Answer: 200.96
Step-by-step explanation:
3.14 times 8^2 = 200.96
Answer:
8 ≤ n/(-4)
Step-by-step explanation:
Let n = the number.
Then n/(-4) = the quotient of the number and negative 4
The inequality is
8 ≤ n/(-4)
Answer:
After 7 hours Tevin will catch Enrique
Step-by-step explanation:
We have given speed of Enrique is 20 mph
Tevin left traveling after 2 hours of Enrique
Speed of Tevin is 28 mph
When Tevin catches Enrique will cover same distance
Let Enrique is traveling for t hours
Then Tevin will travel for t-2 hours
We know that distance = speed × time
So ![20\times t=28\times (t-2)](https://tex.z-dn.net/?f=20%5Ctimes%20t%3D28%5Ctimes%20%28t-2%29)
![20t=28t-56](https://tex.z-dn.net/?f=20t%3D28t-56)
![8t=56](https://tex.z-dn.net/?f=8t%3D56)
t = 7 hours
So after 7 hours Tevin will catch Enrique
Answer: Choice A.
Three fourths of the fish he caught were larger than 22 pounds.
==============================================================
Explanation:
The 1st quartile (Q1) is visually located at the left edge of the box. From the diagram, it's not clear where that left edge is exactly. However, 22 seems to be the closest of the answer choices because the left edge appears to be between 20 and 30.
1/4 of the population is below Q1 while the remaining 3/4 is above this marker.
If Q1 = 22, then statement A saying "Three fourths of the fish he caught were larger than 22 pounds" is correct.
Step-by-step explanation:
You must write formulas regarding the volume and surface area of the given solids.
![\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=\dfrac{4}{3}\pi r^3\\SA=4\pi r^2](https://tex.z-dn.net/?f=%5Cbold%7B%5C%231%5C%20Rectangular%5C%20prism%3A%7D%5C%5C%5C%5CV%3Dlwh%5C%5CSA%3D2lw%2B2lh%2B2wh%3D2%28lw%2Blh%2Bwh%29%5C%5C%5C%5C%5Cbold%7B%5C%232%5C%20Cylinder%3A%7D%5C%5C%5C%5CV%3D%5Cpi%20r%5E2h%5C%5CSA%3D2%5Cpi%20r%5E2%2B2%5Cpi%20rh%3D2%5Cpir%28r%2Bh%29%5C%5C%5C%5C%5Cbold%7B%5C%233%5C%20Sphere%3A%7D%5C%5C%5C%5CV%3D%5Cdfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3%5C%5CSA%3D4%5Cpi%20r%5E2)
![\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\](https://tex.z-dn.net/?f=%5Cbold%7B%5C%234%5C%20Cone%3A%7D%5C%5C%5C%5CV%3D%5Cdfrac%7B1%7D%7B3%7D%5Cpi%20r%5E2h%5C%5C%5C%5C%5Ctext%7Bwe%20need%20calculate%20the%20length%20of%20a%20slant%20length%7D%5C%20l%5C%5C%5Ctext%7Buse%20the%20Pythagorean%20theorem%3A%7D%5C%5C%5C%5Cl%5E2%3Dr%5E2%2Bh%5E2%5Cto%20l%3D%5Csqrt%7Br%5E2%2Bh%5E2%7D%5C%5C%5C%5CSA%3D%5Cpi%20r%5E2%2B%5Cpi%20rl%3D%5Cpi%20r%5E2%2B%5Cpi%20r%5Csqrt%7Br%5E2%2Bh%5E2%7D%3D%5Cpi%20r%28r%2B%5Csqrt%7Br%5E2%2Bh%5E2%7D%29%5C%5C%5C%5C%5Cbold%7B%5C%235%5C%20Rectangular%5C%20Pyramid%3A%7D%5C%5C%5C%5CV%3D%5Cdfrac%7B1%7D%7B3%7Dlwh%5C%5C%5C%5C)
![\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}](https://tex.z-dn.net/?f=%5C%5C%5Ctext%7Bwe%20need%20to%20calculate%20the%20height%20of%20two%20different%20side%20walls%7D%5C%20h_1%5C%20%5Ctext%7Band%7D%5C%20h_2%5C%5C%5Ctext%7Buse%20the%20Pythagorean%20theorem%3A%7D%5C%5C%5C%5Ch_1%5E2%3D%5Cleft%28%5Cdfrac%7Bl%7D%7B2%7D%5Cright%29%5E2%2Bh%5E2%5Cto%20h_1%3D%5Csqrt%7B%5Cleft%28%5Cdfrac%7Bl%7D%7B2%7D%5Cright%29%5E2%2Bh%5E2%7D%3D%5Csqrt%7B%5Cdfrac%7Bl%5E2%7D%7B4%7D%2Bh%5E2%7D%3D%5Csqrt%7B%5Cdfrac%7Bl%5E2%7D%7B4%7D%2B%5Cdfrac%7B4h%5E2%7D%7B4%7D%7D%5C%5C%5C%5Ch_1%3D%5Csqrt%7B%5Cdfrac%7Bl%5E2%2B4h%5E2%7D%7B4%7D%7D%3D%5Cdfrac%7B%5Csqrt%7Bl%5E2%2B4h%5E2%7D%7D%7B%5Csqrt4%7D%3D%5Cdfrac%7B%5Csqrt%7Bl%5E2%2B4h%5E2%7D%7D%7B2%7D)
![\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}](https://tex.z-dn.net/?f=%5C%5C%5C%5Ch_2%5E2%3D%5Cleft%28%5Cdfrac%7Bw%7D%7B2%7D%5Cright%29%5E2%2Bh%5E2%5Cto%20h_2%3D%5Csqrt%7B%5Cleft%28%5Cdfrac%7Bw%7D%7B2%7D%5Cright%29%5E2%2Bh%5E2%7D%3D%5Csqrt%7B%5Cdfrac%7Bw%5E2%7D%7B4%7D%2Bh%5E2%7D%3D%5Csqrt%7B%5Cdfrac%7Bw%5E2%7D%7B4%7D%2B%5Cdfrac%7B4h%5E2%7D%7B4%7D%7D%5C%5C%5C%5Ch_2%3D%5Csqrt%7B%5Cdfrac%7Bw%5E2%2B4h%5E2%7D%7B4%7D%7D%3D%5Cdfrac%7B%5Csqrt%7Bw%5E2%2B4h%5E2%7D%7D%7B%5Csqrt4%7D%3D%5Cdfrac%7B%5Csqrt%7Bw%5E2%2B4h%5E2%7D%7D%7B2%7D)
![SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}](https://tex.z-dn.net/?f=SA%3Dlw%2B2%5Ccdot%5Cdfrac%7Blh_1%7D%7B2%7D%2B2%5Ccdot%5Cdfrac%7Bwh_2%7D%7B2%7D%5C%5C%5C%5CSA%3Dlw%2B2%5C%21%5C%21%5C%21%5C%21%5Cdiagup%5Ccdot%5Cdfrac%7Bl%5Ccdot%5Cfrac%7B%5Csqrt%7Bl%5E2%2B4h%5E2%7D%7D%7B2%7D%7D%7B2%5C%21%5C%21%5C%21%5C%21%5Cdiagup%7D%2B2%5C%21%5C%21%5C%21%5C%21%5Cdiagup%5Ccdot%5Cdfrac%7Bw%5Ccdot%5Cfrac%7B%5Csqrt%7Bw%5E2%2B4h%5E2%7D%7D%7B2%7D%7D%7B2%5C%21%5C%21%5C%21%5C%21%5Cdiagup%7D%5C%5C%5C%5CSA%3Dlw%2B%5Cdfrac%7Bl%5Csqrt%7Bl%5E2%2B4h%5E2%7D%7D%7B2%7D%2B%5Cdfrac%7Bw%5Csqrt%7Bw%5E2%2B4h%5E2%7D%7D%7B2%7D%5C%5C%5C%5CSA%3D%5Cdfrac%7B2lw%7D%7B2%7D%2B%5Cdfrac%7Bl%5Csqrt%7Bl%5E2%2B4h%5E2%7D%7D%7B2%7D%2B%5Cdfrac%7Bw%5Csqrt%7Bw%5E2%2B4h%5E2%7D%7D%7B2%7D%5C%5C%5C%5CSA%3D%5Cdfrac%7B2lw%2Bl%5Csqrt%7Bl%5E2%2B4h%5E2%7D%2Bw%5Csqrt%7Bw%5E2%2B4h%5E2%7D%7D%7B2%7D)