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lakkis [162]
3 years ago
6

To the variation for this equation a=1/2*b*h

Mathematics
1 answer:
natka813 [3]3 years ago
4 0
One variation of the area of the area of a triangle
a= b*h/2
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7.6x = 64 is approximately 3.05.<br> is this the solution?
lisov135 [29]
7.6x = 64
x = 64/7.6
x = <span> <span> <span> 8.4210526316 </span> </span> </span>
So, 3.05 is NOT the solution

3 0
3 years ago
Find the area of the figure​
statuscvo [17]
I just looked it up and it said it is 180
3 0
3 years ago
Describe how the graph of y=|x| and how it is difficult
Ksju [112]
Y= to the absolute function of x?
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4 0
3 years ago
A football player is running the length of a 100 yard long football field. For the first part of the field, he runs at a rate of
notka56 [123]

Answer:

\frac{2}{5}th part of the field is covered in the second sprint.

Step-by-step explanation:

A football player is running the length of a 100 yard long football field.

Let player sprints x yards with the speed = 2 yards per second.

So time taken to cover x yards player will take time = \frac{x}{2} seconds

Now rest distance (100 - x) yards when covered with the speed of 4 yards per second, so time taken to cover this distance = \frac{Distance}{Speed}

= \frac{100-x}{4} seconds

Now total time taken by the player can be represented by the equation

\frac{x}{2}+\frac{100-x}{4}=40

Now we can solve this equation for the value of x.

\frac{2x+100-x}{4}=40

x + 100 = 40×4

x + 100 = 160

x = 160 - 100 = 60 yards

And length of the second part will be = 100 - 60 = 40 yards

Now the fraction of the field covered by the player in second sprint will be

= \frac{40}{100}

= \frac{2}{5} or 40%

Therefore, \frac{2}{5}th part of the field was covered in second sprint.

5 0
3 years ago
Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.
Rzqust [24]

Answer:

a = 6 or 9

Step-by-step explanation:

The "a", "b", and "c" of the quadratic formula are the coefficients of a², a, and the constant term in the given equation:

... a = 2, b = -30, c = 108

Then the quadratic formula tells you the solutions are ...

... (-b ± √(b² -4ac))/(2a)

... = (-(-30) ± √((-30)² -4(2)(108)))/(2(2))

... = (30 ± √(900 -864))/4

... = (30 ± √36)/4

... = (30 ± 6)/4 = {24, 36}/4

... = {6, 9}

_____

The <em>a</em> variable should not be confused with the "a" that is used to name the coefficient of the square of the variable in the quadratic formula. If it is too confusing, rewrite one or the other. For example, you could write ...

... The solution to pa² +qa +r = 0 is ... a = (-q ± √(q²-4pr))/(2p)

where p=2, q=-30, r=108 in the given equation.

5 0
3 years ago
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