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3 years ago

PLEASE help. The problem is to find the indicated sum.

Mathematics

1 answer:

alukav5142 [94]3 years ago

8 0

The indicated sum is 350. The answer is 350 because the E or sigma is making you do (6j+2) ten times. Every time you do 6j+2 you start with j as 1 and add a 1 to j for every time you add. For example, [6(1)+2]+[6(2)+2]+[6(3)+2]...[6(10)+2].

Hope this helps.

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Let m = 7.

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If h(x) = 2x - 1 and g(x) = 3x + 1, what is (h o g)(2)?

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Step-by-step explanation:

g(2) = 3(2) + 1 = 6 + 1 = 7

h(7) = 2(7) - 1 = 14 - 1 = 13

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BELL RINGER #2

NemiM [27]

Answer:

A. 2 4/5

Step-by-step explanation:

To find how many hours she worked for $210, you must get the amount of money she gets in 1 hour.

Because she charges $43 dollars every hour, and fines a fee of $30 flat, we must add both of the amount to get how many she earns in 1 hour.

So:

$45 + $30= $75

She earn $75 in 1 hour.

Next, divide $210 dollars that she earned for working for hour(s) to the amount of money she earned in 1 hour to find how many hours she worked.

So:

$210 ÷ $75= 2.8 hours

The answer is 2.8 hours

Because the given answers is in fraction, we must change the decimal into a fraction.

To change a decimal into a fraction, you must place the decimal over its place value.

Because 8 in the decimal 2.8 is in the tenths place, you must place it over 10

So:

2.8 into a decimal is 2 8/10

Simplify (only simplify if possible):

Divide 8 and 10 to their GCF which is 2.

So:

8 ÷ 2= 4

10 ÷ 2= 5

So the fraction and the answer is now:

2 4/5

I hope this helps! I'm sorry if it's wrong and too complicated.

4 0

3 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago

PLEASE help. The problem is to find the indicated sum.​

PLEASE help. The problem is to find the indicated sum.