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MA_775_DIABLO [31]

2 years ago

6

A water well drilling rig has dug to a height of -15 feet in three hours. If the rig has been running constantly and is currentl

y at a height of -192.5 feet, for how long has the rig been running?
22.25 hours

97.4 hours

38.5 hours

12.8 hours

2 answers:

Svet_ta [14]2 years ago

7 0

Answer: I think its A If you do the math corrcetly

Step-by-step explanation:

Ilia_Sergeevich [38]2 years ago

6 0

A. 22.25 i think it’s a because it you do your math correctly it was give you A

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Which equation does the graph below represent <br> !!!NEED HELP ASAP!!!

ella [17]

Answer:

im pretty sure its -4

Step-by-step explanation:

you cant really hold me to it but its -1/4 or -4

4 0

2 years ago

Read 2 more answers

1. Isaac goes to an amusement park where tickets for the rides cost $10 per sheet and tickets for the shows cost $15 each.

Degger [83]

The answers I have gotten are
A- x+y= $25
B- $95

7 0

3 years ago

Using algebra, solve the inequality<br> writing your answer in set notation.<br> x²-x> 20

Xelga [282]

Answer:

(-4, 0) U (1, ∞)

Step-by-step explanation:

Set each factor EQUAL to zero to find the zeroes (since it is not actually equal to zero, you will use an open circle when graphing and an open bracket when writing in interval notation).

x = 0 x-1 = 0 x + 4 = 0

x = 1 x = -4

Next, choose a value to the far left, between each of the zeroes, and to the far right to evaluate if it makes a true statement when input into the given inequality.

far left (I choose -5): -5(-5 - 1)(-5 + 4) > 0 → (-)(-)(-) > 0 → negative > 0 FALSE

- 4 to 0 (I choose -2): -2(-2 - 1)(-2 + 4) > 0 → (-)(-)(+) > 0 → positive > 0 TRUE

0 to 1 (I choose 0.5): .5(.5 - 1)(.5 + 4) > 0 → (+)(-)(+) > 0 → negative > 0 FALSE

far right (I choose 2): 2(2 - 1)(2 + 4) > 0 → (+)(+)(+) > 0 → positive > 0 TRUE

3 0

8 months ago

Write a solution in Interval Notation - (you don't have to help me on all, 1 or 2 is fine c: )

Gemiola [76]

QUESTION 1

The given inequality is

$|m|-2>0$

We group like terms to get,

$|m|>2$

This implies that,

$-m>2$ or $m>2$ .

We simplify the inequality to get,

$m$ or $m>2$ .

We can write this interval notation to get,

$(-\infty,-2)\cup (2,+\infty)$ .

QUESTION 2

$|x-4|-3\:>\:5$ .

We group like terms to get,

$|x-4|\:>\:5+3$ .

$|x-4|\:>\:8$

We split the absolute value sign to get,

$-(x-4)\:>\:8$ or $x-4\:>\:8$

This implies that,

$x-4\:$ or $x-4\:>\:8$

$x\:$ or $x\:>\:8+4$

$x\:$ or $x\:>\:12$

We can write this interval notation to get,

$(-\infty,-4)\cup (12,+\infty)$ .

QUESTION 3

The given inequality is

$|6+9x|\leq 24$

We split the absolute value sign to obtain,

$-(6+9x)\leq 24$ or $(6+9x)\leq 24$

This simplifies to

$6+9x\ge -24$ and $6+9x\leq 24$

$9x\ge -24-6$ and $9x\leq 24-6$

$9x\ge -30$ and $9x\leq 18$

$x\ge -\frac{10}{3}$ and $x\leq 2$

$-\frac{10}{3}\leq x\leq2$

We write this in interval form to get,

$[-\frac{10}{3},2]$

QUESTION 4

The given inequality is

$|1-5a|>29$

We split the absolute value sign to get,

$-(1-5a)>29$ or $1-5a>29$

This simplifies to,

$1-5a\:$ or $1-5a\:>\:29$

This implies that,

$-5a\:$ or $-5a\:>\:29-1$

$-5a\:$ or $-5a\:>\:28$

$a\:>\:6$ or $a\:$

We write this in interval notation to get,

$(-\infty,-\frac{28}{5})\cup (6,+\infty)$

7 0

2 years ago

Help me what if this answer

Brums [2.3K]

I think you need to make the 6 positive and then add it to 13 which would give you a=19

6 0

2 years ago