Let <em>f(x)</em> = <em>x</em>³ + <em>x</em> - 5. <em>f(x)</em> is a polynomial so it's continuous everywhere on its domain (all real numbers). Since
<em>f</em> (1) = 1³ + 1 - 5 = -3 < 0
and
<em>f</em> (2) = 2³ + 2 - 5 = 5 > 0
it follows by the intermediate value theorem that there at least one number <em>x</em> = <em>c</em> between 1 and 2 for which <em>f(c)</em> = 0.
Answer:
x = 7.39
Step-by-step explanation:
log(x)²=(logx)², find the value of x
log x² = log x * log x
2log x = log x * log x
2 = log x
Take the exponent pf both sides
e² = e^logx
e² = x
x = e²
x = 7.39
<u>Q</u><u>uest</u><u>ion</u><u>:</u>
To Simplify:
118 {121÷(11×11)-(-4)-(3-7)}
<u>Solu</u><u>tion</u>:
↠118 {121÷121+4-(-4)}
↠118 {1+4+4}
↠118 {5+4}
↠118{9}
↠118×9
↠1062
▬▬▬▬▬▬▬▬▬▬▬▬
The upper Solution is done by applying BODMAS
<u>Abou</u><u>t</u><u> </u><u>BODMAS</u><u>:</u>
B→ Bracket
O→ Of
D→ Division
M→ Multiplication
A→ Addition
S→ Subtraction
Answer:
The height of the ball after 3 secs of dropping is 16 feet.
Step-by-step explanation:
Given:
height from which the ball is dropped = 160 foot
Time = t seconds
Function h(t)=160-16t^2.
To Find:
High will the ball be after 3 seconds = ?
Solution:
Here the time ‘t’ is already given to us as 3 secs.
We also have the relationship between the height and time given to us in the question.
So, to find the height at which the ball will be 3 secs after dropping we have to insert 3 secs in palce of ‘t’ as follows:


h(3)=160-144
h(3)=16
Therefore, the height of the ball after 3 secs of dropping is 16 feet.