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Aleksandr [31]
3 years ago
5

the temperature in Portland, Marine, reached -3.5ºF one day last winter. Between with two integers does the temperature lie?

Mathematics
1 answer:
Rzqust [24]3 years ago
3 0

It lies between -3 and -4.

An integer is any whole number, positive or negative.

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The Length of a rectangle exceeds its breadth by 9 cm. if length and breadth are each increased by 3 cm, the area of new rectang
gizmo_the_mogwai [7]

Answer:

Breadth = 8 cm

Length = 17 cm

Step-by-step explanation:

Given rectangle:

Breadth = x cm

Length = x + 9 cm

Area = length * breadth

        = x * (x + 9)

       = x² + 9x

Rectangle with length and breadth increased:

Breadth = x + 3 cm

Length = x + 9 + 3 = x + 12 cm

Area = 84 + (x² + 9x) cm²

(x +3) * (x +12) = 84 + x² +  9x

Use FOIL method

x*x  + x*12 + 3*x + 3*12 = 84 + x² + 9x

      x² + 12x + 3x + 36  = 84 + x² + 9x  {add the like terms}

             x² + 15x  + 36  = 84 + x² + 9x

x² + 15x + 36 - 84 - x² - 9x = 0

      6x - 48 = 0

             6x = 48

             x = 48/6

        x = 8

Breadth = x = 8 cm

Length = x + 9 = 8 +9 = 17 cm

7 0
3 years ago
How many counterexamples do you need to prove a conjecture to be false
Hoochie [10]

Answer:

one , i guess....

because the conjecture is only true if it satisfies all the cases.

3 0
3 years ago
Compute <br><br> I need help
nataly862011 [7]

(\sqrt{3}-\sqrt{6}+\sqrt{12}-\sqrt{24})\times \frac{\sqrt{6}}{2} \\ (\sqrt{3}-\sqrt{6}+2\sqrt{3}-\sqrt{24})\times \frac{\sqrt{6}}{2} \\ (\sqrt{3}-\sqrt{6}+2\sqrt{3}-2\sqrt{6})\times \frac{\sqrt{6}}{2} \\ ((\sqrt{3}+2\sqrt{3})+(-\sqrt{6}-2\sqrt{6}))\times \frac{\sqrt{6}}{2} \\ (3\sqrt{3}-3\sqrt{6})\times \frac{\sqrt{6}}{2} \\ \frac{(3\sqrt{3}-3\sqrt{6})\sqrt{6}}{2} \\ \frac{3(\sqrt{3}-\sqrt{6})\sqrt{6}}{2}

7 0
2 years ago
Shane found gift bags in packs of 8 and bows in packs of 10. If Shane wanted to have the
Sergeeva-Olga [200]

Answer:

5 gift bags

Step-by-step explanation:

8 x 5 = 40

10 x 4 = 40

4 0
3 years ago
The edge of a cube was found to be 15 cm with a possible error in measurement of 0.2 cm. Use differentials to estimate the maxim
Umnica [9.8K]

Answer:

A) Maximum error = 36 cm²

B) Relative Error = 0.0267

C) percentage error = 2.67%

Step-by-step explanation:

Surface area of the cube is;

A(x) = 6x²

Where x is the length of the edge of the cube.

dA/dx = 12x

When x is small like in this case, we can write;

ΔA/Δx ≈ 12x

Thus, ΔA ≈ 12x•Δx

A) Maximum error = 12*15*0.2

Maximum error = 36 cm²

B) Relative Error = Max Error/Surface Area

Surface area = 6 x 15² = 1350

Thus,relative error = 36/1350

Relative Error = 0.0267

C) percentage error = Relative Error x 100%

percentage error = 0.0267 x 100%

percentage error = 2.67%

π

7 0
3 years ago
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