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yan [13]
3 years ago
15

If there are approximately 6 1/2 hours in the school day, how many minutes are in the school day?

Mathematics
2 answers:
natita [175]3 years ago
4 0

390 minutes

Multiply 6 by 60 and then add 30

nadezda [96]3 years ago
3 0

In 6 hours there is 360 minutes and 1/2 hours is 30 minutes so add 30 to 360 to get 390. There are 390 minutes in a school day

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Solve this by substitution
frozen [14]

Answer:

y= 1 and x=-2

Step-by-step explanation:

Since they both equal y, you can equal them to each other:

x + 3 = 2x + 5

Then solve:

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3 years ago
ASAPPP!!!! HELP I WILL GIVE 100POINTS AND BRAINLIST Question 9 (Essay Worth 10 points)
Luda [366]

Answer:

Part A)

The <em>x-</em>intercepts are (-1/4, 0) and (4, 0).

Part B)

The vertex is a maximum because the leading coefficient is negative.

The vertex is (1.875, 72.25).

Part C)

We can plot the zeros and the vertex, and connect them with a curve.

Step-by-step explanation:

The function given is:

f(x)=-16x^2+60x+16

Part A)

To find the <em>x-</em>intercepts of the function, set the function equal to 0 and solve for <em>x: </em>

<em />0=-16x^2+60x+16<em />

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0=4x^2-15x-4

Factor:

0=(4x+1)(x-4)

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Solve for each case:

\displaystyle x=-\frac{1}{4}\text{ or }x=4

Hence, our zeros are:

\displaystyle \left(-\frac{1}{4}, 0\right)\text{ and } \left(4, 0\right)

Part B)

Note that the leading coefficient of our function is negative.

So, our function will be concave down.

Hence, our vertex will be the maximum.

The vertex is given by:

\displaystyle \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)

In this case, <em>a </em>= -16, <em>b</em> = 60, and <em>c</em> = 16.

Find the <em>x-</em>coordinate of the vertex:

\displaystyle x=-\frac{(60)}{2(-16)}=\frac{15}{8}=1.875

Substitute this back into the function to find the <em>y-</em>coordinate:

f(1.875)=-16(1.875)^2+60(1.875)+16=72.25

Hence, our vertex is:

(1.875, 72.25)

Part C)

Since we already determined the zeros and the vertex, we can plot the two zeros and the vertex and draw a curve between the three points.

The graph is shown below. Again, to do this by hand, simply plot the three points and connect them with a parabola. If necessary, we can also find the <em>y-</em>intercept.

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