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Gre4nikov [31]
2 years ago
14

On average, cody runs 4 miles in 38 minutes. How fast should he plan to run a 10-mile race if he maintains this place.

Mathematics
1 answer:
viva [34]2 years ago
6 0
C. 1 hour and 35 minutes
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The number of customers in a grocery store is modeled by the function y -X? + 10x + 50, where y is
vova2212 [387]

Using the vertex of the quadratic function, it is found that:

a) The maximum number of customers in the store is at 12 P.M.

b) 75 customers are in the store at this time.

The number of customers in x hours after 7 AM is given by:

y = -x^2 + 10x + 50

Which is a quadratic equation with coefficients a = -1, b = 10, c = 50

Item a:

The maximum value, considering that a < 0, happens at:

x_v = -\frac{b}{2a}

Hence:

x_v = -\frac{10}{2(-1)} = 5

5 hours after 7 A.M, hence, the maximum number of customers in the store is at 12 P.M.

Item b:

The value is y(5), hence:

y(5) = -(5)^2 + 10(5) + 50 = 75

75 customers are in the store at this time.

A similar problem is given at brainly.com/question/24713268

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2 years ago
Please help me thank u
finlep [7]
It would be A=18 because of the equation a^2+b^2=c^2.
3 0
3 years ago
Which special version of the Pythagorean Theorem can you use to find the length of any square's diagonal, d, using only the leng
kodGreya [7K]
The square's diagonal is the triangle's hypotenuse.

the original Pythagorean theorem is  a^{2} +  b^{2} = c^{2}   where a and b are the two sides and c is the hypotenuse.

that means the Pythagorean theorem for this question is:

s^{2} + s^{2} = d^{2}  or 2( s^{2} )= d^{2}
8 0
3 years ago
Read 2 more answers
Draw the two lines y = 2x + 2/3, and y = 2x + 1/3. They are parallel a
Nonamiya [84]

Answer:

Two line are perpendicular when they are at right angles to each other.

The red line is perpendicular to the blue line in each of these examples:

Perpendicular Example

Step-by-step ex;planation:

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2 years ago
Can anyone help me with this?<br><br> Reduce the fraction
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\bf \cfrac{-20t^5u^2v^3}{48t^7u^4v}\implies \cfrac{-20}{48}\cdot \cfrac{t^5u^2v^3}{t^7u^4v^1}\implies \cfrac{-5}{12}\cdot \cfrac{v^3v^{-1}}{t^7t^{-5}u^4u^{-2}}\implies \cfrac{-5}{12}\cdot \cfrac{v^{3-1}}{t^{7-5}u^{4-2}} \\\\\\ \cfrac{-5}{12}\cdot \cfrac{v^2}{t^2u^2}\implies \cfrac{-5v^2}{12t^2 u^2}

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