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user100 [1]
3 years ago
6

The product of w and 5 is less than or equal to -25

Mathematics
1 answer:
Elenna [48]3 years ago
8 0

Answer:

5w≤-25

w≤-5

Step-by-step explanation:

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12 friends shared 8 small pizzas equally how much pizza did each person get?
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To get the answer you have to do 12 divided by 8
12/8=1.5
So everyone got 1 and 1/2 pizzas per person
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Which of the following equation has both -6 and 6 as possible values of c.... C²=36, c³=216, or none of the above
erastova [34]

Answer: THe equation C²=36 can have both -6 and 6 as possible values since -6 squared is 36 and so is 6 squared

7 0
3 years ago
Please help me solve step by step it's urgent​
NNADVOKAT [17]

Answer:

2i: 169.71

2ii: 0.17L

3a: 4×10⁻⁵

3b: 110011

Step-by-step explanation:

2i. The surface of the top and bottom of the tin is two times (top and bottom) π·r² = 2·π·3² = 18π cm².

The circumference of the circle is 2·π·r = 6π cm².

The area of the material connecting top and bottom is a rectangle of the tin height times the circumference: 6·6π = 36π cm².

This gives a total of  18π + 36π = 54π  cm².

With π approximated by 22/7 the total surface area is 54*22/7 ≈ 169.71.

Notice how the calculation is simple by waiting until the very last moment to substitute π.

2ii. The volume is the area π·r² of the circle times the height of the tin: 9π*6 = 54π cm³ ≈ 169.71 cm³.

Since 1L = 1000 cm³ the volume is 0.16971 litres, which should be rounded to 0.17 L.

3a: If we rewrite P as 36 x 10⁻⁴ and realize that 36/2.25 = 16, then the fraction can be written as

16 x 10⁻⁴⁻⁶ = 16 x 10⁻¹⁰.

The square root of that is taking it to the power of 1/2, so (16x10⁻¹⁰)^0.5 = 4x10⁻⁵ = 0.00004

3b: 1111 1111 is 255 in decimal. 101 is 5 in decimal. 255/5 is 51 in decimal. 51 in binary is 110011.

6 0
3 years ago
Please help me with the below question.
Snezhnost [94]

6a. By the convolution theorem,

L\{t^3\star e^{5t}\} = L\{t^3\} \times L\{e^{5t}\} = \dfrac6{s^4} \times \dfrac1{s-5} = \boxed{\dfrac5{s^4(s-5)}}

6b. Similarly,

L\{e^{3t}\star \cos(t)\} = L\{e^{3t}\} \times L\{\cos(t)\} = \dfrac1{s-3} \times \dfrac s{1+s^2} = \boxed{\dfrac s{(s-3)(s^2+1)}}

7. Take the Laplace transform of both sides, noting that the integral is the convolution of e^t and f(t).

\displaystyle f(t) = 3 - 4 \int_0^t e^\tau f(t - \tau) \, d\tau

\implies \displaystyle F(s) = \dfrac3s - 4 F(s) G(s)

where g(t) = e^t. Then G(s) = \frac1{s-1}, and

F(s) = \dfrac3s - \dfrac4{s-1} F(s) \implies F(s) = \dfrac{\frac3s}{\frac{s+3}{s-1}} = 3\dfrac{s-1}{s(s+3)}

We have the partial fraction decomposition,

\dfrac{s-1}{s(s+3)} = \dfrac13 \left(-\dfrac1s + \dfrac4{s+3}\right)

Then we can easily compute the inverse transform to solve for f(t) :

F(s) = -\dfrac1s + \dfrac4{s+3}

\implies \boxed{f(t) = -1 + 4e^{-3t}}

6 0
1 year ago
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