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3 years ago

a figure in the second quadrant is reflected over the y axis.in which quadrant will reflected figure appear

2 answers:

7 0

So than this figure is in the second quadrant and reflected over the y axis than will appear in the first quadrant sure

hope helped

makvit [3.9K]3 years ago

4 0

Answer:

1st quadrant

Step-by-step explanation:

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Find an exponential function of the form y = ab* that passes through the points (0,4)

IrinaK [193]

Answer:

2nd option

Step-by-step explanation:

given an exponential function in the form y = a $b^{x}$

to find a and b use the coordinate points it passes through

using (0, 4 ) , then

4 = a $b^{0}$ [ $b^{0}$ = 1 ] , that is

a = 4 , so

y = 4 $b^{x}$

using (2, 100 )

100 = 4b² ( divide both sides by 4 )

25 = b² ( take the square root of both sides )

$\sqrt{25}$ = b , that is

b = 5

then exponential function is

y = 4 . $5^{x}$

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2 years ago

A mass weighing 3 lb stretches a spring 3 in. If the mass is pushed upward, contracting the spring a distance of 1 in and then s

Bas_tet [7]

Answer:

Step-by-step explanation:

$m=\frac{3}{32}\\ \gamma =0\text{ damping constant.}\\ k=\frac{3}{\frac{1}{4}}=12\\ F(t)=0\\ \text{so equation of motion is }\\ mu''+\gamma u'+ku=F(t)\\ \frac{3u''}{32}+12u=0\\ u''+128u=0\\ \text{and initial conditions are}\\ u(0)=-\frac{1}{12}\\ u'(0)=2\\ u''+128u=0\\ \text{characterstic equation is}\\ r^2+128=0\\ \text{roots are }\\ r=\pm 8\sqrt{2}i\\ \text{therefore general solution is}\\ u(t)=A\cos8\sqrt{2}t+Bsin8\sqrt{2}t\\ \text{using initial conditions we get}\\ A=-\frac{1}{12}\\ B=\frac{\sqrt{2}}{8}\\ \text{therefore solution is }\\ u(t)=-\frac{1}{12}\cos8\sqrt{2}t+\frac{\sqrt{2}}{8}\sin8\sqrt{2}t\\
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3 0

3 years ago

Laura invested $2000 in a three-year CD that pays 4% compounded annually.

Angelina_Jolie [31]

240 because 2000x0.04=80 X3= 240

5 0

3 years ago

Read 2 more answers

I really need help so please be right .Write the radical in simplest radical form and you must show all steps please. √(5/8)

Reptile [31]

Answer:

√5/6

Step-by-step explanation:

7 0

3 years ago

How do you find the area of the shaded region

Amiraneli [1.4K]

well, first off, let's notice that we have a trapezoid with a rectangle inside it, so the rectangle is really "using up" area that the trapezoid already has.

now, if we just get the area of the trapezoid, and then the area of the rectangle alone, and then subtract that area of the rectangle, the rectangle will in effect be making a hole inside the trapezoid's area, and what's leftover, is the shaded section, that part the hole is not touching.

$\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h&=height\\ a,b&=\stackrel{parallel~sides}{bases} \\\cline{1-2} h&=8\\ a&=16\\ b&=8 \end{cases}\implies A=\cfrac{8(16+8)}{2}\implies A=96 \\\\\\ \stackrel{\textit{area of the rectangle}}{(5\cdot 3)\implies 15}~\hfill \stackrel{~\hfill \textit{shaded area}}{\stackrel{\textit{trapezoid}}{96}~~ - ~~\stackrel{\textit{rectangle}}{15}~~ = ~~81}$

3 0

3 years ago