Look at the attachment

What are the rules for multiplying negative integers

Sergeeva-Olga [200]

Answer:

Negative time a negative equals a positive!

Step-by-step explanation:

When you multiply two negative numbers or two positive numbers then the product is always positive. The same also goes for dividing negative integers.

7 0

3 years ago

Read 2 more answers

Factor Completely x^4-2x^2y^2+y^4

12345 [234]

<h3>Answer :</h3>

$\bf (x+y)^2(x-y)^2$

==============

<h3>Solving Steps:</h3>

===============

$\rightarrow \sf x^4-2x^2y^2+y^4$

breakdown

$\rightarrow \sf x^4-x^2y^2-x^2y^2+y^4$

factor common term

$\rightarrow \sf x^2(x^2-y^2)-y^2(x^2-y^2)$

collect in groups

$\rightarrow \sf (x^2-y^2)(x^2-y^2)$

apply difference between two square method: x² - y² = (x+ y)(x-y)

$\rightarrow \sf (x+y)(x-y) (x+y)(x-y)$

simplify

$\rightarrow \sf (x+y)^2(x-y)^2$

7 0

2 years ago

F(x)=-2(x-1)^2+2 a) standard form b) graph

OverLord2011 [107]

Answer:

a) f(x) = -2x² + 4x

Explanation:

To find the standard form we need to solve the equation as:

$\begin{gathered} f(x)=-2(x-1)^2+2 \\ f(x)=-2(x^2-2x+1)+2 \\ f(x)=-2x^2-2(-2x)-2(1)+2 \\ f(x)=-2x^2+4x-2+2 \\ f(x)=-2x^2+4x \end{gathered}$

So, the standard form is f(x) = -2x² + 4x

To make the graph we will use the initial form because when the equation is written like f(x) = a(x-h)²+k, the coordinate (h,k) is the vertex of the parabola

So, in this case, the vertex of the parabola is the point (1, 2)

On the other hand, we can find a point in the graph. For example, if x is equal to 0, then f(x) is equal to:

f(x) = -2(x-1)² + 2

f(0) = -2(0-1)² + 2

f(0) = -2(-1)² + 2

f(0) = - 2 + 2

f(0) = 0

So, the parabola passes through the point (0,0) and has a vertex in the point (1, 2). Then, the graph is:

4 0

2 years ago

A baseball is thrown into the air with an upward velocity of 30 ft/s. its initial height was 6 ft, and its maximum height is 20.

marissa [1.9K]

Check the picture below.

where is the -16t² coming from? that's Earth's gravity pull in feet.

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{30}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{6}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+30t+6 \\\\[-0.35em] ~\dotfill$

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+30}t\stackrel{\stackrel{c}{\downarrow }}{+6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)$

$\bf \left(-\cfrac{30}{2(-16)}~~,~~6-\cfrac{30^2}{4(-16)} \right)\implies \left( \cfrac{30}{32}~,~6+\cfrac{225}{16} \right)\implies \left(\cfrac{15}{16}~,~\cfrac{321}{16} \right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{\stackrel{\textit{how many}}{\textit{seconds it took}}}{0.9375}~~,~~\stackrel{\stackrel{\textit{how many feet}}{\textit{up it went}}}{20.0625})~\hfill$

7 0

4 years ago

Read 2 more answers

Graph the exponential function. y = 5(2)^x 1.A 2.B 3.D 4.C 5.D

svetlana [45]

EXPONENTIAL FUNCTIONS. Graph each exponential function. Write the asymptote in the blank. 1. f(x) = 2(3'). ~. II -= 0. 2. f(x) = 2,,·5+ 2 ..... c) $750 invested at 6.25% compounded <w!y for 5years. / .D~\g~6·~. A-:. 750 l 1+ "Ole$'). -::-10;).5, (D. •. 7. Write the function and find the principal needed to obtain ...

4 0

4 years ago