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Tcecarenko [31]
3 years ago
15

PLEASE HELP ME!!!!

Mathematics
1 answer:
Nat2105 [25]3 years ago
7 0
8 * 2 * 10 = 160 * 10 = 1600
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The square root of 64 is 8 but put 8 to the 1/2 power
The answer is 4
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2 years ago
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Please help and if you can please explain it well thank you
Ksju [112]

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4 0
3 years ago
Please! Solve f(x)-g(x) if f(x)=1/2x^4 - x^2 and g(x)= 3/4x^4 +3x^3 - x^2
igomit [66]

Answer:

f(x)-g(x)=\frac{-1}{4}x^{4}-3x^{3}

Step-by-step explanation:

This is a subtraction between polynomials, since both functions are polynomials, hence, to solve the problem with need to perform the subtraction, by coefficients.

f(x)= \frac{1}{2}x^{4}+0x^{3}-x^{2}\\\\g(x)=\frac{3}{4}x^{4}+3x^{3}-x^{2}

By performing the subtraction, we have:

f(x)-g(x)=\frac{-1}{4}x^{4}-3x^{3}

6 0
3 years ago
Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist
yuradex [85]

Answer:

0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5)y' + (ln t)y = 6t

and the condition y(1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [(ln t)/(t - 5)]y = 6t/(t - 5)

(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).

6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.

3 0
3 years ago
If c = 4, evaluate the following expression: <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7Bc%7D%20" id="TexFormula1" title=
mash [69]

Answer:

2 is the answer

Step-by-step

6 0
2 years ago
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