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4 years ago

Write an equation to express the value of h+k. explain

Mathematics

1 answer:

mars1129 [50]4 years ago

7 0

You have to be a little more detailed

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The first part of a trip driving at 40 mph was 15 miles longer and took 30 minutes longer than the second part of the trip drivi

kenny6666 [7]

Answer:

65 miles

Step-by-step explanation:

There are two simultaneous equations . Remeber that v=d/t, 1hour = 60 minutes.

1. 40 = (d+15)/(t+0.5)

2. 50 = d/t

solving: d= 25 miles and t=0.5 hours.

The distance traveled is:

D=d+d+15

D=65 miles

5 0

3 years ago

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Which is the better deal ? Option A : 40 trash bags for $8.00 or Option B : 24 trash bags for $3.60

Setler79 [48]

Answer:

They're the same

Step-by-step explanation:

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4 years ago

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HELP ME PLEASE thank you so much to who helps me :)

igor_vitrenko [27]

Answer:

π(12.1)²

Step-by-step explanation:

12.1 is the radius

Formula for base = πr²

= π(12.1)²

3 0

2 years ago

Dave uses 5 gallons of gas to drive 140 miles on the highway. If he travels at the same rate, how many miles can he drive on 7 g

Solnce55 [7]

Answer:

168 miles

Step-by-step explanation:

If he uses 5 gallons to drive 140 miles, that means that he can drive 24 miles for every gallon (140/5=24)

7 times 24 is 168 so he can drive 168 miles on 7 gallons of gas

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3 years ago

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ITS TIMED PLEASE HELP

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)

5 0

4 years ago