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3 years ago

12

carol had a birthday party at a local pizza restaurant the bill came 54.80 not including tip to be 15% how much whould that add

to the bill

1 answer:

luda_lava [24]3 years ago

5 0

It would be 63.02
I hope this helps :)

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turner's mom measured his room to see how long the wallpaper border needed to be. If two walls are 15 feet long and two walls ar

anyanavicka [17]

Answer:

54 feet

Step-by-step explanation:

The perimeter of the room is the sum of wall lengths:

2·(15 feet) + 2·(12 feet) = 30 feet + 24 feet = 54 feet

The border should be cut to 54 feet.

_____

<em>Comment on the question</em>

This seems to be a math problem designed to get you to figure the perimeter of the room. If you were actually installing wallpaper, you might want to cut the border in four (4) pieces, each slightly longer than the corresponding wall. That way any lack of squareness in the corners could be managed more easily, and the possibility of a gap at the seam could be eliminated.

4 0

3 years ago

A hot-air balloon is 160 ftabove the ground when a motorcycle (traveling in a straight line on a horizontal road) passes dire

Charra [1.4K]

Answer:

73.77 ft/s

Step-by-step explanation:

Let's imagine this situation as a right triangle where the distance between the two points is the hypotenuse.

-Applying Pythagoras theorem:

$z^2=x^2+y^2\\\\z=\sqrt{x^2+y^2}\\\\x=dx*t+x_i, x_i=0\\\\y=dy*t+y_i\\\\y_i=160\ ft$

#Take the derivative of the first equation and solve for $dz$ :

$d(z^2=x^2+y^2)\\\\2z*dz=2x*dx+2y*dy\\\\dz=\frac{x*dx+y*dy}{z}\\\\dz=\frac{x*dx+y*dy}{\sqrt{x^2+y^2}}\\\\\\dz=\frac{dx^2*t+dy(dy*t+160)}{\sqrt{(dx*t)^2+(dy*t)}}$

#We then substitute the values given in the question to solve for $dz$ :

$dz=\frac{7*66^2+14(14*7+160)}{\sqrt{(66*7)^2+7*14+160}}\\\\=73.78$

Hence, the rate of change of the distance between the motorcycle and the balloon 7 seconds later is 73.77 ft/s

3 0

3 years ago

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago

If tñ=2ñ+3, find S10

Allisa [31]

Answer:

$S_{10}=280$

Step-by-step explanation:

Given that,

$t_n=2n+3$ ...(1)

We need to find the value of $S_{10}$ .

Put n = 1 to find the first term.

$t_1=2(1)+3=5$

Put n = 2 to find the second term.

$t_2=2(2)+3=7$

Put n = 3 to find the third term.

$t_3=2(3)+3=9$

Put n = 10 to find the tenth term.

$t_{10}=2(10)+3=23$

It means we need to find the sum of 5,7,9,.....,23.

The formula for the sum of n terms is given by :

$S_=\dfrac{n}{2}(a+a_n)$

We have, n = 10, a = 5 and $a_n=23$

So,

$S=\dfrac{10}{2}(5+23)\\\\S_{10}=10\times 28\\\\=280$

So, the value of $S_{10}$ is equal to 280.

3 0

3 years ago

Estimate the perimeter and area of the shaded figure to the nearest tenth.

Maru [420]

Answer:

Perimeter = 18.7 units

Area = 13.5 units²

Step-by-step explanation:

Perimeter of ADEC = AD + DE + EC + AC

Length of AD = 3 units

By applying Pythagoras theorem in ΔDBE,

DE² = DB² + BE²

DE² = 3² + 3²

DE = √18

DE = 4.24 units

Length of EC = 3 units

By applying Pythagoras theorem in ΔABC,

AC² = AB² + BC²

AC² = 6² + 6²

AC = √72

AC = 8.49 units

Perimeter of ADEC = 3 + 4.24 + 3 + 8.49

= 18.73 units

≈ 18.7 units

Area of ADEC = Area of ΔABC - Area of ΔBDE

Area of ΔABC = $\frac{1}{2}(AB)(BC)$

= $\frac{1}{2}(6)(6)$

= 18 units²

Area of ΔBDE = $\frac{1}{2}(BD)(BE)$

= $\frac{1}{2}(3)(3)$

= 4.5 units²

Area of ADEC = 18 - 4.5

= 13.5 units²

6 0

3 years ago