What is the y intercept of the quadratic equation f(x)=(x-8)(x+3)?

Mathematics

1 answer:

ivann1987 [24]3 years ago

7 0

The Y-intercept is when the line touches or crosses the y-axis and so the answer is (0,-24)
hope this helps!

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A rectangle has a height of 4x ^ 3 width of x ^ 3 + 3x ^ 2 + 2x . Express the area of the entire rectangle .

astra-53 [7]

Answer:

$4x^6+12x^5+8x^4$

Step-by-step explanation:

To find the area of a triangle, you need to multiply the width by the height. In this case:

$(4x^3)(x^3+3x^2+2x)=\\4x^6+12x^5+8x^4$

Hope this helps!

6 0

3 years ago

Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete

Katyanochek1 [597]

Answer:

b/a
16a²b²
n¹⁰/(16m⁶)
y⁸/x¹⁰
m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

$a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}$

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

$\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}$

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

$\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2$

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

$\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}$

4. This works the same way the previous problem does.

$\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}$

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

$\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}$