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3 years ago

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Which of the following is not true for adding mixed numbers?

1 answer:

chubhunter [2.5K]3 years ago

5 0

1.which of the following is not true for adding mixed numbers?

A.the denominators of the fractions must be the same in order to add them.

B.you can add mixed numbers by changing them into improper fractions or by using a number line.

C.the sum of the mixed numbers written as an improper fraction is in simplest form.

D.the sum of the mixed numbers must be written in simplest form

The answer is C

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I need help so much please help the attachment below is the question i need help on

leonid [27]

Answer:

$\dfrac{\sqrt[12]{55296}}{2}$

Step-by-step explanation:

Rationalize the denominator, then use a common root for the numerator.

$\dfrac{\sqrt[4]{6}}{\sqrt[3]{2}}=\dfrac{(2\cdot 3)^{\frac{1}{4}}}{2^{\frac{1}{3}}}\\\\=\dfrac{(2\cdot 3)^{\frac{1}{4}}}{2^{\frac{1}{3}}}\cdot\dfrac{2^{\frac{2}{3}}}{2^{\frac{2}{3}}}=\dfrac{2^{\frac{1}{4}+\frac{2}{3}}3^{\frac{1}{4}}}{2}\\\\=\dfrac{2^{\frac{11}{12}}3^{\frac{3}{12}}}{2}=\dfrac{\sqrt[12]{2^{11}3^{3}}}{2}\\\\=\dfrac{\sqrt[12]{55296}}{2}$

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3 years ago

A sandbox has a perimeter of 21 1/2 feet. If the length is 5 1/4 feet, what is the area of the sand box?

LiRa [457]

The perimeter of a rectangle is twice the sum of length and width, so the sum of length and width is half the perimeter, 10 3/4 ft.

The width will be the difference between this and the length, hence 5 1/2 ft.

The area is the product of length and width.
A = length×width
A = (5.25 ft)×(5.5 ft) = 28.875 ft²

The area of the sandbox is 28 7/8 ft².

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3 years ago

can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc

mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

$\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx$
$\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx$
$\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)$

and so on.

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Luden [163]

Answer:

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Step-by-step explanation:

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Anton [14]

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3 years ago