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Natalija [7]
2 years ago
10

Work out the surface area of this solid prism.

Mathematics
1 answer:
zhannawk [14.2K]2 years ago
3 0
THE SURFACE AREA IS 3220CM^2
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Determine the XNY intercepts of the graph of X+2y=-4
podryga [215]

Answer:

x intercepts (-4,0)

y intercepts (0, -2)

4 0
3 years ago
3(m - 1) = 5m +3 -2m please help me solve this step by step asap!
cupoosta [38]

Answer:

0 = 6

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

3(m−1)=5m+3−2m

(3)(m)+(3)(−1)=5m+3+−2m (Distribute)

3m+−3=5m+3+−2m

3m−3=(5m+−2m)+(3) (Combine Like Terms)

3m−3=3m+3

3m−3=3m+3

Step 2: Subtract 3m from both sides.

3m−3−3m=3m+3−3m

−3=3

Step 3: Add 3 to both sides.

−3+3=3+3

0=6

3 0
3 years ago
Which of the following are vertical asymptotes of the function y = 2cot(3x) + 4? Check all that apply. A.x = pi/3 B.x = +/- pi/2
Kisachek [45]
Vertical asymptotes occur when the denominator of a rational is 0, whilst not zeroing out the numerator, making the rational, undefined, in this case

\bf y=2cot(3x)+4\implies y=2\cdot \cfrac{cos(3x)}{sin(3x)}+4\impliedby \textit{if \underline{sin(3x)} turns to 0}\\\\
-------------------------------\\\\
\textit{let's check}
\\\\\\
A)\qquad \cfrac{cos(3x)}{sin\left( 3\frac{\pi }{3} \right)}\implies \cfrac{cos(3x)}{sin\left( \pi \right)}\implies \cfrac{cos(3x)}{0}\impliedby unde f ined

\bf B)\qquad \cfrac{cos(3x)}{sin\left( 3\frac{\pm\pi }{2} \right)}\implies\cfrac{cos(3x)}{sin\left( \frac{\pm3\pi }{2} \right)}\implies \cfrac{cos(3x)}{\pm 1}\implies \pm cos(3x)
\\\\\\
C)\qquad \cfrac{cos(3x)}{sin\left( 3(2\pi )\right)}\implies \cfrac{cos(3x)}{sin(6\pi )}\implies \cfrac{cos(3x)}{0}\impliedby unde f ined
\\\\\\
D)\qquad \cfrac{cos(3x)}{sin(3(0))}\implies \cfrac{cos(3x)}{sin(0)}\implies \cfrac{cos(3x)}{0}\impliedby unde f ined
6 0
3 years ago
Read 2 more answers
Please help help help help​
amid [387]

Answer:

option b is the correct i mean 10

7 0
3 years ago
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1/3 divided by 1/8<br> a, 1/24<br> b. 3/8<br> C. 2 and 2/3<br> d. 24
Degger [83]
The answer is B. 3/8
5 0
2 years ago
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