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2 years ago

Will mark branliest! show work.

Mathematics

1 answer:

Sindrei [870]2 years ago

3 0

Answer:

3/-2........-1.5 as decimal

Step-by-step explanation:

slope formula=y2-y1/x2-x1

6+3/-1-5

9/-6

3/-2..........or-1.5 as deciamal

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A stack of magazines is 4 2/5 inches high. Each magazine is 2/5 inch thick.

oee [108]

Answer:

Equation:(4 2/5)/2/5=? ; estimate: a

little more than 10 magazines; answer:

11 magazines in the stack.

Step-by-step explanation:

7 0

1 year ago

The perimeter of a triangle is 53 feet. One side of the triangle is 12 feet shorter than the second side. The third side is is 4

grandymaker [24]

Let the shortest side be = a, then side b = 2·a and side c = (b + 24)

We are given that a + b + c = 84

Substituting for b and c

a + 2·a + (a + 24) = 84

4·a + 24 = 84

4·a = 84 - 24 = 60

a = 60/4 = 15 feet

b = 2·15 = 30 feet

c = 15 + 24 = 39 feet

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2 years ago

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Vitek1552 [10]

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2 years ago

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What is 1/3 of 21, 2/5 of 40 4/7 of 280

Lubov Fominskaja [6]

$\frac{1}{3} \times 21=\frac{21}{3}=\frac{7 \times 3}{3}=7 \\ \\
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\frac{4}{7} \times 280=\frac{4 \times 280}{7}=\frac{4 \times 40 \times 7}{7}=4 \times 40=160$

7 0

3 years ago

Read 2 more answers

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.

6 0

2 years ago