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olga_2 [115]
3 years ago
14

I'm so stressed Please help I will try to give you BRAINLIEST if you get it right I promise please help​

Mathematics
1 answer:
kicyunya [14]3 years ago
5 0
A: (10,-3) Quadrant IV (4)
B: (8,9) Quadrant I (1)
C: (-3,-2) Quadrant III (3)
D: (4,0) On the x-axis
E: (-7,-4) Quadrant III (3)
F: (8,7) Quadrant I (1)
G: (0,-7) On the y-axis
H: (1,5) Quadrant I (1)
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X - 6 = 8 what is x?
kondor19780726 [428]

Answer:

x = 14

Step-by-step explanation:

  • x - 6 = 8
  • Group like terms.
  • x = 8 + 6
  • x = 14
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3 years ago
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Jackie is selling smoothies at a school fair. She starts the day with $15 in her cash box to provide change to her customers. If
nadezda [96]

the answer would be D

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How many significant numbers are in 343?
Alona [7]

Answer:

there are 3.

Step-by-step explanation:

3, 4 and 3 are all significant numbers

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3 years ago
If s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n st ?
yulyashka [42]

Answer:

A) None

Step-by-step explanation:

1) s^t shoudnt neccesarily be a factor of nst, for example, if s = 3, t = 4, and n = 12, then both s and t are factors of n, but 3^4 = 81 is not a factor of nst = 144.

2) (st)^2 shoudnt neccesarily be a factor of nst. Let s be 4, let t be 6, and let n be 12. Then n is a factor of both s and t, but (st)^2 = 24^2 is not a factor of nst = 12*24. In fact, it is a greater number.

3) Again, s+t isnt necessarily a factor of nst, let s be 2 and t be 3. Then both s and t are factor of n = 12. However 5 = s+t is not a factor of nst = 72.

So, neither of the three options is guaranteed to be a factor of nst. In fact, for s = 4, t = 6, and n = 12, none of the three options are valid.

4 0
3 years ago
A rectangular storage container with a lid is to have a volume of 2 m3. The length of its base is twice the width. Material for
Scilla [17]

Answer:

Dimensions are 2 m by 1 meter by 1 meter,

Minimum cost is $ 18.

Step-by-step explanation:

Let w be the width ( in meters ) of the container,

Since, the length is twice of the width,

So, length of the container = 2w,

Now, if h be the height of the container,

Volume = length × width × height

2 = 2w × w × h

1 = w² × h

\implies h=\frac{1}{w^2}

Since, the area of the base = l × w = 2w × w = 2w²,

Area of the lid = l × w = 2w²,

While the area of the sides = 2hw + 2hl

= 2h( w + l)

= 2\times \frac{1}{w^2}(w+2w)

=\frac{6w}{w^2}

=\frac{6}{w}  

Since, Material for the base costs $1 per m². Material for the sides and lid costs $2 per m²,

So, the total cost,

C(w) = 1\times 2w^2+2\times 2w^2 + 2\times \frac{6}{w}

=2w^2+4w^2+\frac{12}{w}

=6w^2+\frac{12}{w}

Differentiating with respect to w,

C'(w) = 12w -\frac{12}{w^2}

Again differentiating with respect to w,

C''(w) = 12 + \frac{24}{w^3}

For maxima or minima,

C'(w) = 0

\implies 12w -\frac{12}{w^2}=0

\implies 12w^3 - 12=0

w^3-1=0\implies w = 1

For w = 1, C''(w) = positive,

Hence, for width 1 m the cost is minimum,

Therefore, the minimum cost is C(1) = 6(1)²+12 = $ 18,

And, the dimension for which the cost is minimum is,

2 m by 1 meter by 1 meter.

7 0
3 years ago
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