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3 years ago

Solve for z over the imaginary numbers. Please help!

Mathematics

1 answer:

bezimeni [28]3 years ago

5 0

So you subtract the five over to get -5=2e^z, divide by 2, -2.5=e^z, take the natural log of both sides to get z=ln(-2.5) which does not exist

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Answer:10 1/9 oz

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3 years ago

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If $1000 is invested at 8% interest, compounded annually, then after n years the investment is worthan = 1000(1.08)ndollars.(a)

Luba_88 [7]

Answer:

Diverge

Step-by-step explanation:

(a)

1st year: $1000 * 1.08^1 = \$1080$

2nd year: $1000 * 1.08^2 = \$1166.4$

3rd year: $1000 * 1.08^3 = \$1259.71$

4th year: $1000 * 1.08^4 = \$1360.49$

5th year: $1000 * 1.08^5 = \$1469.33$

(b) The sequence is divergent, because if we take the derivative of the function with respect to n year:

$(1000*1.08^n)^{'} = 1000ln(n)*1.08^n$

This is a positive, meaning the slope of the function is positive. If we take the second derivative using product rule

$(1000ln(n)*1.08^n)^{'} = 1000\frac{ln(n)}{n}1.08^n$

This is also positive when n > 0. Therefore, the slope is positive and increasing. This means the sequence diverges.

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3 years ago

A pet store has 15 birds and 75 fish.

Ludmilka [50]

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Someone plz help me

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Step-by-step explanation:

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3 years ago

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zloy xaker [14]

Answer:

$1)\ \ 4h^2-13h+6\\2)\ \ 7x^3y^2-x^2y+1\\3)\ \ -7n+2\\4)\ \ -8m+4$

Step-by-step explanation:

Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.

$2h^2-7h+2h^2-h+6+4h-9h$

$(2h^2+2h^2)+(-7h-h+4h-9h)+(6)$

$4h^2-13h+6$

Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.

$8x^3y^2-7x^2y+8x-4-x^3y^2+2x^2y+4x^2y-8x+5$

$(8x^3y^2-x^3y^2)+(-7x^2y+2x^2y+4x^2y)+(8x-8x)+(-4+5)$

$7x^3y^2-x^2y+1$

Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following ( $a(b+c)=(a)(b)+(a)(c)=ab+ac$ ). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,