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3 years ago

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17. In a speech class, students must give a final speech for their exam. The speech can be 15 minutes, or within that time by 4

minutes. Write an absolute value equation that represents the minimum and maximum lengths of the final speech

1 answer:

tekilochka [14]3 years ago

7 0

Answer:

| t − 15 | = 4

Step-by-step explanation:

absolute value is weird and i see no use for it

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A sphere and a cylinder have the same radius and height. The volume of the cylinder is 21m. what is the volume of the sphere.

Goshia [24]

Answer:

The volume of the sphere is 14m³

Step-by-step explanation:

Given

Volume of the cylinder = $21m^3$

Required

Volume of the sphere

Given that the volume of the cylinder is 21, the first step is to solve for the radius of the cylinder;

<em>Using the volume formula of a cylinder</em>

The formula goes thus

$V = \pi r^2h$

Substitute 21 for V; this gives

$21 = \pi r^2h$

Divide both sides by h

$\frac{21}{h} = \frac{\pi r^2h}{h}$

$\frac{21}{h} = \pi r^2$

The next step is to solve for the volume of the sphere using the following formula;

$V = \frac{4}{3}\pi r^3$

Divide both sides by r

$\frac{V}{r} = \frac{4}{3r}\pi r^3$

Expand Expression

$\frac{V}{r} = \frac{4}{3}\pi r^2$

Substitute $\frac{21}{h} = \pi r^2$

$\frac{V}{r} = \frac{4}{3} * \frac{21}{h}$

$\frac{V}{r} = \frac{84}{3h}$

$\frac{V}{r} = \frac{28}{h}$

Multiply both sided by r

$r * \frac{V}{r} = \frac{28}{h} * r$

$V = \frac{28r}{h}$ ------ equation 1

From the question, we were given that the height of the cylinder and the sphere have equal value;

This implies that the height of the cylinder equals the diameter of the sphere. In other words

$h = D$ , where D represents diameter of the sphere

Recall that $D = 2r$

So, $h = D = 2r$

$h = 2r$

Substitute 2r for h in equation 1

$V = \frac{28r}{2r}$

$V = \frac{28}{2}$

$V = 14$

Hence, the volume of the sphere is 14m³

4 0

3 years ago

Read 2 more answers

Write three numbers grater than 20 that have 9 as a factor

Burka [1]

27, 36, 45 are greater than 20 and multiples of 9.

5 0

3 years ago

between 1995 and 2006, the population of samburg, usa(in thousands) can be modeled by f(x) = 0.35x(squared) - 2.1x+15.8 where x=

Juli2301 [7.4K]

$\bf f(x)=0.35x^2-2.1x+15.8\qquad 
\begin{cases}
x=\textit{year since 1995}\\
f(x)=\textit{population amount}
\end{cases}$

the equation is a quadratic one, and it has a positive coefficient on the leading term, meaning, is opening upwards, so it has a "burrow" for the vertex.

the minimum or lowest point for a quadratic opening upwards is, well, the vertex point :), the "x" value is the year, the "y" or f(x) value is the population, we're asked for the year, or the x-coordinate of the vertex

well $\bf \begin{array}{llll}
f(x)=&0.35x^2&-2.1x&+15.8\\
&\quad \uparrow &\quad \uparrow&\uparrow \\
&\quad a&\quad b &c
\end{array}
\\\\

\\\\
\qquad \textit{vertex of a parabola}\\ \quad \\
\qquad 

\left(\boxed{-\cfrac{{{ b}}}{2{{ a}}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

5 0

3 years ago

) You are finally well!  But now your child is sick!  You know there have been mistakes in orders so you are on the internet t

rjkz [21]

Answer:

The doctor's dosage was not appropriate.

The right dosage is from 56.625 mg to 113.25 mg of Antibiotic to be given every 6 hour for the child with weight 25 lbs.

Step-by-step explanation:

It is given that 20 to 40 mg/kg/day is the recommended dosage.

Now, the doctor's order is 150 mg of antibiotic to be given every 6 hours. And my child weighs 25 lbs.

Now, 1 lb is equivalent to 0.453 kg.

So, my child's weight is (25 × 0.453) = 11.325 kg.

So, the doctor's order is 150 mg of antibiotic to be given every 6 hours for a child of weight 11.325 kg.

Hence, the dosage is [(150 × 4) ÷ 11.325] = 52.98 mg/kg/day.

So, this is not within the limit of 20 to 40 mg/kg/day.

Therefore, the doctor's dosage was not appropriate.

Now, let the appropriate dosage is x mg per every 6 hours for a child with weight 25 lbs i.e. 11.325 kg.

So, $20 \leq \frac{4x}{11.325} \leq 40$

⇒ $20\leq 0.3532x \leq 40$

⇒ $56.625 \leq x \leq 113.25$

So, the right dosage is from 56.625 mg to 113.25 mg of antibiotic to be given every 6 hours for the child with weight 25 lbs. (Answer)

5 0

2 years ago

Can someone help me with this please !!

pogonyaev

add all amd divide by the 40 ???let me know if im wrong

7 0

3 years ago

Read 2 more answers