}{m}=\frac{1}{\frac{1}{20}}=20$ .
The standard deviation is: $\sigma=\sqrt{variance}= \sqrt{\frac{1}{(m)^{2}} } =\frac{1}{m} =\frac{1}{\frac{1}{20}} =20$

Compute the probability that a person is willing to commute more than 25 miles as follows:

$P(X>25)=\int\limits^{\infty}_{25} {\frac{1}{20} e^{-\frac{1}{20}x}} \, dx \\=\frac{1}{20}|20e^{-\frac{1}{20}x}|^{\infty}_{25}\\=|e^{-\frac{1}{20}x}|^{\infty}_{25}\\=e^{-\frac{1}{20}\times25}\\=0.2865$

Thus, the probability that a person is willing to commute more than 25 miles is 0.2865.

7 0

3 years ago

Please someone answer ASAP!!

Nina [5.8K]

Answer:

<h2> Please put some pictures so we can answer your question,so you will not mad or angry that we're answering this question.</h2>

8 0

2 years ago

Martin decided to buy 2 bags of grapes weighing 5 and two-thirds pounds each, instead of the 3 bags weighing 4 and one-fifth pou

Juli2301 [7.4K]

Answer: He ended up with about 1 pound less grapes

Step-by-step explanation:

2 bags of grapes weighing 5 and two-thirds pounds each will give a total weight of:

= 2 × 5 2/3

= 2 × 17/3

= 34/3

= 11 1/3 pounds

3 bags weighing 4 and one-fifth pounds each, will give a weight of:

= 3 × 4 1/5

= 3 × 21/5

= 63/5

= 12 3/5

He ended up buying lesser grapes. This was about 12 3/5 - 11 1/3 = 1 4/15 less. This means that he ended up with about 1 pound less grapes

3 0

2 years ago

Read 2 more answers

Point C divides segment AB so that AC:AB is 2:9. Which statement is NOT true?

jeka94

Answer:

A. Point C divides segment AB so that AC:CB is 1:3

or either c

5 0

3 years ago

3х - 30 = у 7y - 6 = 3х

mars1129 [50]

Answer:

Step-by-step explanation:

This one is simple substitution... at least, substitution is the easiest method. The first equation is 3x – 30 = y and the second is 7y – 6 = 3x

As I look, I see 3 ways to use substitution to solve this:

substitute 7y – 6 for 3x
substitute 3x – 30 for y
solve 3x – 30 = y for 3x and make it equal to 7y – 6

We're going to only use 1 method for the sake of time. Try the other two on your own. Assuming you don't make any mistakes, they will work.

Method 1:

3x – 30 = y

7y – 6 = 3x — initial system of equations

7y – 6 – 30 = y — substitute 7y – 6 for 3x

7y – 6 – 30 = y — marking like terms, bold for constants, underlined for variables

7y – 36 = y — combining the constants and simplifying

Here, you could diverge into multiple paths: add 36 to both sides, subtract y from both sides, divide by 6 OR subtract 7y from both sides and divide by –6 . For the sake of time, I'm subtracting 7y, though I don't like dealing with negatives.

7y – 7y – 36 = y – 7y — subtract 7y from both sides

–36 = –6y — simplify

–36 ÷ –6 = –6y ÷ –6 — divide by –6 on both sides

y = 6 — simplify

Again, we can diverge here: substitute y into 3x – 30 = y or substitute y into 7y – 6 = 3x

I'm going to choose 3x – 30 = y but it will work either way, should you take the time (if you have it) to chase down every path this problem can take.

3x – 30 = y — initial equation

3x – 30 = 6 — substitute 6 for y

3x – 30 + 30 = 6 + 30 — add 30 to both sides to isolate 3x

3x = 36 — simplify the expression

3x ÷ 3 = 36 ÷ 3 — divide both sides by 3 to isolate x

x = 12 — simplify

So, we have x = 12 and y = 6 . We know they work for 3x – 30 = y but not if they work for 7y – 6 = 3x . Let's substitute those in to see if (12, 6) really is the solution point.

7y – 6 = 3x — original equation

7(6) – 6 ≟ 3(12) — substitute 6 for y and 12 for x

42 – 6 ≟ 36 — simplify by multiplying

36 = 36 ✔ — simplify by combining like terms on left side

Success! It works! We have found our solution!

I hope this helps increase your understanding of the concept. Have a great day!

8 0

2 years ago