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Effectus [21]
3 years ago
7

PLEASE HELPP!! (10 points)

Mathematics
2 answers:
yulyashka [42]3 years ago
7 0

Answer:

The answer is the 4th option: 0.58

Step-by-step explanation:

Absolute value is how far away that number is from the numberline.

I hope this helped and if it did I would appreciate it if you marked me Brainliest. Thank you and have a nice day!

marishachu [46]3 years ago
5 0

Answer:

D.0.58

Step-by-step explanation:

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The dimensions of a conical funnel are shown below:
jasenka [17]
Volume of a conical funnel = π r² h/3
v = 3.14 * (2in)² * 6in/3
v = 3.14 * 4in² * 2in
v = 25.12 in³

25.12 in³ ÷ 10 in³ per minute = 2.512 minutes

It will take 2.51 minutes for all the liquid to pass through the nozzle.
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3 years ago
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5w&lt;15 what is w?<br><br> a. w&lt;75<br> b. w&lt;10<br> c. w&lt;5<br> d. &lt;3
timama [110]

Step-by-step explanation:

5w < 15 \\  \frac{5w}{5}  <  \frac{15}{5}  \\ w < 3

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4 years ago
3. The zeros of a quadratic function are 2
Sophie [7]

Answer:

A (3,0)

Step-by-step explanation:

The vertex is half way between the zeros

Add the zeros together and divide by 2

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natita [175]

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C

Step-by-step explanation:

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3 years ago
The length s(t)s(t)s, (, t, )of the side of the base of a square prism is decreasing at a rate of 777 kilometers per minute and
gavmur [86]

Answer:

\frac{dA_{s}}{dt} = -148\,\frac{km^{2}}{min}

Step-by-step explanation:

The surface area of the square prism is obtained by using the following formula:

A_{s} (t) = 4\cdot l(t)\cdot h(t) + 2\cdot [l(t)]^{2}

The rate of change of the surface area can be found by deriving the function with respect to time:

\frac{dA_{s}}{dt} = 4\cdot [h(t)\cdot \frac{dl}{dt} + l(t)\cdot \frac{dh}{dt}] + 2\cdot l(t)\cdot \frac{dl}{dt}

Known variables are summarized below:

h(t) = 9\,km

l(t) = 4\,km

\frac{dh}{dt} = 10\,\frac{km}{min}

\frac{dl}{dt} = -7\,\frac{km}{min}

The rate of change is:

\frac{dA_{s}}{dt} = 4\cdot [(9\,km)\cdot (-7\,\frac{km}{min} )+(4\,km)\cdot (10\,\frac{km}{min} )] + 2\cdot (4\,km)\cdot (-7\,\frac{km}{min} )

\frac{dA_{s}}{dt} = -148\,\frac{km^{2}}{min}

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