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diamong [38]
3 years ago
12

What is -7(1+5b)+6 .......

Mathematics
1 answer:
Licemer1 [7]3 years ago
5 0
-1+5b is your answer.
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The base of a cone can be
Brut [27]
The base of a cone is something circular, like a circle or an ellipse. 
The rest of the options (square, triangle, and pentagon) are all based of pyramids, not cones. Cones have to have a circular base because otherwise they'd be pyramids.
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Multiply untegers and divide integers
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Multiply untegers and divide integers?
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4 years ago
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Determine if a slope of two consecutive sides are parallel or neither? ​
inessss [21]

Answer:

No

Step-by-step explanation:

Two consecutive sides means that the two sides are connected in a shape by a vertex. If the sides were parallel, they would have to be right on top of each other, so two consecutive sides cannot be parallel.

3 0
4 years ago
For the given term, find the binomial raised to the power, whose expansion it came from: 220(5x)3(−6y)9.
sukhopar [10]
Answer: (5x - 6y)^{12}
Explanation: For a general binomial expansion, (x + y)^{n}, we know that the powers have to add up to the initial power. This means that the power of x and power of y have to add up to n. This is the binomial theorem.

To further demonstrate this, let's use:
(x + y)^{4}

We can easily expand this. Using Pascal's Triangle, we get:
(x + y)^{4} = x^{4} \cdot y^{0} + 4x^{3} \cdot y^{1} + 6x^{2} \cdot y^{2} + 4x^{1} \cdot y^{3} + x^{0} \cdot y^{4}

As we progress along the expansion, we can see that in each term, the summation of each power remains constant, namely 4.

It doesn't matter what term the binomials are, because the power summation will never change.
This is why we can say that it is raised to the 12th power, and the binomial is:
(5x - 6y).

Thus, we get: (5x - 6y)^{12}
5 0
3 years ago
Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt
ser-zykov [4K]

Answer:

I(t)=\frac{1}{3}(1-e^{-30t})

Step-by-step explanation:

We are given that

\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}

R=150 ohm

L=5 H

V(t)=10 V

P=\frac{R}{L}=\frac{150}{5}=30

I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}

I(t)\times I.F=\int e^{30 t}\times 10 dt+C

I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C

I(t)=\frac{1}{3}+Ce^{-30 t}

I(0)=0

Substitute t=0

0=\frac{1}{3}+C

C=-\frac{1}{3}

Substitute the values

I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}

I(t)=\frac{1}{3}(1-e^{-30t})

5 0
3 years ago
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