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Tatiana [17]
2 years ago
13

D%7B1%20%2B%20tan%20%5Calpha%20%7D%20" id="TexFormula1" title=" \frac{cot \alpha }{1 + cot \alpha } = \frac{1}{1 + tan \alpha } " alt=" \frac{cot \alpha }{1 + cot \alpha } = \frac{1}{1 + tan \alpha } " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
strojnjashka [21]2 years ago
8 0
Im not sure what its asking.. whats the question?
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Solve the following inequality. 2(P + 1) > 7 + P
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Step-by-step explanation: screenshot

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True or false? <br>Help will make brainliest !​
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True is the correct answer.

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What are the roots of the quadratic equation 2x2+7x+4=0? Select all that apply.
skad [1K]

Answer:

option A and B

x=\frac{-7+\sqrt{17}} {4}

and

x=\frac{-7-\sqrt{17}} {4}

Step-by-step explanation:

we have

2x^2+7x+4=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

2x^2+7x+4=0

so

a=2\\b=7\\c=4

substitute in the formula

x=\frac{-7\pm\sqrt{7^{2}-4(2)(4)}} {2(2)}

x=\frac{-7\pm\sqrt{17}} {4}

so

x=\frac{-7+\sqrt{17}} {4}

and

x=\frac{-7-\sqrt{17}} {4}

7 0
3 years ago
Approximate f by a Taylor polynomial with degree n at the number a. Step 1 The Taylor polynomial with degree n = 3 is T3(x) = f(
MAVERICK [17]

Answer:

If you center the series at x=1

T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + \frac{56}{3} e^2(x-1)^3 + R(x)

Where R(x) is the error.

Step-by-step explanation:

From the information given we know that

f(x) = e^{2x^2}

f'(x) = 4x e^{2x^2}   (This comes from the chain rule )

f^{(2)}(x) = 4e^{2x^2} (4x^2+1)   (This comes from the chain rule and the product rule)

f^{(3)}(x) = 16xe^{2x^2}(4x^2 + 3)  (This comes from the chain rule and the product rule)

If you center the series at x=1  then

T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + \frac{56}{3} e^2(x-1)^3 + R(x)

Where R(x) is the error.

3 0
3 years ago
What's 6 / 3 and 13 divided by 1at 3
Korvikt [17]

Answer:

5

Step-by-step explanation:

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3 years ago
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