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ziro4ka [17]
3 years ago
12

Show that the number 6 is a rational number by finding a ratio of two integers equal to the number.

Mathematics
1 answer:
nydimaria [60]3 years ago
4 0

I think this is right

-2 :-4

is the answer

Hope this helps : )

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Suppose an airline policy states that all baggage must be box shaped with a sum of​ length, width, and height not exceeding 114
NISA [10]

Answer:

Step-by-step explanation:

Represent the length of one side of the base be s and the height by h.  Then the volume of the box is V = s^2*h; this is to be maximized.

The constraints are as follows:  2s + h = 114 in.  Solving for h, we get 114 - 2s = h.

Substituting 114 - 2s for h in the volume formula, we obtain:

V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)

This is to be maximized.  To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:

dV

----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2

ds

Simplifying this, we get dV/ds = -4s^2 + 114s = 0.  Then either s = 28.5 or s = 0.

Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in

and the volume is V = s^2(h) = 46,298.25 in^3

7 0
3 years ago
Solve for the variable in the following inequality<br><br> 18 ≥ 6x
baherus [9]

Answer:

3 <u>> </u>x

Step-by-step explanation:

(Sorry if im wrong)

Divide both sides by 6

18/6 <u>></u> 6x/6

3 <u>></u> x

6 0
3 years ago
4 heyyyyyyy help me please
dolphi86 [110]

Answer:

16 runs

Step-by-step explanation:

16-6=10

3 0
3 years ago
A 24-ounce can of tomato sauce costs $3.60. What is the cost per ounces?
cluponka [151]
.15 cents. You do the money divided by the ounces or weight.
6 0
3 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
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