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timama [110]
3 years ago
7

4 - |-2|= what is the answer please

Mathematics
1 answer:
Gelneren [198K]3 years ago
5 0
The answer is 2 hope it helps
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Four more than the quotient of a number and 3 is at least 9
Natali [406]
The correct answer is 15

 <span>4 + (x/3) = 9 </span>

<span>(x/3) = 9 - 4 </span>

<span>(x/3) = 5 </span>

<span>x = 15 </span>
7 0
3 years ago
Help!!
lora16 [44]
False

(5,5)
y < x + 3
5 < 5 + 3
5 < 8 (correct)

(5,5)
y > -2x - 1
5 > -2(5) - 1
5 > -10 - 1
5 > -11 (correct)

so (5,5) is a solution..and there are many more 
8 0
3 years ago
Which is most applicable when continuous inspection is used? I-chart, P-chart, c-chart C-bar chart
Zepler [3.9K]

Answer:

I - Chart

Step-by-step explanation:

The charts are used for representation of inspected data in to structure way which helps understand the changes easily. I chart is usually used when there is continuous inspection for the data. This type of chart used to monitor the process when measuring data at regular intervals.

8 0
3 years ago
Consider the equation 3p – 7 + p = 13. What is the resulting equation after the first step in the solution?
riadik2000 [5.3K]
4p-7=13
After the first step which is combining like terms, so add the 3p and the single p together.

Next, the negative 7 turns to a positive when we exterminate it from one half of the equation and add it towards the second part.
4p-7=13
+7. +7
4p=20

Divide both sides by 4 to get the p alone.

4p=20
/4. /4

P=5
That is the answer to the entire equation, again though the first step answer is at the very top, i just felt like solving it for you as well.


3 0
4 years ago
Read 2 more answers
Please Help - Suppose f(t)=6t−8−−−−√.
Gnoma [55]

(a)\\\\\\\text{Given that,}~~\\\\ f(t) =\dfrac 6{\sqrt{t-8}}\\\\\\\\\implies f'(t) = 6\left[\dfrac{\left(\sqrt{t-8}\right) \cdot 0 - \dfrac 1{2\sqrt{t-8 }}}{\left(\sqrt{t-8}\right)^2}\right]\\\\\\\\\implies f'(t) = -6\left(\dfrac {\dfrac 1{2\sqrt{t-8}}}{t-8}\right)\\\\\\\implies f'(t) = -6\left(\dfrac{1}{2\sqrt{t-8} (t-8)}\right)\\\\\\\implies f'(t) =-\dfrac 3{(t-8)^{\tfrac 32}}

(b)

\text{Given that,}\\\\y=f(t)\\\\\text{Slope of  y,} ~~ f'(t) =-\dfrac 3{(t-8)^{\tfrac 32}}\\\\\text{At point  (33, 6/5)}\\\\\\f'(t) = -\dfrac 3{(33-8)^{\tfrac 32}} = - \dfrac 3{ (25)^{ \tfrac 32}} = - \dfrac 3{5^3} = -\dfrac 3{125}\\\\\\\text{Equation with given points,}\\\\y - \dfrac 65 = -\dfrac 3{125} ( t - 33)\\\\\\\implies y =-\dfrac 3{125} t +\dfrac{99}{125} + \dfrac 65\\\\\\\implies y = -\dfrac 3{125} t +\dfrac{249}{125}

8 0
2 years ago
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