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Dominik [7]
3 years ago
11

Laneka owns a cake shop she is currently preparing cakes for two anniversary parties the first cake has three small tears and on

e medium tear and will serve a total of 100 guests the second one has three small tears into medium tears and will serve a total of 140 cast represent the situation with a system of equations
Mathematics
1 answer:
stiv31 [10]3 years ago
3 0

Answer:

potato

Step-by-step explanation:

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Find the value of the integral that converges.<br> ∫^-5_-[infinity] x^-2 dx.
Bingel [31]

Answer:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

Step-by-step explanation:

For this case we want to find the following integral:

\int_{-\infty}^{-5} x^{-2}dx

And we can solve the integral on this way:

\int_{-\infty}^{-5} x^{-2}dx= \frac{x^{-2+1}}{-2+1} \Big|_{-\infty}^{-5}

\int_{-\infty}^{-5} x^{-2}dx= -\frac{1}{x} \Big|_{-\infty}^{-5}

And if we evaluate the integral using the fundamental theorem of calculus we got:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

8 0
3 years ago
Ok here it is ..........
SIZIF [17.4K]
The answer is y=3x-4
7 0
3 years ago
Read 2 more answers
Write a polynomial of least degree with rational coefficients, a leading coefficient of 1, and zeros at 3 and sqr root 7. Expand
Tresset [83]

Answer:

(x-3)(x^2-7) = x^3 -3x^2 -7x + 21

just change the sign of the root and put x in front of it, then multiply the factors all together

(x-3)(x-sqr7)(x+sqr7)    roots come in conjugate pairs to eliminate irrational coefficients

(x-sqr7)(x+sqr7) = x^2-7.  similar to (a-b)(a+b) = a^2-b^2

Step-by-step explanation:

6 0
3 years ago
What is the answer to this question?
harina [27]

Answer:

f(g(0) )= -1

Step-by-step explanation:

Find g(0) = -1

Then find f(-1) = 1

f(g(0) )= -1

4 0
2 years ago
Given: ∠BCD ≅ ∠EDC and ∠BDC ≅ ∠ECD Prove: Δ BCD ≅ Δ EDC
never [62]
It would be bisecting angles
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3 years ago
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