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3 years ago

Which one is rong Scenario Equation Graph or Table

Mathematics

2 answers:

Kay [80]3 years ago

7 0

The scenario is wrong. according to the three other tools, the hot air balloon would be increasing in altitude, not descending.

melisa1 [442]3 years ago

6 0

Scenario is incorrect

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Look at this cone:<br><br>,,

Assoli18 [71]

Slant height=l=3ft
Radius=r=2ft

We know

$\boxed{\sf \star TSA_{(Cone)}=\pi r(r+\ell)}$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=2 \dfrac{22}{7}\times 2(2+3)$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=\dfrac{44}{7}(5)$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=\dfrac{220}{7}$

$\\ \sf\longmapsto TSA_{(Old\:Cone)}=31.4ft^2$

Now

New slant height =2(3)=6cm
New radius=2(2)=4cm

$\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{22}{7}\times 4(4+6)$

$\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{88}{7}(10)$

$\\ \sf\longmapsto TSA_{(New\:Cone)}=\dfrac{880}{7}$

$\\ \sf\longmapsto TSA_{(New\:Cone)}=125.7cm^2$

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cone)}}{TSA_{(Old\:Cone)}}=\dfrac{125.7}{31.4}$

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cone)}}{TSA_{(Old\:Cone)}}=\dfrac{4}{1}$

$\\ \sf\longmapsto\underline{\boxed{\bf{ {TSA_{(New\:Cone)}}:{TSA_{(Old\:Cone)}}=4:1}}}$

7 0

3 years ago

X+y=3 3y+x=5 Substitution

jeka57 [31]

Y=1 and x=2 because 2+1=3 and 2(x) plus 3(y) is 5

3 0

3 years ago

Rationalize the denominator and simplify.

cestrela7 [59]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3151018

——————————

$\mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}$

Multiply and divide by the conjugate of the denominator, which is $\mathsf{4\sqrt{6}+3\sqrt{2}:}$

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}$

Expand those products and eliminate the brackets:

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}$

$\mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}}$

$\mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}}$

$\mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

4 0

3 years ago

Can someone please help me with my calc homework? Anyone who is good at Radical Expressions pls help.

Lesechka [4]

1. x
2. x
3. -x^2
4.
5. x
6. x^3
7. x
8. |x|
9. x-2
10. x/3
11. x/2
12. x+2

3 0

3 years ago

2x| if X is less than 0 <br> I NEED HELP EVERYONE THIS IS DUE

trasher [3.6K]

Answer:

-2x

Step-by-step explanation:

1. Flip it around and then it turns Negative

2. The answer will be -2x!

7 0

2 years ago