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Tcecarenko [31]
3 years ago
9

URGET!!! IM TAKING A TEST. DUE TODAY

Mathematics
1 answer:
Novay_Z [31]3 years ago
7 0

Answer:

15x \geq 345

x = number of hours she works

Step-by-step explanation:

First: find the inequality symbol

If she wants to earn at least 345, she wants to make 345 or more

that means you use the greater than or equal symbol (\geq)

Next, if she makes 15 dollars an hour, x has to be the number of hours she works. The left side of the inequality would be $15* number of hours, or 15x

Thus, your inequality would be 15x \geq 345

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Solve the inequality
mafiozo [28]

Answer:

\large\boxed{q\geq-8}

Step-by-step explanation:

7q-5\leq10q+19\qquad\text{add 5 to both sides}\\\\7q\leq10q+24\qquad\text{subtract 10q from both sides}\\\\-3q\leq24\qquad\text{change the signs}\\\\3q\geq-24\qquad\text{divide both sides by 3}\\\\q\geq-8

5 0
3 years ago
The half-life is a substance is 375 years. If 70 grams is present now, how much will be present in 500 years?
Lynna [10]

Answer:

27.76 grams will be present in 500 years

Step-by-step explanation:

The given formula is A=A_{o}e^{kt} , where A is the value of the substance in t years, and A_{o} is the initial value

∵ The half-life is a substance is 375 years

- Substitute A by \frac{1}{2}A_{o} and t by 375 to find the value of k

∴ \frac{1}{2}A_{o}=A_{o}e^{375k}

- Divide both sides by A_{o}

∴ \frac{1}{2}=e^{375k}

- Insert ㏑ in both sides

∴ ㏑( \frac{1}{2} ) = ㏑ ( e^{375k} )

- Remember ㏑ ( e^{n} ) = n

∵ ㏑ ( e^{375k} ) = 375 k

∴ ㏑( \frac{1}{2} ) = 375 k

- Divide both sides by 375

∴ k ≈ -0.00185

∴  A=A_{o}e^{-0.00185t}

∵ 70 grams is present now

- That means the initial value is 70 grams

∴ A_{o} = 70

∵ The time is 500 years

∴ t = 500

- Substitute the values of A_{o} and t in the formula

∵ A=70e^{-0.00185(500)}

∴ A = 27.76

∴ 27.76 grams will be present in 500 years

3 0
3 years ago
Evaluate the expression 3x+2
scoray [572]

It's  3x+2 because 3 ÷2 is 6 so it's  3x+2

3 0
4 years ago
Which point lies in the solution set of the inequality 1/2(x+2)+3y<8?
nasty-shy [4]

Note: Options are missing. So, the general solution of the given inequality is shown below.

Given:

The inequality is:

\dfrac{1}{2}(x+2)+3y

To find:

The point lies in the solution set of the given inequality.

Solution:

We have,

\dfrac{1}{2}(x+2)+3y

It can be written as:

\dfrac{1}{2}(x)+\dfrac{1}{2}(2)+3y

\dfrac{1}{2}x+1+3y

\dfrac{1}{2}x+3y

\dfrac{1}{2}x+3y

All the points that satisfy the above inequality are in the solution set of the given inequality.

For example (0,0).

\dfrac{1}{2}(0)+3(0)

0

This statement is true. It means (0,0) is in the solution set.

For example (0,3).

\dfrac{1}{2}(0)+3(3)

9

This statement is false. It means (0,3) is not in the solution set.

The graph of the inequality \dfrac{1}{2}x+3y is shown below.

All the points in the shaded region are in the solution set but the points on the boundary line are not in the solution set.

8 0
3 years ago
A bag has 5 red marbles, 6 blue marbles and 4 black marbles. What is the probability of picking a
n200080 [17]

A bag has 5 red marbles, 6 blue marbles and 4 black marbles

so total no. of marbles = 5 + 6 + 4 = 15

now probability of picking a black marble = 4/15

after replacement again probability of picking black marble is 4/15

it is given that black marble is picked two times

so probability of picking black marble two times or picking , replacing , and then picking is 4/15*4/15 = 16/225

5 0
3 years ago
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