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Afina-wow [57]
3 years ago
9

Which of the following systems of linear inequalities is represented by the solution graphed below?

Mathematics
1 answer:
sattari [20]3 years ago
5 0
Answer is A, y us smaller/equal to 2 and y is smaller/equal to X
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The table below shows the distance d(t) in meters that an object travels in t seconds:
Usimov [2.4K]
The answer is B because its going 90m/s and and in 2 seconds its traveled 180 meters which if you add to 60 gives you 240
7 0
3 years ago
Read 2 more answers
A rectangular area adjacent to a river is fenced​ in; no fence is needed on the river side. The enclosed area is 1500 square fee
ZanzabumX [31]

Answer:

a) C(x) = 15000/x + 6x +80

b) Domain of C(x)  { R x>0 }

Step-by-step explanation:

We have:  

Enclosed area = 1500 ft²   = x*y      from which     y  =  1500 / x    (a) where x is perpendicular to the river

Cost = cost of sides of fenced area perpendicular to the river  + cost of side parallel to river + cost of 4 post then

Cost = 10*y + 2*3*x + 4*20 and accoding to (a)  y = 1500/x

Then

C(x)  = 10* ( 1500/x ) + 6*x + 80

C(x) = 15000/x + 6x +80

Domain of C(x)  { R x>0 }

5 0
2 years ago
Select the correct answer.
Tpy6a [65]
When i do it, with the cone formula w volume, i get 678.567 but that’s none of your choices
5 0
3 years ago
What is the RANGE and the IQR for 17,18,22,17,16,21,19,and16
Delvig [45]

Answer:

so 1- 2 is  negative 3 so minus tha by 3 than 4x2

Step-by-step explanation:

8 0
2 years ago
In this triangle, what is the value of x?
nalin [4]

Answer: x=56.5\ km

Step-by-step explanation:

Given the right triangle in the image, you need to remember the following identity:

cos\alpha=\frac{adjacent}{hypotenuse}

Observe the triangle. You can identify that:

\alpha=28\°\\adjacent=x\\hypotenuse=64

Then, knowing these values, you can substitute them into  cos\alpha=\frac{adjacent}{hypotenuse}:

cos(28\°)=\frac{x}{64}

Finally, you have to solve for "x".

Therefore, the value of "x" rounded to the nearest tenth is:

64*cos(28\°)=x\\x=56.5\ km

8 0
3 years ago
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