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3 years ago

Are all parallelograms quadrilaterals

Mathematics

2 answers:

Leni [432]3 years ago

7 0

Answer:

yeah i think so

Step-by-step explanation:

vitfil [10]3 years ago

3 0

Answer:

im pretty sure they are

Step-by-step explanation: i may be wrong

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A rectangle has an area of 24 square unit. The width is 5 units less than the length. What is the length, in units, of the recta

vlabodo [156]

L-5=w and (L)(W)=24

From there you can use elimination or substitution to solve... You should get
L=8
W=3

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3 years ago

*please help*<br><br> I just need someone to check this for me.. am i doing this right?

BigorU [14]

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3 years ago

What would you do first to solve this problem? 7x4 [ (6+3) -4]

mr Goodwill [35]

Answer:

do whats in the parenthesis first

Step-by-step explanation:

so like this

7x^4[(6+3)-4]

7x^4[9-4]

7x^4+5

If brainiest is earned its greatly appreciated

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3 years ago

Read 2 more answers

Good morning guys can you guys plz help on my math :D

Softa [21]

Answer:

Question 1:

Question 2:

False

Question 3:

Question 4:

Step-by-step explanation:

<em>Please let me know if these are wrong is that I can fix them.</em>

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3 years ago

Help finding derivative

marusya05 [52]

$\bf \displaystyle \cfrac{d}{dx}\left[ \int\limits_{\sqrt{x}}^{\frac{\pi }{2}}~[cos(t^2)]dt \right]\implies \cfrac{d}{dx}\left[ -\int\limits_{\frac{\pi }{2}}^{\sqrt{x}}~[cos(t^2)]dt \right]\\\\\\ \cfrac{d}{dx}\left[ -\int\limits_{\frac{\pi }{2}}^{x^{\frac{1}{2}}}~[cos(t^2)]dt \right]\\\\
-------------------------------\\\\
u=x^{\frac{1}{2}}\implies \cfrac{du}{dx}=\cfrac{1}{2\sqrt{x}}\\\\\\ \textit{and now, let's use the \underline{2nd fundamental theorem of calculus}}\\\\
-------------------------------\\\\$

$\bf \displaystyle \cfrac{d}{dx}\left[ -\int\limits_{\frac{\pi }{2}}^{u}~[cos(t^2)]dt \right]\impliedby \cfrac{df}{dx}=\cfrac{df}{du}\cdot \cfrac{du}{dx}
\\\\\\
\cfrac{df}{dx}=-cos(u^2)\cdot \cfrac{du}{dx}\implies \cfrac{df}{dx}=-cos(u^2)\cdot \cfrac{1}{2\sqrt{x}}
\\\\\\
\cfrac{df}{dx}=-cos[(\sqrt{x})^2]\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{df}{dx}=-\cfrac{cos(x)}{2\sqrt{x}}$

6 0

3 years ago

Are all parallelograms quadrilaterals​

Are all parallelograms quadrilaterals