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3 years ago

All quadratic equations can be solved by factoring. True or false

Mathematics

1 answer:

MaRussiya [10]3 years ago

3 0

Answer:

false all the quadratic equations cannot be solved by factorising there is also graphical method to get the solution

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Simplify thefollowing algebraic expression 6(2y +8) -2 (3y

telo118 [61]

6(y+8) is the answer but you didn’t finish the equation so

5 0

3 years ago

Any1 know the answer to this need help ASAP

lukranit [14]

Answer:

D, -77 5/8

Step-by-step explanation:

6 3/4 = 27/4 = 6.75

6.75 x -11.5 = -77.625 or -77 5/8

6 0

3 years ago

Read 2 more answers

Identify the range of the function. A) (2, 8) B) [2, 8] C) (−5, 9) D) [−5, 9]

Ivanshal [37]

Answer:

The answer is "Option A"

Step-by-step explanation:

The domain is the collection of the value, which belongs to the separate variable (horizontal axis). So, to find a region with a graph, it must search for the function, which starts and end. And at all these levels we are searching at x-values.

Its starting point is (2,9) and the ending point is (8,3). Therefore, x= 2 to x=8 is the domain.

5 0

3 years ago

WILL GIVE BRAINLIEST

In-s [12.5K]

Answer:

C. 3⁶

Step-by-step explanation:

Exponents with the same base follow this exponential rule for division:

xᵃ/xᵇ = xᵃ⁻ᵇ

When the bases are the same, we can subtract the exponents.

The bases are the same in this problem: 3. Therefore, we can subtract the exponents from each other:

3⁸/3²
3⁸⁻²
3⁶

The final and correct answer is C. 3⁶.

8 0

2 years ago

Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course. 30% of the SAT prep students were

zloy xaker [14]

Answer:

The required probability is 0.927

Step-by-step explanation:

Consider the provided information.

Surveys indicate that 5% of the students who took the SATs had enrolled in an SAT prep course.

That means 95% of students didn't enrolled in SAT prep course.

Let P(SAT) represents the enrolled in SAT prep course.

P(SAT)=0.05 and P(not SAT) = 0.95

30% of the SAT prep students were admitted to their first choice college, as were 20% of the other students.

P(F) represents the first choice college.

The probability he didn't take an SAT prep course is:

$P[\text{not SAT} |P(F)]=\dfrac{P(\text{not SAT})\cap P(F) }{P(F)}$

Substitute the respective values.

$P[\text{not SAT} |P(F)]=\dfrac{0.95\times0.20 }{0.05\times0.30+0.95\times0.20}$

$P[\text{not SAT} |P(F)]\approx0.927$

Hence, the required probability is 0.927

6 0

3 years ago