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nlexa [21]
3 years ago
13

Given f(x) = x^2+4, does there exist an x-intercept between [-2, 3]. Why or why not. Be sure to include in your discussion the I

ntermediate Value Theorem
Mathematics
1 answer:
Anna35 [415]3 years ago
7 0
The x-intercepts are the values of x for which f(x) = 0.
x^2 + 4 = 0
x^2 = -4
x= ±√(-4), which are not real numbers. f(x) does not have an x-intercept.
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3 years ago
Please can someone help? As i'm not confident this is right for factorising 4b^2 + 8b + 3 ​
SIZIF [17.4K]

Answer:

your answer is correct

Step-by-step explanation:

4b² + 8b + 3

Consider the factors of the product of the coefficient of the b² term and the constant term which sum to give the coefficient of the b- term.

product = 4 × 3 = 12 and sum = + 8

The factors are 2 and 6

Use these factors to split the b- term

4b² + 2b + 6b + 3 ( factor the first/second and third/fourth terms )

= 2b(2b + 1) + 3(2b + 1) ← factor out (2b + 1) from each term

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3 years ago
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OleMash [197]

Answer:

i can help you one sec

Step-by-step explanation:

6 0
3 years ago
2) 2x2 – 32 and 3x2 + 6x – 24
ankoles [38]

4-32=-28 && i cant help with the other one

6 0
3 years ago
Solve for t<br><br> -18 = 6t<br><br><br> t = -3<br> t = -1/3<br> t = 12<br> t = 24
larisa86 [58]

Answer:

t=-3

Step-by-step explanation:

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-18÷6=-3

t=-3

5 0
3 years ago
Read 2 more answers
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