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Sunny_sXe [5.5K]
3 years ago
12

Round your answer to the nearest hundredth. ? А C 3 70° B

Mathematics
1 answer:
Anika [276]3 years ago
7 0
It would be 100 degrees
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Use the net to find the surface area of the regular pyramid
Alja [10]

Answer:

151.9 if the side length of the pentagon/ base of each triangle is 6 m

Step-by-step explanation:

if the side length of the pentagon/ base of each triangle is 6 m then  it is a pretty simple question, we just need to add the surface area of the base pentagon and each triangle.


We have the area of the base, so we just need the triangles.  The area of a triangle is .5bh, where the base of a triangle here is one side of the pentagon and the height is that indicated red 6. so that means one triangle has an area of .5*6*6 or 18.  There are 5 triangles total so that means that with all the triangles there is an area of 90.  Adding that to 61.9 gets us 151.9.  let me know if you need any more help.  

8 0
3 years ago
Help? Please. Thanks???
Naya [18.7K]

The height is represented by d.

d = 10 feet

10 = -  16 {t}^{2}  - 7t + 61 \\ 0 =  - 16 {t}^{2}  - 7t + 51

Use quadratic formula:

t = 1.58, -2.02 (reject)

1.58 seconds

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3 years ago
Use intercepts to graph 3y = -5x -30
disa [49]
The final solution is y=-5/3x-10
3 0
3 years ago
Finding the difference between 827 and 513
koban [17]
Answer: 314

Take each number place and subtract them from each other.
4 0
3 years ago
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Solve 2cos ²y -siny -1=0 for 0° ≤y≤360°
natima [27]

 

\displaystyle\bf\\2cos^2y -sin\,y -1=0~~~for~~0^o\leq y \leq360^o\\\\cos^2y=1-sin^2y\\\\2(1-sin^2y) -sin\,y -1=0\\\\2-2sin^2y-sin\,y-1=0\\\\-2sin^2y-sin\,y+2-1=0\\\\-2sin^2y-sin\,y+1=0~~~\Big|\times(-1)\\\\2sin^2y+sin\,y-1=0

.

\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}

 

 

5 0
3 years ago
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