I WILL MARK BRAINLIEST

The absolute value of a number is always positive or 0. A. True B. False

Serggg [28]

A. True. True. True. True. True.

5 0

3 years ago

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HELP WILL GIVE BRAINLIEST

CaHeK987 [17]

Answer:

I = 575.0, .08, 3.0

I= 138, i think

Step-by-step explanation:

4 0

3 years ago

Answer please I’m dying from math

charle [14.2K]

Answer:

$\huge\boxed{\text{D)} \ 15x^4 + 2x^3 - 8x^2 - 22x - 15}$

Step-by-step explanation:

We can solve this multiplication of polynomials by understanding how to multiply these large terms.

To multiply two polynomials together, we must multiply each term by each term in the other polynomial. Each term should be multiplied by another one until it's multiplied by all of the terms in the other expression.

We can do this by focusing on one term in the first polynomial and multiplying it by all the terms in the second polynomial. We'd then repeat this for the remaining terms in the second polynomial.

Let's first start by multiplying the first term of the first polynomial, $3x^2$ , by all of the terms in the second polynomial. ( $5x^2+4x+5$ )

$3x^2 \cdot 5x^2 = 15x^4$
$3x^2 \cdot 4x = 12x^3$
$3x^2 \cdot 5 = 15x^2$

Now, we can add up all these expressions to get the first part of our polynomial. Ordering by exponent, our expression is now

$\displaystyle 15x^4 + 12x^3 + 15x^2$

Now let's do the same with the second term ( $-2x$ ) and the third term ( $-3$ ).

$-2x \cdot 5x^2 = -10x^3$
$-2x \cdot 4x = -8x^2$
$-2x \cdot 5 = -10x$

Adding on to our original expression: $\displaystyle 15x^4 + 12x^3 - 10x^3 + 15x^2 - 8x^2 - 10x$

$-3 \cdot 5x^2 = -15x^2$
$-3 \cdot 4x = -12x$
$-3 \cdot 5 = -15$

Adding on to our original expression: $\displaystyle 15x^4 + 12x^3 - 10x^3 + 15x^2 - 8x^2 - 15x^2 - 10x - 12x - 15$

Phew, that's one big polynomial! We can simplify it by combining like terms. We can combine terms that share the same exponent and combine them via their coefficients.

$12x^3 - 10x^3 = 2x^3$
$15x^2 - 8x^2 - 15x^2 = -8x^2$
$-10x - 12x = -22x$

This simplifies our expression down to $15x^4 + 2x^3 - 8x^2 - 22x - 15$ .

Hope this helped!

7 0

3 years ago

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On an interval of [0, 2π), can the sine and cosine values of a radian measure ever be equal? If so, enter the radian measure(s)

Lisa [10]

Answer:

Yes, they are equal in the values (in radians):

π/4, 5π/4

If cos(x) and sin(x) are defined to you as nonnegative functions (in terms of lengths), then 3π/4 and 7π/4 are also included

Step-by-step explanation:

Remember that odd multiples of 45° are special angles, with the same sine and cosine values (you can prove this, for example, by considering a right triangle with an angle of 45° and hypotenuse with length 1, and finding the trigonometric ratios).

The radian measure of 45° corresponds to π/4, hence the odd multiples on the interval [0, 2π) are π/4, 3π/4, 5π/4, 7π/4.

If you define sin(x) and cos(x) using the cartesian coordinate system (via unit circle), then cos(3π/4)=-sin(3π/4) and cos(7π/4)=-sin(7π/4). In this case, only π/4 and 5π/4 are valid choices.

6 0

4 years ago

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in s

aleksandrvk [35]

This is going to be a fourth degree polynomial because if x = 4, one of the factors is x - 4; if x = -2, then one of the factors is x + 2; and by the conjugate root theorem, if x = -1 + 2i, then -1-2i HAS to also be a root. If x = -1+2i, then the factor is (x-(-1+2i)). Simplfying that gives us (x+1-2i). Likewise, if x = -1-2i, then the factor is (x-(-1-2i)). Simplifying that gives us (x+1+2i). We will FOIL those complex factors first. Doing that we have $x^2+x-2ix+x+1-2i+2ix+2i-4i^2$ . Once we simplify that down it's much easier to deal with than what it looks like right now. $x^2+2x+1-4i^2$ . Since i^2 = -1, what we have in the end is $x^2+2x+5$ . Now we will multiply that by x-4. $(x^2+2x+5)(x-4)=x^3-2x^2-3x-20$ . Now finally we will multiply in the last factor of x+2. $x^4-7x^2-26x-40$ is your final fourth degree polynomial.

7 0

3 years ago

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