Answer:9787987987987987899999999888888888888888888'
Step-by-step explanation:
that that that 14523095480239845023984203984230482309423094823029834
<h2><u>Math</u><u> </u><u>Problem</u><u> </u></h2>
<u>If</u><u> </u><u>die</u><u> </u><u>is</u><u> </u><u>rolled</u><u> </u><u>two</u><u> </u><u>times</u><u>,</u><u> </u><u>find</u><u> </u><u>the</u><u> </u><u>probably</u><u> </u><u>of</u><u> </u><u>getting</u><u> </u><u>"</u><u>1'twice</u><u>.</u>
<h2><u>Math</u><u> </u><u>Answer</u><u> </u></h2>
<u>
</u>
<u>Feel</u><u> </u><u>fre</u><u> </u><u>e</u><u> </u><u>too</u><u> </u><u>ask</u><u> </u><u>me</u><u> </u>
<u>Please</u><u> </u><u>understand</u><u> </u><u>it</u>
<u>#BrainliestBunch</u>
If <em>x</em> = -1, you have
2(-1) + 3 cos(-1) + <em>e</em> ⁻¹ ≈ -0.0112136 < 0
and if <em>x</em> = 0, you have
2(0) + 3 cos(0) + <em>e</em> ⁰ = 4 > 0
The function <em>f(x)</em> = 2<em>x</em> + 3 cos(<em>x</em>) + <em>eˣ</em> is continuous over the real numbers, so the intermediate value theorem applies, and it says that there is some -1 < <em>c</em> < 0 such that <em>f(c)</em> = 0.
I believe it is 2 but I am not sure