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3 years ago

Select the point that is a solution to the system of inequalities. y< x² + 3 y>x^2 -4

Mathematics

1 answer:

ivanzaharov [21]3 years ago

6 0

Answer:

D seems to be the only correct answer

Step-by-step explanation:

Full algebraic walkthrough in your other question. You can see that point D lies between the two functions in the picture to check the answer.

I did it also algebraicly in my head.

(would really, reallly appreciate the brainliest)

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Convert 65 mm to meters

Ymorist [56]

0.065 meters is equal to 65 millimeters

7 0

4 years ago

A company wishes to manufacture a box with a volume of 2424 cubic feet that is open on top and is twice as long as it is wide. f

olganol [36]

Let x = the width
Let 2x = the length
Let h = the height
then vol = x*2x*h. So we have 2x^2*h = 24
h=24/(2*x^2)=12/x^2
Surface area: two ends + 1 bottom + 2 sides (no top)
S.A. = 2(x*h) + 1(2x*x) + 2(2x*h)
S.A. = 2xh + 2x^2 + 4xh S.A. = 2x^2 + 6xh
Replace h with 12/x^2
S.A = 2x^2 + 6x(12/x^2)
S.A = 2x^2 + 6(12/x)
S.A = 2x^2 + (72/x)
Graph this equation to find the value of x for minimum material
Min surface area when x = 3.0 is the width then
2(3) = 6 is the length
Find the height:
h=12/(3.0)^2
h=1.33
Box dimensions for min surface area: 3.0 by 6 by 1.33; much better numbers
Check the vol of these dimensions: 3.0*6*1.33 ~ 24
graphic attachment

3 0

3 years ago

Find the equation of the line below. (2, 10) ) (1, 5)

svetlana [45]

Answer:

y=5x

Step-by-step explanation:

(y2-y1)/(x2-x1)

(5-10)/(1-2) = -5/-1

m=5

y=mx+b

(2,10)

10=5(2)+b

10=10+b

b=0

8 0

3 years ago

Write a number with one decimal place ,that is bgger than 6 4/5 but smaller than 7

Ann [662]

Answer:

6 9/10

Step-by-step explanation:

3 0

3 years ago

Suppose that, in addition to edge capacities, a flow network has vertex capacities. That is each vertex has a limit l./ on how m

storchak [24]

Answer:

See explanation and answer below.

Step-by-step explanation:

The tranformation

For this case we need to construct G' dividing making a division for each vertex v of G into 3 edges that on this case are $v_1, v_2 and l(v)$ .

We assume that the edges from the begin are the incoming edges of $v_1$ and all the outgoing edges from v are outgoing edges from $v_2$

We need to construct $G' = (V', E')$ with capacity function a' and we need to satisfy the follwoing:

For every $v \in V$ we create 2 vertices $v_1, v_2 \in V'$

Now we can add a new edge asscoiated to $v_1, v_2 \in E'$ with the condition $a' (v_1,v_2) = l(v)$

Now for each edges $(u,v)\in E$ we can create the following edge $( u_r, v_1) \in E'$ and the capacity is given by: $a' (u_r, v_1) = a (u,v)$

And for this case we can see this:

$|V'| = 2|V|, |E'|= |E| +|V|$

Now we assume that x is the flow who belongs to G respect vertex capabilities. We can create a flow function x' who belongs to G' with the following steps:

For every edge $(u,v) \in G$ we can assume that $x' (u_r ,v_1) = x(u,v)$

Then for each vertex $u \in V -t$ and we can define $x\(u_1,u_r) = \sum_{v \in V} x(u,v)$ and $x' (t_1,t_2) = \sum_{v \in V} x(v,t)$

And after see that the capacity constraint on this case would be satisfied since for every edge in G' on the form $(u_r, u_1)$ we have a corresponding edge in G because:

$u \in V -(s,t)$ we have that:

$x' (u_1, u_r) = \sum_{v \in V} x(u,v) \leq l(u) = a' (u_1, u_r)$

$x' (t_1,t_2) = \sum_{v \in V} x(v,t) \leq (t) = a' (t_1,t_2)$

And with this we have the maximization problem solved.

We assume that we have K vertices using the max scale algorithm.

6 0

3 years ago