i need help with this one

Answer:

There would be 104 rooms on 4 floors in total.

Step-by-step explanation:

If there were 26 rooms on 2 floors, then you multiply 26 x 4, the result would be 104 rooms on 4 floors in total.

Answer:

52 rooms

Step-by-step explanation:

you could set up a proportion to solve this

26 rooms/ 2 floors = ?/4 floors

to find the answer for the question mark you can divide the amount of rooms by floors and get 13.

With the 13 you can multiply that by 4 to get 52 rooms. You basically used the 13 as the amount of increase

i hope this helped

if you have anymore questions feel free to ask again:3

Answer:

e): m=18.50d + 25

Step-by-step explanation:

Answer:I drew a quadratic function with a vertex and axis of symmetry, hope this helps

A vertical line going through any given point would pass through the x-coordinate.

In this case, the x-coordinate is 3. Therefore, the vertical line that would pass through the point (3, 5) is x = 3.

Hope this helps!

Answer:

  180 cm^2

Step-by-step explanation:

For similar figures the ratio of areas is the square of the ratio of linear dimensions.

We take "tripled in size" to mean that the linear dimensions of the larger rectangle are 3 times those of the smaller rectangle. That means the area of the larger rectangle will be 3² = 9 times the area of the smaller one.

  Area = (3²)(20 cm²) = 180 cm² . . . . area after tripled in size

Answer:

120 square feet     ( 13.333 square yards)

Step-by-step explanation:

8 ft   x   15 feet   = 120  ft^2    

Answer:

42 Pages

Step-by-step explanation:

7x5=35

7+35=42

Answer:

D&E, A&B

Step-by-step explanation:

It's pretty confusing but I know any of the answers with "c" in it is incorrect. The other options seem correct.

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

Recall that the projection of a vector [tex]u[/tex] onto [tex]v[/tex] is [tex]\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v[/tex].

Identify the vectors:

[tex]u=\langle-10,-7\rangle[/tex]

[tex]v=\langle-8,4\rangle[/tex]

Compute the dot product:

[tex]u\cdot v=(-10*-8)+(-7*4)=80+(-28)=52[/tex]

Find the square of the magnitude of vector v:

[tex]||v||^2=\sqrt{(-8)^2+(4)^2}^2=64+16=80[/tex]

Find the projection of vector u onto v:

[tex]\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v\\\\proj_vu=\biggr(\frac{52}{80}\biggr)\langle-8,4\rangle\\\\proj_vu=\biggr\langle\frac{-416}{80} ,\frac{208}{80}\biggr\rangle\\\\proj_vu=\biggr\langle\frac{-26}{5} ,\frac{13}{5}\biggr\rangle\\\\proj_vu=\langle-5.2,2.6\rangle[/tex]

Thus, B is the correct answer

Problem 2

Treat the football and wind as vectors:

Football: [tex]u=\langle42\cos172^\circ,42\sin172^\circ\rangle[/tex]

Wind: [tex]v=\langle13\cos345^\circ,13\sin345^\circ\rangle[/tex]

Add the vectors: [tex]u+v=\langle42\cos172^\circ+13\cos345^\circ,42\sin172^\circ+13\sin345^\circ\rangle\approx\langle-29.034,2.481\rangle[/tex]

Find the magnitude of the resultant vector:

[tex]||u+v||=\sqrt{(-29.034)^2+(2.481)^2}\approx29.14[/tex]

Find the direction of the resultant vector:

[tex]\displaystyle \theta=tan^{-1}\biggr(\frac{2.841}{-29.034}\biggr)\approx -5^\circ[/tex]

Because our resultant vector is in Quadrant II, the true direction angle is 6° clockwise from the negative axis. This means that our true direction angle is [tex]180^\circ-5^\circ=175^\circ[/tex]

Thus, C is the correct answer

Problem 3

We identify the initial point to be [tex]R(-2,12)[/tex] and the terminal point to be [tex]S(-7,6)[/tex]. The vector in component form can be found by subtracting the initial point from the terminal point:

[tex]v=\langle-7-(-2),6-12\rangle=\langle-7+2,-6\rangle=\langle-5,-6\rangle[/tex]

Next, we find the magnitude of the vector:

[tex]||v||=\sqrt{(-5)^2+(-6)^2}=\sqrt{25+36}=\sqrt{61}\approx7.81[/tex]

And finally, we find the direction of the vector:

[tex]\displaystyle \theta=tan^{-1}\biggr(\frac{6}{5}\biggr)\approx50.194^\circ[/tex]

Keep in mind that since our vector is in Quadrant III, our direction angle also needs to be in Quadrant III, so the true direction angle is [tex]180^\circ+50.194^\circ=230.194^\circ[/tex].

Thus, A is the correct answer

Problem 4

Add the vectors:

[tex]v_1+v_2=\langle-60,3\rangle+\langle4,14\rangle=\langle-60+4,3+14\rangle=\langle-56,17\rangle[/tex]

Determine the magnitude of the vector:

[tex]||v_1+v_2||=\sqrt{(-56)^2+(17)^2}=\sqrt{3136+289}=\sqrt{3425}\approx58.524[/tex]

Find the direction of the vector:

[tex]\displaystyle\theta=tan^{-1}\biggr(\frac{17}{-56} \biggr)\approx-17^\circ[/tex]

Because our vector is in Quadrant II, then the direction angle we found is a reference angle, telling us the true direction angle is 17° clockwise from the negative x-axis, so the true direction angle is [tex]180^\circ-17^\circ=163^\circ[/tex]

Thus, A is the correct answer

Problem 5

A vector in trigonometric form is represented as [tex]w=||w||(\cos\theta i+\sin\theta i)[/tex] where [tex]||w||[/tex] is the magnitude of vector [tex]w[/tex] and [tex]\theta[/tex] is the direction of vector [tex]w[/tex].

Magnitude: [tex]||w||=\sqrt{(-16)^2+(-63)^2}=\sqrt{256+3969}=\sqrt{4225}=65[/tex]

Direction: [tex]\displaystyle \theta=tan^{-1}\biggr(\frac{-63}{-16}\biggr)\approx75.75^\circ[/tex]

As our vector is in Quadrant III, our true direction angle will be 75.75° counterclockwise from the negative x-axis, so our true direction angle will be [tex]180^\circ+75.75^\circ=255.75^\circ[/tex].

This means that our vector in trigonometric form is [tex]w=65(\cos255.75^\circ i+\sin255.75^\circ j)[/tex]

Thus, C is the correct answer

Problem 6

Write the vectors in trigonometric form:

[tex]u=\langle40\cos30^\circ,40\sin30^\circ\rangle\\v=\langle50\cos140^\circ,50\sin140^\circ\rangle[/tex]

Add the vectors:

[tex]u+v=\langle40\cos30^\circ+50\cos140^\circ,40\sin30^\circ+50\sin140^\circ\rangle\approx\langle-3.661,52.139\rangle[/tex]

Find the magnitude of the resultant vector:

[tex]||u+v||=\sqrt{3.661^2+52.139^2}\approx52.268[/tex]

Find the direction of the resultant vector:

[tex]\displaystyle\theta=tan^{-1}\biggr(\frac{52.139}{-3.661} \biggr)\approx-86^\circ[/tex]

Because our resultant vector is in Quadrant II, then our true direction angle will be 86° clockwise from the negative x-axis. So, our true direction angle is [tex]180^\circ-86^\circ=94^\circ[/tex].

Thus, B is the correct answer

17.29162

[tex]\sqrt{18^2-5^2}[/tex]

__________________________________________

[tex]17.29162[/tex]

I hope this helps you

:)

Answer:

[tex]x=-7,-3[/tex]

Step-by-step explanation:

Completing the square means you are changing the polynomial so that it is a perfect square trinomial. Since the middle term is 10, we need the last term to be 25 since 5+5=10, thus, 5*5=25:

[tex]x^2+10x+21=0\\x^2+10x+21+4=0+4\\x^2+10x+25=4\\(x+5)^2=4\\x+5=\pm2\\x=-7,-3[/tex]

As you can see, because we change [tex]x^2+10x+21[/tex] to [tex]x^2+10x+25[/tex] by adding 4 to both sides, we were able to reduce it to [tex](x+5)^2[/tex], a perfect square, and solve the rest of the problem.

Answer:

28.26 ft squared

Step-by-step explanation:

Area is pi*r^2

3.14*(3)^2

3.14*9

28.26

Answer:

I Think the answer is B I'm sorry if I'm wrong

Answer:

from my side it's blurry

Margin of error = 0.8226401

Confidence interval = (9.17736,10.82264)

We have given data for one week recorded for 6 plants.

With a sample standard deviation of 1 inch and a sample mean of 10 inches.

For this sample, it is given that,

Sample size = n = 6

Sample SD = s = 1

Sample mean = x = 10

At a 0.10 significance level, the critical point is with t-distribution (n-1)df=5 df,

t*  =2.015048

Margin of error = E=2.015048*s/ √n = 0.8226401

Confidence interval

= (9.17736,10.82264)

Learn more about Margin of error here: https://brainly.com/question/10218601

#SPJ4

Step-by-step explanation:

[tex]see \: if \: 10 \: are \: left \: then \: we \: hve \: spent \\ 250 - 10 = 240 \: dollars \\ [/tex]

hence

[tex]six \: flasks \: cost \: 240 \: dollars \\ so \: if \: we \: divide \: 240 \: by \: 6 \: we \: get \: 40 \: \\ [/tex]

so one flask costs 40$

Answer:

choice for your help today 12 m is not answering me, sorry I think 25m

The equation that can be used to determine the number of bean bags in the instuctor's box is b = 17 x 4.

The number of bean bags in the instructor's box is 68.

Multplication is a mathematcial operation that can be used to determine the product of two or more numbers. The sign used to represent multiplication is ×.

To learn more about multiplication, please check: https://brainly.com/question/13814687

Answer:

30%

Step-by-step explanation:

The sign means that x can be equal to or LESS than -3. The only ones are -3 and -5. 0.17+0.13=0.30

Answer:

[tex]5x^{3} y^3 \sqrt{15y}[/tex] is the right answer and other are wrong

Answer:

a) 3

b) 1/3

Step-by-step explanation:

I've never used Brainly math before, but I can help you out with the work for these problems.

Part A

3(x + 1) - x = 15

3x + 3 - x = 15

4x + 3 = 15

4x = 12

x = 3

Part B

5(x + 2) - 2x = 11

5x + 10 - 2x = 11

3x + 10 = 11

3x = 1

x = 1/3

Hope this helps!

Answer:

0.25

Step-by-step explanation:

Every number has equal probability here

Answer:

Step-by-step explanation:

The usual English meanings of the words "reduction" and "enlargement" apply to geometrical figures.

Figure A'B'C'D' is smaller than figure ABCD, so represents a reduction.

__

Every coordinate of every image point is half as far from the origin in A'B'C'D' as in the original figure ABCD. For example, the x-coordinate of C' is 2, and the x-coordinae of C is 4. The dilation factor is the ratio of these values:

  k = Cx'/Cx = 2/4 = 1/2

The scale factor is 1/2.

Step-by-step explanation:

First find out the value of X and second find the value of Y and where is figure

Step-by-step explanation:

2. since there are 15 shirt

the probability of her wearing a blue shirt is 5/15

the probability of her wearing a blue shirt is 4/15

Answer:

A. evaluate progress

Step-by-step explanation:

common sense

Answer:

Depends on how many points the star has. For example, a 5 pointed star will have 5 lines of symmetry. Knowing this, you should be able to count using a diagram or drawing of the star