The measurement of ∠B is 168°
Given,
In the question:
Angle A and Angle B are supplementary angles.
If mA (x - 9)° and m/B= (7x + 21)°
To find the measure of ∠B
Now, According to the question;
What are supplementary angles?
Two angles are Supplementary when they add up to 180 degrees.
Now, ∠A+∠B = 180°
So,
⇒(x - 9)˚+(7x + 21)°=180°
=> 8x +12 = 180°
=> 8x = 180° - 12
=> 8x = 168
=> x = 168/8
=> x = 21
For the measure of ∠B :
∠B = (7x + 21)
∠B = 7 x 21 + 21
∠B = 168°
Hence, The measurement of ∠B is 168°
Learn more about Supplementary angles at:
brainly.com/question/13045673
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6% chance he is certified
Molly has a 54%chance
60% chance
If you do them out of 100%
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)
Answer:
1. 27.5 in.^2
2. 360 in.^2
Step-by-step explanation:
1.
A = bh/2
A = (4 + 7)(5)/2
A = 27.5 in.^2
2.
A = bh/2
A = (35 + 13)(15)/2
A = 360 in.^2
Answer:
0.4
Step-by-step explanation:
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