What is the answer of -3 2/3 + 1.1 as a fraction?

1 answer:

7 0

$- 3 \frac{2}{3} + 1.1 = - \frac{11}{3} + \frac{11}{10} = - \frac{110}{30} + \frac{33}{30} \\ = -\frac{77}{30} = -2 \frac{17}{30}$

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What are the intercepts of 6x + 8y = 24?

Elden [556K]

6x + 8y = 24

x intercept is the value when y =0 & y intercept is the value when x=0

for y=0 ==> 6x+0 =24 ==> x=4 for x=0==> 0+8y =24 ==> y=3

(4,0) & (0,3) (ANSWER d)

5 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.