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3 years ago

What is a minimum - pre cal

Mathematics

1 answer:

UkoKoshka [18]3 years ago

4 0

Answer: y-value of the vertex of a ∪-shaped (positive) parabola

<u>Step-by-step explanation:</u>

For a quadratic: it is the y-value of the vertex of a positive parabola.

<em>If it is a negative parabola ( ∩-shaped), it is the maximum.</em>

For a cubic: it is the y-value of all of the vertices of the ∪-shaped sections of the graph. These are typically referred to as the "relative minima" when given an interval.

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Item 5

Sedbober [7]

Answer: 300.3858/ maybe 300.385

Step-by-step explanation: 4 2/7 as decimal= 4.29 x (-18) = -77.22 x -3.89= 300.3858 (3.89 = decimal form of -3 8/9) 300.3858 x 1 = 300.3858

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3 years ago

In the computer club, 45% of the

EastWind [94]

Answer: 40

Step-by-step explanation: Do 18/.45 then to check your work you can do 40 times .45

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3 years ago

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It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp

Semmy [17]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$ , $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , $(a^m)^{n}$ = $a^{mn}$

Simplifying the product of the first 2 terms

$\frac{a^{p^2+pq} }{a^{pq+q^2} }$ × $\frac{a^{q^2+qr} }{a^{qr+r^2} }$

= $a^{p^2-q^2}$ × $a^{q^2-r^2}$

= $a^{p^2-r^2}$

Simplifying the third term

5( $(a^p+r)^{p-r}$

= 5 $a^{(p+r)(p-r)}$ = 5 $a^{(p^2-r^2)}$

Performing the division, that is

$\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} }$ ← cancel $a^{(p^2-r^2)}$ on numerator/ denominator leaves

= $\frac{1}{5}$

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3 years ago

How do i do this help me pls

IceJOKER [234]

Answer:

Brackets always means multiply, you multiply the power and the 4 with the number power outside

That gives you 16v^18

Step-by-step explanation:

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3 years ago

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andreev551 [17]

Answer: i believe the answers are A and C

Step-by-step explanation:

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