HELP ME THIS IS MISSING HOMEWORK AND I DONT WANT ANY FILES OK!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

aliya0001 [1]3 years ago

8 0

Answer:

4a) $\frac{2}{3}$ · (5 + 7)

b) 3 ÷ (3/5)

c) ( $\frac{2}{3}$ - $\frac{1}{2}$ ) ÷ 3

Step-by-step explanation:

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Jackie scored 81 and 87 on her first two quizzes . Write and solve a compound inequality to find the possible values for a third

asambeis [7]

Answer:

The third score must be larger than or equal to 72, and smaller than or equal 87

Step-by-step explanation:

Let's name "x" the third quiz score for which we need to find the values to get the desired average.

Recalling that average grade for three quizzes is the addition of the values on each, divided by the number of quizzes (3), we have the following expression for the average:

$average=\frac{81+87+x}{3}$

SInce we want this average to be in between 80 and 85, we write the following double inequality using the symbols that include equal sign since we are requested the average to be between 80 and 85 inclusive:

$80\leq average\leq 85\\80\leq \frac{81+87+x}{3} \leq 85\\80\leq \frac{168+x}{3} \leq 85$

Now we can proceed to solve for the unknown "x" treating each inaquality at a time:

$80\leq \frac{168+x}{3}\\3*80\leq 168+x\\240\leq 168+x\\240-168\leq x\\72\leq x$

This inequality tells us that the score in the third quiz must be larger than or equal to 72.

Now we study the second inequality to find the other restriction on "x":

$\frac{168+x}{3} \leq 85\\168+x\leq 85 *3\\168+x\leq 255\\x\leq 255-168\\x\leq 87$

This ine $72\leq x} \leq 87$ quality tells us that the score in the third test must be smaller than or equal to 87 to reach the goal.

Therefore to obtained the requested condition for the average, the third score must be larger than or equal to 72, and smaller than or equal 87:

4 0

3 years ago

What is the solution set of x2 + 5x + 1 = 0?

DerKrebs [107]

Answer:

$\huge\boxed{\left\{-\dfrac{5+\sqrt{21}}{2};\ \dfrac{-5+\sqrt{21}}{2}\right\}}$

Step-by-step explanation:

$x^2+5x+1=0\\\\\text{Use the quadratic formula:}\\\\\text{For}\ ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\\text{if}\ \Delta < 0,\ \text{then the equation has no real solution}\\\text{if}\ \Delta=0,\ \text{then the equation has one real solution}\ x=\dfrac{-b}{2a}\\\text{if}\ \Delta>0,\ \text{then the equation has two real solution}\ x=\dfrac{-b\pm\sqrt{\Delta}}{2a}$