The high temperature today is 80 degrees. If the high temperature tomorrow is supposed to

Mathematics

1 answer:

wlad13 [49]2 years ago

4 0

Answer:

Step-by-step explanation:

80 x 0.05 = 4

80 - 4 = 76

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Whole-wheat bread costs $2.06/lb. How much would 3.8 lb of whole-wheat bread cost?

Xelga [282]

Whole-wheat bread = $2.06/lb

We need the cost of 3.8 lbs of whole-wheat bread.

So, we simply need to multiply the cost with the amount needed.

(cost per pound) × (amount of bread)

($2.06) × (3.8)

Simplify.

$7.828

Round to the nearest tenths.

$7.83

So, 3.8 pounds of whole wheat bread costs $7.83

~Hope I helped!~

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3 years ago

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Is there a proportional relationship between x and y? Explain.

Vlad1618 [11]

I don’t think so but divide the top one by the bottom one for every one of them and if it is all the same answer they are proportional x/y

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2 years ago

Solve the following problems Answer: Explanatoin:

NemiM [27]

Answer:
1) 121.7
2) 131.8
3) 106.2
4) 88.5
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6) 110.2

Explanation:
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3 years ago

Use series to verify that <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

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2 years ago

Best estimate for 7/8-6/11

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Ummm we are allll gonna be dead cause Biden be da president

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