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4 years ago

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An engineer is planning a new water pipe installation. The circular pipe has a diameter of d=20\text{ cm}d=20 cmd, equals, 20, s

pace, c, m. What is the area AAA of the circular cross section of this pipe?

1 answer:

aivan3 [116]4 years ago

4 0

Answer: The answer is 314.28 cm² (approx.).

Step-by-step explanation: Given that an engineer is going to install a new water pipe. The diameter of this circular pipe is, d = 20 cm.

We need to find the area 'A' of the circular cross-section of the pipe.

Given, diameter of the circular section is

$\textup{d}=20~\textup{cm}.$

So, the radius of the circular cross-section will be

$\textup{r}=\dfrac{\textup{d}}{2}=\dfrac{20}{2}=10~\textup{cm}.$

Therefore, cross-sectional area of the pipe is

$\textup{A}=\pi \textup{r}^2=\dfrac{22}{7}(10)^2=\dfrac{2200}{7}=314\dfrac{2}{7}=314.28~.~.~.~\textup{cm}^2.$

Thus, the answer is 314.28 cm² (approx.).

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Find sin X, sin Z, cos X, and cos Z. Write each answer as a simplified fraction.

just olya [345]

Answer:

Find the answers below

Step-by-step explanation:

Using m<X as the reference angle

Opposite YZ = 7

Adjacent XY = 10

Hypotenuse XZ = √149

Using the SOH CAH TOA identity

sinX = opp/hyp

sinX =YZ/XZ

sinX = 7/√149

For cos X

cos X = adj/hyp

cos X =10/√149

Using m<Z as reference angle;

Opposite XY = 10

Adjacent YZ = 7

Hypotenuse XZ = √149

Using the SOH CAH TOA identity

sinZ = opp/hyp

sinZ =10/√149

sinZ = 7/√149

For cos Z

cosZ = 7/√149

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3 years ago

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Galina-37 [17]

Answer:

Step-by-step explanation:

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3 years ago

Simplify the following

den301095 [7]

Answer:

1) 11 $\sqrt{3}$

2) 2 $\sqrt{2}$

3) $20\sqrt{3} + 15\sqrt{2}$

4) $53 + 12\sqrt{10}$

5) -2

6) $7\sqrt{2} - 5\sqrt{3}$

Step-by-step explanation:

1) 2 $\sqrt{12}$ + 3 $\sqrt{48}$ - $\sqrt{75}$

=(2 × 2 $\sqrt{3}$ )+ (3 × 4 $\sqrt{3}$ ) - 5 $\sqrt{3}$

= 4 $\sqrt{3}$ + 12 $\sqrt{3}$ - 5 $\sqrt{3}$

= 11 $\sqrt{3}$

2) 4 $\sqrt{8}$ -2 $\sqrt{98}$ + $\sqrt{128}$

= (4 × 2 $\sqrt{2}$ ) - (2 × 7 $\sqrt{2}$ ) + 8 $\sqrt{2}$

= 8 $\sqrt{2}$ - 14 $\sqrt{2}$ +8 $\sqrt{2}$

= 2 $\sqrt{2}$

3) 5 $\sqrt{12\\}$ - 3 $\sqrt{18} + 4 \sqrt{72} +2\sqrt{75}$

= 5× $2\sqrt{3}$ - 3× $3\sqrt{2}$ + 4× $6\sqrt{2}$ + 2× $5\sqrt{3}$

= $10\sqrt{3} - 9\sqrt{2} +24\sqrt{2} +10\sqrt{3}$

= $20\sqrt{3} + 15\sqrt{2}$

4) $(2\sqrt{2} + 3\sqrt{5} )^{2}$

= $8 + 12\sqrt{10} + 45$

= $53 + 12\sqrt{10}$

5) $(1+\sqrt{3} ) (1-\sqrt{3} )$

= $1 - 3$

= -2

6) $(2\sqrt{6} -1) (\sqrt{3} -\sqrt{2} )$

= $2\sqrt{18}-2\sqrt{12} -\sqrt{3} +\sqrt{2}$

= 2× $3\sqrt{2}$ - 2× $2\sqrt{3}$ - $\sqrt{3} + \sqrt{2}$

= $6\sqrt{2} - 4\sqrt{3} -\sqrt{3} +\sqrt{2}$

= $7\sqrt{2} - 5\sqrt{3}$

Hope the working out is clear and will help you. :)

5 0

3 years ago

Read 2 more answers

A manufacturer of matches randomly and independently puts 23 matches in each box of matches produced. The company knows that one

d1i1m1o1n [39]

Answer:

0.9855 or 98.55%.

Step-by-step explanation:

The probability of each individual match being flawed is p = 0.008. The probability that a matchbox will have one or fewer matches with a flaw is the same as the probability of a matchbox having exactly one or exactly zero matches with a flaw:

$P(X\leq 1)=P(X=0)+P(X=1)\\P(X\leq 1)=(1-p)^{23}+23*(1-p)^{23-1}*p\\P(X\leq 1)=(1-0.008)^{23}+23*(1-0.008)^{23-1}*0.008\\P(X\leq 1)=0.8313+0.1542\\P(X\leq 1)=0.9855$

The probability that a matchbox will have one or fewer matches with a flaw is 0.9855 or 98.55%.

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3 years ago

Find g^−1 (x) when g(x) = 3/5 x −9.<br><br>5/3x − 1/9<br>5/3 x + 9<br>−3/5 x+ 9<br>5/3 x + 15

Cerrena [4.2K]

Answer:

5/3 x + 15

Step-by-step explanation:

The inverse of a function is the reflection across the line y = x. As such, algebraically it is found by switching y and x in the equation and isolating to solve for y.

g(x) = 3/5 x - 9

x = 3/5 y - 9

x + 9 = 3/5 y

5(x+9) = 3y

(5x + 45)/3 = y

5/3x + 15 = y

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3 years ago