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2 years ago

Dev has 9 shells. Zoe has 55 shells. Zoe gives some shells to Dev. Now Zoe has 3 times as many as Dev.

2 answers:

8 0

Answer:let the initial number of shells that dev has be A, and initial number of shells that zoe has be B. let the number of shells that zoe gives to dev be x. after giiving x shells zoe is left with 3 times the number of shell as that of dev. therefore number of shells with zoe = 3×number of shells with dev.

Step-by-step explanation:

PtichkaEL [24]2 years ago

4 0

Might not be correct but I try to explain in my best ways(: question not from safari! let the initial number of shells that dev has be A, and initial number of shells that zoe has be B. let the number of shells that zoe gives to dev be x. after giiving x shells zoe is left with 3 times the number of shell as that of dev. therefore number of shells with zoe = 3×number of shells with dev.

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. A(0,8), B(6,5), C(-3,2)

Keith_Richards [23]

Answer:

Step-by-step explanation:

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3 years ago

Please, I need help in this ??

nignag [31]

Answer:

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

Step-by-step explanation:

$\int\frac{x^{4}}{x^{4} -1}dx$

Adding and Subtracting 1 to the Numerator

$\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx$

Dividing Numerator seperately by $x^{4} - 1$

$\int 1 + \frac{1}{x^{4}-1 }\, dx$

Here integral of 1 is x +c1 (where c1 is constant of integration

$x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx$ ----------------------------------(1)

We apply method of partial fractions to perform the integral

$\frac{1}{(x-1)(x+1)(x^{2}+1)}$ = $\frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}$ ------------------------------------------(2)

$\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}$

1 = $A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)$ -------------------------(3)

Substitute x= 1 , -1 , i in equation (3)

1 = A(1+1)(1+1)

A = $\frac{1}{4}$

1 = B(-1-1)(1+1)

B = $-\frac{1}{4}$

1 = C(i-1)(i+1)

C = $-\frac{1}{2}$

Substituting A, B, C in equation (2)

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }$

On integration

Here $\int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx$

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\frac{1}{4} ln(x-1)$ - $\frac{1}{4} ln(x+1)$ - $\frac{1}{2} arctanx$ + c2---------------------------------------(4)

Substitute equation (4) back in equation (1) we get

$x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2$

Here c1 + c2 can be added to another and written as c

Therefore,

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

4 0

3 years ago

Bytecoin is a new cryptocurrency that is currently valued at $243 per coin. You want to make an investment and purchase 1 Byteco

Digiron [165]

Answer:

Step-by-step explanation:

The growth rate of the coin is exponential. We would apply the formula for exponential growth which is expressed as

y = ab^x

y = b(1 + r)^ t

Where

y represents the value of the coin after x months.

x represents the number of months.

a represents the initial value of the coin.

b represents rate of growth.

From the information given,

a = 243

b = 1 + 15/100 = 1 + 0.15 = 1.15

Therefore, the exponential expression to determine the number of months, x it will take for the coin to attain a certain value, y is expressed as

y = 243(1.15)^x

If y = 1000, it means that

1000 = 243(1.15)^x

1000/243 = 1.15^x

4.115 = 1.15^x

Taking log of both sides, it becomes

log 4.115 = xlog1.15

0.614 = 0.061x

x = 0.614/0.061

x = 10 months

It will take 10 months

7 0

3 years ago

Please help me <br>by the way this is a 2 mark question from mathswatch

BigorU [14]

Answer:

y = - 13, y = 13

Step-by-step explanation:

Given

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y = ± $\sqrt{169}$ = ± 13

Since 13 × 13 = 169 and - 13 × - 13 = 169

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2 years ago

Ok Here’s My. Homework Guccigamegurl

Delvig [45]

Answer:

its in order OK

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3 years ago