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2 years ago

Hurry please im running out of timeill give brainliest

Mathematics

1 answer:

Ipatiy [6.2K]2 years ago

3 0

Answer:

Step-by-step explanation:

16 ticks, the pointer is on 12. there fore it is the second answer.

2 3/4 lb

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Yesterday, Julie’s bank account balance was –$10.28. She deposited some money into her account so that her balance is now $39.89

Natalija [7]

Since we're starting with a negative we're going to disregard the negative sign and add the numbers together: $10.28 + $39.89 = $50.17. :)

4 0

3 years ago

Read 2 more answers

5 increased by half a number is 11

just olya [345]

Answer:

Step-by-step explanation:

let the unknown number is x.

As said in the question, we add half of x to 5 and their sum equals to 11.

steps:

write the equation.
do the L.C.M.
transfer 2 to the other side of the equal sign
keep x as the subject and do the subtraction on the other side of the equal sign

$5 + \frac{1}{2}x = 11\\\frac{10+x}{2} = 11\\10 + x = 22\\x = 22-10\\x = 12$

4 0

3 years ago

3A: Determine the value of angle 1 and provide a reason why. *

Otrada [13]

The answer is 45°

By determining the angles of triangle ACB (A=30°, B=45°, C=105° (total of 180°)) and knowing that angle 1 correlates to angle 4 ((1)°=(4)°), we can find angle 1, which is 45°, equal to angle 4.

8 0

3 years ago

Factor the expression 30a+35

kherson [118]

<h3>Answer is 5(6a+7)</h3>

=====================================

The GCF is 5, which can be factored out of the expression

Note how

30a = 5*6a

35 = 5*7

So we see the common factor between the two terms. You use the distribution property to pull out that 5. If you were do distribute the 5 back in, you'd get

5(6a+7) = 5*6a+5*7 = 30a+35

5 0

3 years ago

Read 2 more answers

Solve this differential equation using power series and indicial roots about (0,0):

Ivanshal [37]

Let $y=\displaystyle\sum_{k\ge0}a_kx^k$ , so that

$y'=\displaystyle\sum_{k\ge1}ka_kx^{k-1}$
$y''=\displaystyle\sum_{k\ge2}k(k-1)a_kx^{k-2}$

Substituting into the ODE gives

$\displaystyle3x\sum_{k\ge2}k(k-1)a_kx^{k-2}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0$
$\displaystyle3\sum_{k\ge2}k(k-1)a_kx^{k-1}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0$

The first series starts with a linear term, while the other two start with a constant. Extract the first term from each of the latter two series:

$\displaystyle6\sum_{k\ge1}ka_kx^{k-1}=6\sum_{k\ge2}ka_kx^{k-1}+6a_1$
$\displaystyle\sum_{k\ge0}a_kx^k=\sum_{k\ge1}a_kx^k+a_0$

Finally, to get the series to start at the same index, shift the index of the first two series by replacing $k$ with $k+1$ . Then the ODE becomes

$\displaystyle3\sum_{k\ge1}k(k+1)a_{k+1}x^k+6\sum_{k\ge1}(k+1)a_{k+1}x^k+\sum_{k\ge1}a_kx^k+6a_1+a_0=0$

which can be consolidated to get

$\displaystyle\sum_{k\ge1}\bigg[(3k(k+1)+6(k+1))a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0$
$\displaystyle\sum_{k\ge1}\bigg[3(k+1)(k+2)a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0$

You're fixing the solution so that it contains the origin, which means

$y(0)=\displaystyle\sum_{k\ge0}a_kx^k=a_0=0$

which in turn means $a_1=0$ . With the given recurrence, it follows that $a_k=0$ for all $k\ge2$ , so the solution would be $y=0$ . This is to be expected, since $x=0$ is clearly a singular point for the ODE.

8 0

3 years ago