Read the statement, and identify the expressions that are equivalent.

The combined area of the 3 windowpanes and frame shown below is 924 in.2. The frame is of uniform width. What is the side length

wariber [46]

Answer:

Option A. $x=10\ in$

Step-by-step explanation:

we know that

The combined area of the 3 windowpanes and frame is equal to

$(3x+12)(x+12)=924\\ \\3x^{2}+36x+ 12x+144=924\\ \\ 3x^{2}+48x-780=0$

using a graphing tool to solve the quadratic equation

The solution is $x=10\ in$

see the attached figure

5 0

3 years ago

Question: Find The Sum Of The Integers From -6 To 58 Of Ss. O 1,500 O 1,734 O 1,690 O 1,621

REY [17]

Answer:

The Sum Of The Integers From -6 To 58 is <u>1690.</u>

Step-by-step explanation:

Given,

$a=-6$

$T_n=58$

We have to find out the sum of integers from -6 To 58.

Firstly we will find out the total number of terms that is 'n'.

Here $a_1=-6\ and\ a_2=-5$

$\therefore d=a_2-a_1=-5-(-6)=-5+6=1$

Now we use the formula of A.P.

$T_n=a+(n-1)d$

On substituting the values, we get;

$58=-6+(n-1)1\\\\n-1=58+6\\\\n-1=64\\\\n=64+1=65$

So there are 65 terms in between -6 To 58.

That means we have to find the sum of 65 terms in between -6 To 58.

Now we use the formula of Sum of n_terms.

$S_n=\frac{n}{2}(2a-(n-1)d)$

On substituting the values, we get;

$S_{65}=\frac{65}{2}(2\times-6+(65-1)1)\\\\S_{65}=\frac{65}{2}(-12+64)\\\\S_{65}=\frac{65}{2}\times52\\\\S_{65}=65\times26=1690$

Hence The Sum Of The Integers From -6 To 58 is <u>1690.</u>

5 0

3 years ago

If the radius of the circle is 9, find the area of the shaded region. Round your

zmey [24]

Answer:

we dont know the shaded area...

Step-by-step explanation:

6 0

3 years ago

Khairul deposited $ 600 in a bank at the end of 2010 and another $ 400 in the same bank at the end of 2011. The bank offers simp

castortr0y [4]

Answer:

bat puro math pede history lng

5 0

3 years ago

1. A sine function has the following key features:

andrew11 [14]

Problem 1

See the attached image (figure 1)

16pi seems like a typo. I'm going to assume that it's a fraction and it is 1/(6pi)
f = 1/(6pi) = frequency
T = 1/f = 1/(1/(6pi)) = 6pi
Amplitude = 2
a = 2
b = 2pi/T = 2pi/(6pi) = 1/3
Midline: y = 3
d = 3

The function is
y = a*sin(bx-c)+d
y = 2*sin(1/3*x-0)+3
y = 2*sin(x/3)+3

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

Problem 2

See the attached image (figure 2)

T = 12 is the period
a = 4 is the amplitude
b = 2pi/T = 2pi/12 = pi/6
y = 1 is the midline so d = 1
The y intercept is (0,1) which is the midline, which indicates no phase shifts have occurred so c = 0

The function is
y = a*sin(bx-c)+d
y = 4*sin((pi/6)x-0)+1
y = 4*sin((pi/6)x)+1

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

Problem 3

See the attached image (figure 3)

Period = 4pi
T = 4pi
b = 2pi/T = 2pi/(4pi) = 1/2 = 0.5
Amplitude = 2
a = 2
Midline: y = 3
d = 3
y-intercept: (0,3)
The function is a reflection of its parent function over the x-axis, so 'a' is negative meaning a = -2 instead of a = 2

The function is
y = a*sin(bx-c)+d
y = -2*sin(0.5x-0)+3
y = -2*sin(0.5x)+3

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

Problem 4

See the attached image (figure 4)

a = 10 which is half of the distance between the highest and lowest points
T = 8 is the period
b = 2pi/T = 2pi/8 = pi/4
c = -pi/2 is the phase shift since its really a cosine graph
d = 0 is the midline

The function is
y = a*sin(bx-c)+d
y = 10*sin((pi/4)*x+(-pi/2))+0
y = 10*sin((pi/4)*x+pi/2)

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

Problem 5

See the attached image (figure 5)

a = 2 is the amplitude since it bobs up and down this distance from the midline
T = 8 seconds is the period (double that of the time it takes for it to go from the highest to the lowest point)
b = 2pi/T = 2pi/8 = pi/4
c = 0 is the phase shift as the buoy starts at normal depth of 20 meters
d = 20 is the midline

The function is
y = a*sin(bx-c)+d
y = 2*sin((pi/4)x-0)+20
y = 2*sin((pi/4)x)+20

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

8 0

3 years ago

Read 2 more answers