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mr Goodwill [35]
3 years ago
11

Mary had a 3-inch piece of ribbon. She cut it in half. How long were they new pieces

Mathematics
2 answers:
allochka39001 [22]3 years ago
7 0

Answer:

1 and a half inches long

Because that 3 divided by 2 is 1.5.

Novay_Z [31]3 years ago
3 0
1 and a half inches!!! Hope this helped!
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PLZ answer ASAP!!!!! Due at midnight!!
Lapatulllka [165]
Answer: D

Good luck !
3 0
2 years ago
Read 2 more answers
If <img src="https://tex.z-dn.net/?f=%5Cmathrm%20%7By%20%3D%20%28x%20%2B%20%5Csqrt%7B1%2Bx%5E%7B2%7D%7D%29%5E%7Bm%7D%7D" id="Tex
Harman [31]

Answer:

See below for proof.

Step-by-step explanation:

<u>Given</u>:

y=\left(x+\sqrt{1+x^2}\right)^m

<u>First derivative</u>

\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If  $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}

<u />

<u />\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If  $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}

<u />

\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1}  \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}

<u>Second derivative</u>

<u />

\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If  $y=uv$  then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}

\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}

\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}

\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m

\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m

\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}

              = \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)

<u>Proof</u>

  (x^2+1)y_2+xy_1-m^2y

= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m

= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m

= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]

= \left(x+\sqrt{1+x^2}\right)^m\left[0]

= 0

8 0
1 year ago
A rectangle has an area of (12m-30n) square units. Its width is 6 units. Factor the expression for the area to find an expressio
garik1379 [7]

Answer: 2m - 5n

<u>Explanation:</u>

Using long division

  <u>    2m - 5n  </u>

6 ) 12m - 30n


width*length = area

6(2m - 5n) = 12m - 30n


5 0
3 years ago
In 2015, there were about 22 million teenagers (aged 13–17) in the United States. They each sent an average of 900 text messages
Lady bird [3.3K]

Answer: 24000

Step-by-step explanation:

Divide the 22 million by the 900 texts a month to get 24,444 then round to 24,000

4 0
3 years ago
Please help me this is due midnight ​
rjkz [21]

Answer:

Wait does it mean the actual real life distance or the distance on paper?

Step-by-step explanation:

If its real like just google it lol.

5 0
2 years ago
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