A triangle has side lengths of 7 millimeters, 6 millimeters, and 7 millimeters. Is this triangle equilateral?

Mathematics

1 answer:

Anna007 [38]3 years ago

8 0

Answer: no

Step-by-step explanation:

Because an equilateral triangle has its all sides equal...

But in this case, two of its sides are equal.... And the third side is different.....

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Cuanto es 3/2 en una docena de huevos?

Nady [450]

Mejor respuesta
equivale a doce unidades, por ejemplo:
tengo 12 huevos = tengo una docena de huevos, pero tambien se pueden acumular, por ejemplo:
tengo 24 canicas = tengo 2 docenas
y para sacar las docenas solo divides el numero o las unidades entre 12
UnDeR · hace 7 años

8 0

2 years ago

Which procedure correctly simplifies the expression? ( 4 x − 3 ) − 2 ( x + 5 ) A. Step 1: 4 x − 2 x − 3 + 5 Step 2: 2 x + 2 B. S

olya-2409 [2.1K]

Answer:

Step 1: 4 x − 3 − 2 x − 10

Step 2: 2 x − 13

Step-by-step explanation:

Given the expression;

(4 x − 3) − 2 (x + 5)

The following steps/ procedure are to be taken when simplifying the expression.

open the parenthesis

(4 x − 3) − 2 (x + 5)

= 4x-3 -2(x)-2(5)

= 4x-3-2x-10

collect the like terms

= 4x-2x-3-10

simplify the resulting expression

= 2x-13

Hence the procedure that correctly simplifies the expression are:

Step 1: 4 x − 3 − 2 x − 10

Step 2: 2 x − 13

6 0

2 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 $y = e^x$

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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.