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3 years ago

And help me with B C D

Mathematics

1 answer:

WARRIOR [948]3 years ago

3 0

B has the equation -2 + 7 = -5
C has the equation -7 + 5 = -2
D has the equation 5 + (-7) = -2

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A cell phone bill provider offers a plan that costs $20 per month plus $0.10 per text message sent or received. A comparable pla

dybincka [34]

Answer:

100 texts

Step-by-step explanation:

20+.1x=30

30-20=10

10 dollar difference in base prices.

10/.1 (to find out how many texts)

100 texts

7 0

3 years ago

Use the function F(3) F(x) =(1/5)^x A.)1/15 B.)1/125 C.)3/5 D.)3/125

nalin [4]

Here is your answer

B) 1/125

REASON:

Given,

$F (x)={(1/5)}^{x}$

for F (3)

Putting x=3, in above function we get

$F (3)= {(1/5)}^{3}= 1/125$

HOPE IT IS USEFUL

6 0

3 years ago

HELP!!! Anyone know how to solve I know the answer is (-3) but I don’t know how to get there. Thanks!!

Dima020 [189]

Answer: v = -3

=========================================

Work Shown:

$5 = \sqrt{-2-6v}+1\\\\5-1 = \sqrt{-2-6v}\\\\4 = \sqrt{-2-6v}\\\\4^2 = \left(\sqrt{-2-6v}\right)^2 \ \text{ .... square both sides}\\\\16 = -2-6v\\\\16+2 = -6v\\\\18 = -6v\\\\-6v = 18\\\\v = 18 \div (-6)\\\\v = -3$

As you can see, we don't square both sides until the square root portion is fully isolated or on its own side. So it happens after we subtract 1 from both sides.

Technically, you are able to square both sides without first isolating the square root. But that would mean you'd have to use the FOIL rule and things would get a bit messier than they have to be. Not to mention that the square root term wouldn't fully go away (so you'd have to square again later down the line).

------------------

Checking the answer:

Replace every copy of v with -3. Simplify both sides. We should end up with the same number on each side

$5 = \sqrt{-2-6v}+1\\\\5 = \sqrt{-2-6(-3)}+1\\\\5 = \sqrt{-2+18}+1\\\\5 = \sqrt{16}+1\\\\5 = 4+1\\\\5 = 5\\\\$

The answer is confirmed.

4 0

3 years ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago

Marcus earns $8.40 per hour. He works for 26 hours each week, 48 weeks each year. If he earns over $10,000, then Marcus has to p

MAVERICK [17]

We have been given that Marcus works for 26 hours each week and 48 weeks each year.

Thus in a year he works for $26 \times 48 = 1248$ hours.

Now, we have been given that Marcus earns $8.40 per hour.

Therefore, in 1248 hours, he earns

$1248 \times 8.40\\
\\
=\$ 10483.2$

Marcus has to pay tax if he earns more than $10,000. Now we have calculated that he earns $ 10483.2 which is higher than $10,000.

Therefore, Marcus has to to pat tax.

6 0

3 years ago

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