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3 years ago

What is the solution set of x2 – 10 = 30x?

Mathematics

2 answers:

kobusy [5.1K]3 years ago

3 0

Answer:

15+sqrt(235), 15-sqrt(235)

Step-by-step explanation:

X^2-30x-10=0

Use quadratic formula (b+-sqrt(b^2-4(a)(c)))/(2(a))

a=1, b=-30, c=-10

lesantik [10]3 years ago

3 0

Answer:

Step-by-step explanation:

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Number of Text Messages a Day A random sample of n = 755 US cell phone users age 18 and older in May 2011 found that the average

Alexeev081 [22]

Answer:

The 95% confidence interval is (29.54 - 53.46}

Step-by-step explanation:

given data:

$\hat X = 41.5$

Se = 6.1

n = 755

a) best estimate $\hat x = 41.5$

b) at 95% confidence interval

$\alpha = 1- 0.95 = 0.05$

$\alpha /2 = 0.025$

$z_{\alpha/2} = z_{0.025} = 1.96$

at 95% confidence interval for [/tex]\mu[/tex]

$\hat x \pm z_{\alpha /2} \times Se$

$41.5 \pm 1.96\times 6.1$

$41.5 \pm 11.96$

The 95% confidence interval is (29.54 - 53.46}

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3 years ago

what are the values of m∠abc, m∠cbf, and m∠dbe in the given figure? Explain or show your work. A good 55 points will go to the o

vesna_86 [32]

Step-by-step explanation:

m∠abc = m∠ebf = 58°

reason: vertical angles

m∠cbf = 180-m∠ebf = 180-58 = 122°

reason: Supplementary angles

m∠dbe = 90-m∠ebf = 90-58 = 32°

reason: complementary angles

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3 years ago

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uranmaximum [27]

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2 years ago

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mina [271]

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3 years ago

Given that (x+y+z)(xy+xz+yz)=18 and that x^2(y+z)+y^2(x+z)+z^2(x+y)=6 for real numbers x, y, and z, what is the value of xyz?

worty [1.4K]

Answer:

Step-by-step explanation:

$(x+y+z)(xy+xz+yz)=18$ Equation 1

$x^2(y+z)+y^2(x+z)+z^2(x+y)=6$ Equation 2

What is the value of $xyz$ where each variable represents a real number?

Let's expand equation 1:

$(x+y+z)(xy+xz+yz)=18$

$x(xy)+x(xz)+x(yz)+y(xy)+y(xz)+y(yz)+z(xy)+z(xz)+z(yz)=18$

Simplify each term if can:

$x^2y+x^2z+xyz+y^2x+xyz+y^2z+xyz+z^2x+z^2y=18$

See if we can factor a little to get some of the left hand side of equation 2:

The first two terms have $x^2$ and if I factored $x^2$ from first two terms I would have $x^2(y+z)$ which is the first term of left hand side of equation 2.

So let's see what happens if we gather the terms together that have the same variable squared together.

$x^2y+x^2z+y^2x+y^2z+z^2y+z^2x+xyz+xyz+xyz=18$

Factor the variable squared terms out of each binomial pairing:

$x^2(y+z)+y^2(x+z)+z^2(y+x)+xyz+xyz+xyz=18$

Replace the sum of those first three terms with what it equals which is 6 from the equation 2:

$6+xyz+xyz+xyz=18$

Combine like terms:

$6+3xyz=18$

Subtract 6 on both sides:

$3xyz=12$

Divide both sides by 3:

$xyz=4$

4 0

3 years ago