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1 year ago

30 divided (3x5)divided 2 + 1?

Mathematics

2 answers:

solong [7]1 year ago

5 0

Answer:

Step-by-step explanation:

30÷(3×5)÷2+1

30÷15÷2+1

2÷2+1

1+1

Hope dis helps

guapka [62]1 year ago

3 0

Answer

Explain

3 x 5 = 15 so 30 divided by 15 equals 2. Now 2 divided by 2 equals 1. Now 1 + 1 = 2

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A polynomial divided by a polynomial is a polynomial. always, sometimes, or never true?

Artemon [7]

I think it is always true. I am not sure but I hope I'm right

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2 years ago

The Medicare tax rate is 1.45% how much would be withheld form someone paycheck for Medicare with wages of $1,738 for that pay p

Tcecarenko [31]

Answer: $\$25.201$

Step-by-step explanation:

Given

The Medicare tax rate is 1.45%

The wages of a person is $1738

The amount which is withheld from the person paycheck is

$\Rightarrow 1738\times 1.45\%\\\Rightarrow 1738\times 0.0145\\\Rightarrow \$25.201$

7 0

2 years ago

Enter the explicit rule for the sequence

Softa [21]

Let's take a look at the first few numbers in the sequence based on the given rule:

$a_{1}=6\\ a_{2}= \frac{1}{4}(6)\\ a_3= \frac{1}{4}\big( \frac{1}{4}\big)6= \big(\frac{1}{4}\big)^2(6)\\ a_{4}= \frac{1}{4} \big(\frac{1}{4}\big)^26= \big(\frac{1}{4}\big)^3(6)$

Inspecting this pattern it seems like the power $\frac{1}{4}$ is being raised to is always one less than the number of the sequence, so if we were on the nth number in the sequence, that part of the expression would be $\big(\frac{1}{4} \big)^{n-1}$ . We also know that we'll be multiplying whatever we get from that by 6, so we can write the full explicit rule for our sequence as

$a_n=6\big( \frac{1}{4} \big)^{n-1}$

Where $a_n$ is the nth number in our sequence.

8 0

2 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

2 years ago

PLEASE HELP!!!<br><br> Which line is the best model for the data in the scatter plot?

fomenos

Answer:

top left

Step-by-step explanation:

it goes through the most dots

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2 years ago