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3 years ago

1. Calvin goes shopping over the weekend and buys a new pair of shoes, a t-shirt, a pair of

Mathematics

1 answer:

Alja [10]3 years ago

7 0

A. 2s+s+1/2s+9=65
B. 3.5s+9=65
C. Subtract 9 from both sides to get: 3.5s=56 and then divide both sides by 3.5 to get s (the cost of the shirt) which is s=16
D. Shoes cost $32, shirt costs $16, and the sunglasses cost $8

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Explain (in words) what you would do to find the missing value in the equation. You do not find the missing number! Equation: ?

maks197457 [2]

Answer:

The missing value = $23\frac{1}{24}$

Step-by-step explanation:

Given - Equation: ? + 59 5/8= 82 2/3

To find - Explain (in words) what you would do to find the missing value in

the equation. You do not find the missing number.

Proof -

Let the missing number = x

So, the equation becomes x + $59\frac{5}{8}$ = $82\frac{2}{3}$

⇒x = $82\frac{2}{3}$ - $59\frac{5}{8}$

= $\frac{248}{3} - \frac{477}{8}$

= $\frac{1984 - 1431}{24}$

= $\frac{553}{24}$

= $23\frac{1}{24}$

⇒x = $23\frac{1}{24}$

∴ we get

The missing value = $23\frac{1}{24}$

Explanation -

Here we have given the ? with addition of number that is equation to some number.

We just have to subtract the two numbers to find the missing value , as we do above.

5 0

2 years ago

Function or no a function

marishachu [46]

Answer:

Not a function

Step-by-step explanation:

If it was a function there should be only one values for every x, in this case there is two values for x=0. Doesn't pass the vertical test.

7 0

2 years ago

Read 2 more answers

For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

I'm assuming you're talking about the indefinite integral

$\displaystyle\int e^{-x^2/2}\,\mathrm dx$

and that your question is whether the substitution $u=\dfrac x{\sqrt2}$ would work. Well, let's check it out:

$u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}$
$\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du$
$=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du$

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried $u=\sqrt t$ next? Then $\mathrm du=\dfrac{\mathrm dt}{2\sqrt t}$ , giving

$=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt$

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

$\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt$

By the fundamental theorem of calculus, taking the derivative of both sides yields

$\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}$

and so the antiderivative would be

$\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)$

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.

3 0

3 years ago

If the median goes to a roller coaster. Would it be faster if it is higher or lower?? Asap plz need help and in English I just n

Kobotan [32]

The answer for this question is higher

5 0

3 years ago

Please help me 1 2 3 or 4

Firlakuza [10]

Answer:

Step-by-step explanation:

branliest pls

8 0

3 years ago