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2 years ago

What is this I need help−4(8k+5)−5(9k+4)

Mathematics

2 answers:

erica [24]2 years ago

6 0

Answer:

-77k-40

Step-by-step explanation:

Distribute:

-32k-20-45k-20

Combine like terms

-77k-40

dusya [7]2 years ago

3 0

Answer:

-77k - 40

Step-by-step explanation:

-4(8k + 5) - 5(9k + 4)

~Distribute both sides

-32k - 20 - 45k - 20

~Combine like terms

-77k - 40

Best of Luck!

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If sin (theta) is in quadrant 3 then what is tan (theta)?

Irina-Kira [14]

Answer:

positive

Step-by-step explanation:

two negatives make a positive

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Consider the parabola given by the equation: f(x) = 4x² - 6x - 8 Find the following for this parabola: A) The vertex: Preview B)

jeyben [28]

Answer:

The vertex: $(\frac{3}{4},-\frac{41}{4} )$

The vertical intercept is: $y=-8$

The coordinates of the two intercepts of the parabola are $(\frac{3+\sqrt{41} }{4} , 0)$ and $(\frac{3-\sqrt{41} }{4} , 0)$

Step-by-step explanation:

To find the vertex of the parabola $4x^2-6x-8$ you need to:

1. Find the coefficients a, b, and c of the parabola equation

$a=4, b=-6, \:and \:c=-8$

2. You can apply this formula to find x-coordinate of the vertex

$x=-\frac{b}{2a}$ , so

$x=-\frac{-6}{2\cdot 4}\\x=\frac{3}{4}$

3. To find the y-coordinate of the vertex you use the parabola equation and x-coordinate of the vertex ( $f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c$ )

$f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c\\f(\frac{3}{4})=4\cdot (\frac{3}{4})^2-6\cdot (\frac{3}{4})-8\\y=\frac{-41}{4}$

To find the vertical intercept you need to evaluate x = 0 into the parabola equation

$f(x)=4x^2-6x-8\\f(0)=4(0)^2-6\cdot 0-0\\f(0)=-8$

To find the coordinates of the two intercepts of the parabola you need to solve the parabola by completing the square

$\mathrm{Add\:}8\mathrm{\:to\:both\:sides}$

$x^2-6x-8+8=0+8$

$\mathrm{Simplify}$

$4x^2-6x=8$

$\mathrm{Divide\:both\:sides\:by\:}4$

$\frac{4x^2-6x}{4}=\frac{8}{4}\\x^2-\frac{3x}{2}=2$

$\mathrm{Write\:equation\:in\:the\:form:\:\:}x^2+2ax+a^2=\left(x+a\right)^2$

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=2+\left(-\frac{3}{4}\right)^2\\x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{41}{16}$

$\left(x-\frac{3}{4}\right)^2=\frac{41}{16}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

$x_1=\frac{\sqrt{41}+3}{4},\:x_2=\frac{-\sqrt{41}+3}{4}$

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