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3 years ago

8

Find the product of −35∙−1.5

1 answer:

9966 [12]3 years ago

6 0

17.5 bc normally it would be that for the answer but even with negatives it still is bc a negative times a negative equals a positive

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Answer:

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Step-by-step explanation:

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Write an equation of a line in point-slope form that has a slope of -2 and passes through (5, -1).

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3|3t+1|=2|6t+3| Need hep

mars1129 [50]

Answer:

your answer would be t=-1 or t=1

Step-by-step explanation:

-(2t-3) = 3t-2

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4 years ago

Find the discriminant and the number of real roots for this equation. x^2 + 2x + 8 = 0

ehidna [41]

Step-by-step explanation:

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3 years ago

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Lubov Fominskaja [6]

Answer:

The graph of the function $f(x)=\frac{1}{2}x^{2}-4x+5$ has a minimum located at (4,-3)

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

$f(x)=a(x-h)^{2}+k$

where

a is a coefficient

(h,k) is the vertex of the parabola

If a > 0 the parabola open upward and the vertex is a minimum

If a < 0 the parabola open downward and the vertex is a maximum

In this problem

The coefficient a must be positive, because we need to find a minimum

therefore

Check the option C and the option D

Option C

we have

$f(x)=\frac{1}{2}x^{2}-4x+5$

Convert to vertex form

$f(x)-5=\frac{1}{2}x^{2}-4x$

Factor the leading coefficient

$f(x)-5=\frac{1}{2}(x^{2}-8x)$

$f(x)-5+8=\frac{1}{2}(x^{2}-8x+16)$

$f(x)+3=\frac{1}{2}(x^{2}-8x+16)$

$f(x)+3=\frac{1}{2}(x-4)^{2}$

$f(x)=\frac{1}{2}(x-4)^{2}-3$

The vertex is the point (4,-3) ( is a minimum)

therefore

The graph of the function $f(x)=\frac{1}{2}x^{2}-4x+5$ has a minimum located at (4,-3)

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4 years ago