HW
Due Tuesday, December 7, 2021
1. Tracy's is having a 25% off sale. Dana wants to buy a coat that has
an original price of $83. What is the sale price?

Answer:

6000 hope this helps

if the question is 5,422 then the round figure is 5000

but the question is 5,821 its above 5500 will be 6000

Answer:

The x-coordinate of this point would be 6.

Step-by-step explanation:

Answer:

w

Step-by-step explanation:

w

Answer:

12 x^5y^12

Step-by-step explanation:

3x^2y^5 * 4x^3y^7

Multiply the constants

3*4 = 12

Multiply the x terms

x^2 * x^3

We know a^b* a^c = a^(b+c)

x^(2+3) = x^5

Multiply the y terms

y^6 * y^7 = y&(5+7) = y^12

Put them back together

12 x^5y^12

Answer:

[tex]\frac{3}{4} =\frac{1}{4} (3)[/tex]

Step-by-step explanation:

like that?

Answer:

3/4

Step-by-step explanation:

Step 1:  Write an expression

A fourth is 1/4

3 fourths means 3 one forths

1/4 + 1/4 + 1/4

3/4

Answer: 3/4

Answer:

option a 5......

...

I hope it's correct

Answer: 101

Step-by-step explanation:

.

Answer:

x =5

Step-by-step explanation:

hope this helps you

please mark as brainliest

Answer:15

Step-by-step explanation:6x +2y=30

2(3x+y) =30

3x+y=30÷2

3x+y=15

Answer:

I believe the red one is the first image then the blue then the green because they show the prime sign

Step-by-step explanation:

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

[tex] \displaystyle \large{ cos \theta = - \frac{1}{2} }[/tex]

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

[tex] \displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }[/tex]

Find Q3

[tex] \displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }[/tex]

Both values are apart of the interval. Hence,

[tex] \displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }[/tex]

Second Question

Isolate sin(4 theta).

[tex] \displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }[/tex]

Rationalize the denominator.

[tex] \displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }[/tex]

The problem here is 4 beside theta. What we are going to do is to expand the interval.

[tex] \displaystyle \large{0 \leqslant \theta < 2 \pi}[/tex]

Multiply whole by 4.

[tex] \displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}[/tex]

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

[tex] \displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }[/tex]

Find Q4

[tex] \displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }[/tex]

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

[tex] \displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}[/tex]

Hence:-

For Q3

[tex] \displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }[/tex]

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

[tex] \displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }[/tex]

Therefore:-

[tex] \displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }[/tex]

Then we divide all these values by 4.

[tex] \displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }[/tex]

Let me know if you have any questions!

9514 1404 393

Answer:

  B) (4,–1)

Step-by-step explanation:

A graph can help you figure this out. The point (4, -1) is in the doubly-shaded area, so is a solution to this system.

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

Explanation:

Choice A can be ruled out because

[tex]y \le x+1\\\\5 \le 0+1\\\\5 \le 1\\\\[/tex]

which is false.

Choices C and D are similar, so they can be ruled out as well.

---------------------

Choice B is the answer because it makes both inequalities true.

If we plug the coordinates of (4,-1) into the first inequality, we get

[tex]y > -x\\\\-1 > -4[/tex]

which is a true statement because -1 is to the right of -4 on the number line.

Now try the other inequality

[tex]y \le x+1\\\\-1 \le 4+1\\\\-1 \le 5\\\\[/tex]

that's true also. So again, (4,-1) makes both inequalities true, and that's why choice B is the answer.

Solve all of the equations below and add all of the solutions to those equations to get your answer.

10x4x2=

8x4x2=

10x8=

(10x8)-(7x6)=

7x2x2=

6x2x2=

7x6=

Answer:

1217

Step-by-step explanation:

Given

[tex]Subjects = 2353[/tex]

[tex]Improvement = 1136[/tex]

Required

The number that didn't improve --- missing from the question.

Let the unimproved be represented as x.

So, we have:

[tex]x + Improvement = Subjects[/tex]

Substitute known values

[tex]x + 1136 = 2353[/tex]

Solve for x

[tex]x =- 1136 + 2353[/tex]

[tex]x =1217[/tex]

Answer:

[tex](a+b)^{2} =a^{2}+2ab+b^{2}[/tex]

[tex](1)w^{2}+2(3)(1)w+3^{2}\\\\=(w+3)^{2}\\\\=(w+3)(w+3)[/tex]

Therefore, [tex]w^{2} +6w+9[/tex] makes a perfect squared.

Answer:

I believe it is (C).

This runner’s time was slower than the mean time by 0.75 standard deviations.

ED2021

Answer:

B) This runner’s time was faster than the mean time by 1.08 standard deviations.

Step-by-step explanation:

The runner is FASTER because of the negative z score

Answer:

Step-by-step explanation:

This is the Law of Cosines. Use the following formula:

[tex]5.1^2=3.3^2+3.3^2-2(3.3)(3.3)cos(x)[/tex] and simplify to

26.01 = 10.89 + 10.89 - 21.78cos(x) and a bit more to

4.23 = -21.78cos(x) and finally to

-.194214876 = cos(x) and use the 2nd button along with the cos button to find that the missing angle is 101.2 degrees

Answer:

$6.60

Step-by-step explanation:

12% = 0.12

12% of $55 = 0.12 x 55

=6.6

So, $6.60

Sh363gdbzej3yve truly urban urge Heydyeh hdye

Work Shown:

30 min = 30/60 = 0.5 hours

distance = rate*time

rate = distance/time

rate = (2450 miles)/(0.5 hours)

rate = (2450/0.5) mph

rate = 4900 mph

For the sake of comparison, a typical commercial passenger jet can reach max speeds of about 600 mph.

Answer:

x = -4

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

-1/3x+3 = x-1

Step-by-step explanation:

The solution is (3,-2)

Check and see if the point solves the equation

-1/3x+3 = x-1

-1/3(3) +3 = 3-1

-1+3 = 3-1

2=2 yes

Answer:

C

Step-by-step explanation:

the answer is

10² ( ten square )

step by step :

10² = 10 × 10

= 100

Answer: 100

Step-by-step explanation: 10*10=100

Answer:

8000 and 12000 respectively

Step-by-step explanation:

Let the amount invested in first account be x and y be the amount invested in the second account.

ATQ, x+y=20000 and 1280=(7/100)*x+(6/100)*y

x=8000 and y=12000.

Answer:

3

Step-by-step explanation:

The dilation factor is 3

Answer:

-3/2 ±1/2 sqrt(5) = x

Step-by-step explanation:

f(x) = 2x^2 + 6x + 2

Set the function equal to 0

0 = 2x^2 + 6x + 2

Factor out 2

0 = 2(x^2 + 3x + 1)

Divide by 2

0 = (x^2 + 3x + 1)

Subtract 1 from each side

-1 = x^2 +3x

Complete the square by dividing the x coefficient by 2 and then squaring

(3/2)^2 = 9/4

-1 +9/4 = x^2 +3x+9/4

-4/4+9/4 = (x+3/2) ^2

5/4 = (x+3/2) ^2

Taking the square root of each side

± sqrt(5/4) = sqrt( (x+3/2) ^2)

±1/2 sqrt(5) =  (x+3/2)

Subtract 3/2 from each side

-3/2 ±1/2 sqrt(5) = x

Answer:

1/2

Step-by-step explanation:

2/5 + 1/10

Get a common denominator of 10

2/5 * 2/2 + 1/10

4/10 + 1/10

5/10

simplify

Divide the top and bottom by 5

1/2

Answer:

[tex] \frac{1}{2} [/tex]

Step-by-step explanation:

[tex] \frac{2}{5} + \frac{1}{10} \\ \frac{2 \times 2}{5 \times 2} + \frac{1}{10} \\ \frac{4}{10} + \frac{1}{10} \\ \frac{5}{10} \\ \frac{5 \div 5}{10 \div 5} \\ = \frac{1}{2} [/tex]

A well formatted table of the distribution is attached below :

Answer:

0.124

0.733

0.408

Step-by-step explanation:

Using the table Given :

1.) P(AP Statistics) = 90 / 725 = 0.124

2.) P(12th grade ; Precalculus or AP Statistics) = (100 + 65) / 225 = 165 /225 = 0.733

3.) P(Algebra 11 | 11th grade) = P(Algebara11 n 11th grade) / P(11th grade) = 100 / 245 = 0.408

The differential equation

[tex]y^{(4)}-n^2y'' = g(x)[/tex]

has characteristic equation

r ⁴ - n ² r ² = r ² (r ² - n ²) = r ² (r - n) (r + n) = 0

with roots r = 0 (multiplicity 2), r = -1, and r = 1, so the characteristic solution is

[tex]y_c=C_1+C_2x+C_3e^{-nx}+C_4e^{nx}[/tex]

For the non-homogeneous equation, reduce the order by substituting u(x) = y''(x), so that u''(x) is the 4th derivative of y, and

[tex]u''-n^2u = g(x)[/tex]

Solve for u by using the method of variation of parameters. Note that the characteristic equation now only admits the two exponential solutions found earlier; I denote them by u₁ and u₂. Now we look for a particular solution of the form

[tex]u_p = u_1z_1 + u_2z_2[/tex]

where

[tex]\displaystyle z_1(x) = -\int\frac{u_2(x)g(x)}{W(u_1(x),u_2(x))}\,\mathrm dx[/tex]

[tex]\displaystyle z_2(x) = \int\frac{u_1(x)g(x)}{W(u_1(x),u_2(x))}\,\mathrm dx[/tex]

where W (u₁, u₂) is the Wronskian of u₁ and u₂. We have

[tex]W(u_1(x),u_2(x)) = \begin{vmatrix}e^{-nx}&e^{nx}\\-ne^{-nx}&ne^{nx}\end{vmatrix} = 2n[/tex]

and so

[tex]\displaystyle z_1(x) = -\frac1{2n}\int e^{nx}g(x)\,\mathrm dx[/tex]

[tex]\displaystyle z_2(x) = \frac1{2n}\int e^{-nx}g(x)\,\mathrm dx[/tex]

So we have

[tex]\displaystyle u_p = -\frac1{2n}e^{-nx}\int_0^x e^{n\xi}g(\xi)\,\mathrm d\xi + \frac1{2n}e^{nx}\int_0^xe^{-n\xi}g(\xi)\,\mathrm d\xi[/tex]

and hence

[tex]u(x)=C_1e^{-nx}+C_2e^{nx}+u_p(x)[/tex]

Finally, integrate both sides twice to solve for y :

[tex]\displaystyle y(x)=C_1+C_2x+C_3e^{-nx}+C_4e^{nx}+\int_0^x\int_0^\omega u_p(\xi)\,\mathrm d\xi\,\mathrm d\omega[/tex]