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2 years ago

7

a map with a scale of 0.25 inches = 30 miles, the distance between two cities is 8.3 inches. Find the actual distance between th

e cities.

1 answer:

Crank2 years ago

5 0

Step-by-step explanation:

the answer to your question is400 miles

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Help? please.........

UkoKoshka [18]

Answer: 60°

Step-by-step explanation: A straight angle is 180°

180°- 120° = 60°

8 0

2 years ago

there 240 students in the middle school band. The band director is dividing the students into groups of 10. Into how many groups

defon

Answer:

Step-by-step explanation:

1 group = 10 students

x group = 240 students.

1/x = 10/240 Cross multiply

10x = 240 Divide by 10

10x/10=240/10 Do the division

x = 24

8 0

2 years ago

1.) Find the value of the discriminant and the the number of real solutions of

ruslelena [56]

Hi there!

When we have an equation standard form...
$a {x}^{2} + bx + c = 0$
...the formula of the discriminant is
D = b^2 - 4ac

When
D > 0 we have two real solutions
D = 0 we have one real solutions
D < 0 we don't have real solutions

1.) Find the value of the discriminant and the the number of real solutions of
x^2-8x+7=0

Plug in the values from the equation into the formula of the discriminant

$( - 8) {}^{2} - 4 \times 1 \times 7 = 64 - 28 = 36$

D > 0 and therefore we have two real solutions.

2.) Find the value of the discriminant and the number of real solutions of
2x^2+4x+2=0

Again, plug in the values from the equation into the formula of the discriminant.

${4}^{2} - 4 \times 2 \times 2 = 16 - 16 = 0$

D = 0 and therefore we have one real solution.

~ Hope this helps you.

4 0

2 years ago

Read 2 more answers

Which shows 23/5 as a mixed number?

satela [25.4K]

Answer:4 3/5

Step-by-step explanation:

3 0

3 years ago

Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6

san4es73 [151]

Answer:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

Step-by-step explanation:

Given

$f(x)= \frac{9}{3x+ 2}$

$c = 6$

The geometric series centered at c is of the form:

$\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.$

Where:

$a \to$ first term

$r - c \to$ common ratio

We have to write

$f(x)= \frac{9}{3x+ 2}$

In the following form:

$\frac{a}{1 - r}$

So, we have:

$f(x)= \frac{9}{3x+ 2}$

Rewrite as:

$f(x) = \frac{9}{3x - 18 + 18 +2}$

$f(x) = \frac{9}{3x - 18 + 20}$

Factorize

$f(x) = \frac{1}{\frac{1}{9}(3x + 2)}$

Open bracket

$f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}$

Rewrite as:

$f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}$

Collect like terms

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}$

Take LCM

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}$

$f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}$

So, we have:

$f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}$

By comparison with: $\frac{a}{1 - r}$

$a = 1$

$r = -\frac{1}{3}x + \frac{7}{9}$

$r = -\frac{1}{3}(x - \frac{7}{3})$

At c = 6, we have:

$r = -\frac{1}{3}(x - \frac{7}{3}+6-6)$

Take LCM

$r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)$

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n$

Substitute 1 for a

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n$

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n$

Substitute the expression for r

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n$

Expand

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]$

Further expand:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................$

The power series converges when:

$\frac{1}{3}|x - \frac{7}{3}| < 1$

Multiply both sides by 3

$|x - \frac{7}{3}|$

Expand the absolute inequality

$-3 < x - \frac{7}{3}$

Solve for x

$\frac{7}{3} -3 < x$

Take LCM

$\frac{7-9}{3} < x$

$-\frac{2}{3} < x$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

6 0

2 years ago