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2 years ago

Find the surface area of a square pyramid with side length 4 mi and slant height 5 mi.

Mathematics

1 answer:

andriy [413]2 years ago

3 0

Answer:

SA = 56 sq mi

Step-by-step explanation:

SA = area of base + 1/2(perimeter of base)(slant height)

SA = 16 + 1/2(16)(5)

SA = 16 + 8(5)

SA = 16 + 40

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Gabe rolled 14 strikes out of 70 attempts. What percent of Gabe's attempts were strikes?

Airida [17]

Answer:

20%

Step-by-step explanation:

14/70x100=20%

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3 years ago

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Answer:

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2 years ago

What are the factors of 2x2 + 3x – 54? Select two options. 2x – 9 2x – 6 2x + 6 x – 6 x + 6

andrew-mc [135]

Answer:

2x − 9 and x + 6

Step-by-step explanation:

2x² + 3x − 54

Factor using the AC method.

2 × -54 = -108

Factors of -108 that add up to 3 are +12 and -9.

Divide by 2 and reduce: 12/2 = 6/1, -9/2 = -9/2.

Therefore, the factors are x + 6 and 2x − 9.

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3 years ago

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

5 0

3 years ago

QUICK!

pochemuha

So,

In order to find numbers that are in between 3/8 and 6/7, Lucy must make common denominators first.

LCM of 8 and 7: 56

Multiply each fraction by the appropriate form of 1:
$\frac{3}{8} * \frac{7}{7}\ AND \ \frac{6}{7} * \frac{8}{8}$

$\frac{21}{56}\ AND\ \frac{48}{56}$

We can immediately see that any fraction with a numerator between 21 and 48 and a denominator of 56 will fit into the criteria Lucy needs.

The correct option is B.

4 0

3 years ago

Read 2 more answers