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2 years ago

Which of these is the same time as 9:47

Mathematics

2 answers:

Afina-wow [57]2 years ago

8 0

No image or info added?

Anna35 [415]2 years ago

7 0

Answer:

No image, no info?

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Find the volume of the solid bounded by z = 2 - x2 - y2 and z = 1. Express your answer as a decimal rounded to the hundredths pl

butalik [34]

Answer:

The answer is " $(\frac{\pi}{2})$ ".

Step-by-step explanation:

$z = 2 - x^2 - y^2.........(1) \\\\z = 1.............(2)$

Let add equation 1 and 2:

Using formula: $x^2+y^2=1 \\\\$

convert to polar coordinates

$r=2$

$\theta \varepsilon (0=\pi)=z\\\\V=\int^{2\pi}_{\theta=0}\int^{1}_{\pi=0}\int^{z_{2}}_{z_1} r \ dr \d \theta\\\\$

$=\int^{2\pi}_{0}\int^{1}_{0} (Z_2-z_1)r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} 1- (-1-x^2-y^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (+1 \pm r^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (-r^3 + r) \ dr \d \theta\\\\=\int^{2\pi}_{0} (-\frac{r^4}{4}+\frac{r^2}{1})^{1}_{0} \d \theta\\\\=\int^{2\pi}_{0} (\frac{1}{4}) \d \theta\\\\=(\frac{2 \pi}{4}) \\\\=(\frac{\pi}{2}) \\\\$

8 0

2 years ago

The perimeter of a kite is 51 meters. If one side measures 18 meters find the measure of the opposite side.

puteri [66]

Let n and m be the sides of the kite
perimeter = 2(m+n)
51 = 2(18+m)
51/2 = 18 + m
25.5 - 18 = m
m = 7.5m

4 0

2 years ago

Which angle is supplementary

Romashka-Z-Leto [24]

Supplementary angle is 180 degrees

6 0

2 years ago

Solve the equation. Check for extraneous solutions. 6|6-4x|=8x+4

MA_775_DIABLO [31]

Use PEMDAS
Parentheses
Exponents
Multiply
Divide
Add
Subtract

8 0

3 years ago

The mean annual income for adult women in one city is $28,520 and the standard deviation of the incomes is $5600. The distributi

pshichka [43]

Answer:

$E(\bar{x}) = 28520$

Standard error of mean = 689

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $28,520

Standard Deviation, σ = $5600

Mean of sampling distribution =

$E(\bar{x}) = \mu = 28520$

As per Central Limit Theorem, if the sample size is large enough, then the sampling distribution of the sample means follow approximately a normal distribution.

Sample size, n = 66

Since the sample size is large, we can use normal distribution for approximation.

Standard error of mean =

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{5600}{\sqrt{66}} = 689.31 \approx 689$

7 0

2 years ago