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3 years ago

Find the value of X to the nearest tenth (giving brainliest and thanks to all!)

Mathematics

2 answers:

Nina [5.8K]3 years ago

5 0

Answer:

Step-by-step explanation:

deff fn [24]3 years ago

3 0

The answer to this question is 12.96 but you round up to 13 so its c

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Given that√7=2.65 and √70=8.37,find the value of

dolphi86 [110]

Answer:

A) 0.265

B) 0.0265

C) 0.837

D) 0.0837

E) 0.00265

F) 0.00837

Step-by-step explanation:

We are given;

√7 = 2.65 and √70 = 8.37

A) √0.07 can be rewritten as;

√(7 × 1/100)

Let's deal with the digits in the bracket.

Square root of 100 is 10. Thus;

√(7 × 1/100) = (1/10)√7 = (1/10) × 2.65 = 0.265

B) √0.0007

Rewrite to get;

√(7 × 1/10000)

Square root of 1/10000 is 1/100

Thus;

√(7 × 1/10000) = (1/100)√7 = (1/100) × 2.65 = 0.0265

C) √0.7

Like above;

√0.7 = √(70 × (1/100))

>> (1/10)√70 = (1/10) × 8.37 = 0.837

D) √0.007

Like above;

Rewrite to get;

√(70 × 1/10000)

Square root of 1/10000 is 1/100

Thus;

√(70 × 1/10000) = (1/100)√70 = (1/100) × 8.37 = 0.0837

E) √0.000007

Rewritten to;

√(7 × (1/1000000))

√(1/1000000) = 1/1000

Thus; √(7 × (1/1000000)) = 1/1000 × √7 = 1/1000 × 2.65 = 0.00265

F)√0.00007

Rewritten to;

√(70 × (1/1000000))

√(1/1000000) = 1/1000

Thus; √(70 × (1/1000000)) = 1/1000 × √70 = 1/1000 × 8.37 = 0.00837

8 0

2 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

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