Point B is the image of point A when point A is rotated about the origin. What is known about point A and B?

On a coordinate plane, point A is (negative 3, 4) and point B is (4, 3).
Point B is the result of a 180° rotation.
Point A and B have the same x-coordinate.
Point B is the result of a 90° counterclockwise rotation.
Point A and point B are the same distance from the origin.

Answer:

12

Step-by-step explanation:

3x+7 = 12

Answer:

18

Step-by-step explanation:

Answer:

Step-by-step explanation:

Tangent from the point outside the circle are equal

WX  = XY

7x - 29 = 2x + 16 {Add 29 to both sides}

7x  = 2x + 16 +29

7x = 2x + 45 {Subtract 2x from both sides}

7x - 2x = 45

5x = 45     {Divide both sides by 5}

x = 45/5

x = 9

WX = 7x - 29

       = 7*9 - 29

        = 63 - 29

WX = 34

Answer:

x=50

Step-by-step explanation:

corresponding angles

Answer:

[tex]y=-2x+2[/tex]

Step-by-step explanation:

[tex]y=mx+b[/tex] is the slope intercept

[tex]m[/tex] is the slope

[tex]b[/tex] is the y-intercept

The slope in the graph is 4/2 or 2 simplified. From point (-2,6), you have to go down 4 and right 2 to get to the point (0,2)

The y-intercept is 2 because the line crosses the y axis at exactly 2.

Hope this helps

Answer:

184 m²

Step-by-step explanation:

Surface Area = 2(10*2) + 2(10*6) + 2(2*6)

Surface Area = 2(20 + 60 + 12)

Surface Area = 2(92)

Surface Area = 184 m²

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

Answer:

10 cm³

Step-by-step explanation:

The total surface area of a cube = 6 cm²

The formula for the surface area of the cube is :

A = 6a², a is the side of the cube

6 = 6a²

a = 1 cm

Now, the length of the formed cuboid = 2 cm

Breadth = 1 cm

Height = 1 cm

Required volume of the formed cubiod is :

V = 2(lb+bh+hl)

Put all the values,

V = 2(2(1)+1(1)+2(1))

V = 10 cm³

So, the required volume of the formed cuboid is equal to 10 cm³

Answer:

971 m to nearest metre

Step-by-step explanation:

training distance = 5*2*(260 + 480) m  

Diagonal distance = √(2602 + 4802) m

Heavy rain training distance = 5*(260 + 480 + √(2602 + 4802)) m

Difference between these:

5 × 2  × ( 260 + 480 ) - 5 ×( 260 + 480  + √ 260 ^ 2 + 480 ^ 2 )

   = 970 . 5311872087638744

difference = 971 m to nearest metre

Hope this answer helps you :)

Have a great day

Mark brainliest

Answer:

I'm sorry but this is very wordy and hard for people to understand.

please reconsider fixing it and submitting it again for assistance.

Answer:

I am not completly sure, cause the answer might be a bit more complex but, for the spinner on the left 4/8 and the right 1/8 chances

Step-by-step explanation:

Answer: 5/8 is the correct answer.

Step-by-step explanation: The other dude is completely wrong.

Answer:

6 weeks = 48 kg

12 weeks = 21 kg

Step-by-step explanation:

6 weeks:

4.5 x 6 = 27 kg

75 - 27 = 48 kg

12 weeks:

4.5 x 12 = 54 kg

75 - 54 = 21 kg

Hope this helps!

Answer:

angle 8

Step-by-step explanation:

not sure but if correct 1 heart pls and comment

Answer:           A is Correct

Step-by-step explanation:

angle 1 is opposite of angle 3 which means its also 105°

Answer:

Least:0.078

In between: 0.78

Greatest: 8.7

Step-by-step explanation:

Answer:

f(n) = 6n + 12

Step-by-step explanation:

There is a common difference in consecutive number of seats, that is

42 - 36 = 36 - 30 = 30 - 24 = 24 - 18 = 6

This indicates the sequence is arithmetic with nth term

[tex]a_{n}[/tex] = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 18 and d = 6 , then

f(n) = 18 + 6(n - 1) = 18 + 6n - 6 = 6n + 12

goodStep-by-step explanation:

Answer:

Area of four walls is 3960 square feet.

Step-by-step explanation:

length, L = 30 ft

Width, W = 50 ft

height, H = 6 ft

The area of walls is

= 2 (L W + W H + H L)

= 2 (30 x 50 + 50 x 6 + 6 x 30)

= 2 (1500 + 300 + 180)

= 3960 square feet

Answer/Step-by-step explanation:

How much tile would he need to cover the floor?

Answer: 1500 Ft^2

If the height of the basement is 6 ft, how much wall area is there to paint?

Answer: 960 Ft^2

Answer:

x = 8/k

Step-by-step explanation:

kx - 3 = 5

Add 3 to each side

kx -3+3= 5+3

kx = 8

Divide by k

kx/k = 8/k

x = 8/k

Answer:

x = 8 / k

Step-by-step explanation:

Let's solve for x.

kx − 3 = 5

Step 1: Add 3 to both sides.

kx −3 +3 = 5 + 3

kx = 8

Step 2: Divide both sides by k.

kx / k = 8k

x = 8 / k

Answer:

Step-by-step explanation:

I'll show you how to do the first one; the other are exactly the same, so pay attention.

The formula for arc length is

[tex]AL=\frac{\theta}{360}*2\pi r[/tex] where θ is the central angle's measure. It just so happens that the measure of the central angle is the same as the measure of the arc it intercepts. Our arc shows a measure of 40°; this measure is NOT the same as the length. Measures are in degrees while length is in inches, or cm, or meters, etc. Going off that info, our central angle measures 40°. Filling in the formula and using 3.1415 for π:

[tex]AL=\frac{40}{360}*2(3.1415)(3)[/tex]. I'm going to reduce that fraction a bit (and I'll use the same reduction in the Area of a sector coming up next):

[tex]AL=\frac{1}{9}*2(3.1415)(3)[/tex] which makes

AL = 2.09 units. Now for Area of the Sector. The formula is almost identical, but instead uses the idea that the area of a circle is πr²:

[tex]A_s=\frac{\theta}{360}*\pi r^2[/tex] where θ is, again, the measure of the central angle (which is the same as the measure of the arc it intercepts). Filling in:

[tex]A_s=\frac{1}{9}*(3.1415)(3)^2[/tex] which simplifies a bit to

[tex]A_s=\frac{1}{9}*(3.1415)(9)[/tex]. As you can see, the 9's cancel each other out, leaving you with

[tex]A_s=3.14[/tex] units²

Answer:

Step-by-step explanation:

[tex]Arc \ Length = 2 \pi r \frac{ \theta}{360}[/tex]

[tex]Sector \ Area = \pi r^2 \frac{ \theta }{360}[/tex]

[tex]1) Arc \ Length = 2 \pi \times 3 \times \frac{40}{360} = 2\pi \times 3 \times \frac{1}{9} = \frac{2}{3} \pi = 2.09[/tex]

  [tex]Sector \ Area = \pi \times 9 \times \frac{40}{360} = 9\pi \times \frac{1}{9} = \pi = 3.14[/tex]

[tex]2) Arc \ Length = 2 \pi \times 5 \times \frac{88}{360}= 7.68[/tex]

  [tex]Sector \ Area = \pi \times 25 \times \frac{88}{360} = 19.2[/tex]

[tex]3) Arc \ Length = 2 \pi \times 6 \times \frac{260}{360} = 27.22[/tex]

  [tex]Sector \ Area = \pi \times 36 \times \frac{260}{360} = 81.68[/tex]

Answer:

a. (3.83952, 5.16049)

b. (7.32, 8.88)

c. The difference in the range of the confidence intervals suggest that taking the zinc loz-enges reduces the overall duration of cold symptoms

d. The assumptions required for the validity validity of a t-test include that the data is sampled from a source that have data that are normally distributed

Step-by-step explanation:

The given parameters are;

The number of participants that took zinc lozenges, n₁ = 25

The number of participants that took a placebo, n₂ = 23

The mean duration of the symptoms for the zinc lozenge group, [tex]\overline x_1[/tex] = 4.5 days

The standard deviation of overall duration, s₁ = 1.6 days

The mean duration of the symptoms for the placebo group, [tex]\overline x_2[/tex] = 8.1 days

The standard deviation, s₂ = 1.8 days

a. The 95% confidence interval if everyone in the population were to take zinc lozenges is given as follows;

[tex]CI=\bar{x}_1\pm t_{\alpha/2} \times \dfrac{s_1}{\sqrt{n_1}}[/tex]

n₁ = 25, the degrees of freedom (df) = n₁ - 1 = 24

The t-value for 95% confidence interval, with df = 24, with t = 2.064

Therefore, we have;

[tex]CI_{zl}=4.5 \pm 2.064 \times \dfrac{1.6}{\sqrt{25}}[/tex]

[tex]CI_{zl}[/tex] = 4.5 ± 0.66048

[tex]CI_{zl}[/tex] = (3.83952, 5.16049)

b. The 95% confidence interval if everyone in the population were to take the placebo is given as follows;

[tex]CI_p=\bar{x_2}\pm t_{\alpha/2} \times \dfrac{s_2}{\sqrt{n_2}}[/tex]

n₂ = 23, the degrees of freedom (df) = n₂ - 1 = 22

The t-value for 95% confidence interval, with df = 22, with t = 2.074

Therefore, we have;

[tex]CI_p =8.1 \pm 2.074 \times \dfrac{1.8}{\sqrt{23}}[/tex]

[tex]CI_p[/tex] = 8.1 ± 0.7784

[tex]CI_p[/tex] = (7.32, 8.88)

c. Based on the difference in the range of the 95% confidence interval for the mean duration of cold symptoms of the group that take zinc lozenges, (3.83952, 5.16049), and the mean of the group that are on a placebo, (7.32, 8.88), there is sufficient statistical evidence to suggest that there is a difference between the mean and taking zinc lozenges reduces the overall duration of cold symptoms more than if taking placebo

d. The assumptions required for validity when carrying out a t-test include;

i) The measurement scale is ordinal or continuous

ii) The sample is a simple random sample

iii) The data gives a normal distribution curve

iv) The sample size is reasonably large

v) The variance are homogeneous

Among these assumptions, the data must be randomly sampled from the population of interest and that the data variables follow a normal distribution

The common assumptions made when doing a t-test include those regarding the scale of measurement, random sampling, normality of data distribution, adequacy of sample size and equality of variance in standard deviation

Answer:

try 144

Step-by-step explanation:

m<E is basically just ot asking you to find the angle of E, i think :').

And since all angles equal to 180, 180-36=144

(hope ots right)

Answer:

Click the second one, I truly hope that worked! :)

Answer:

Step-by-step explanation:

multiply both sides by 8

do over and write g=8x5

solve 8x5=40

g=40

Answer:

180

Step-by-step explanation:

hope this answer be right :)

Answer:

-1456

-864

4

-11

-5,25

-1,93

Step-by-step explanation:

Answer:

Step-by-step explanation:

Rate of change is:

Answer:

The amplitude is 5

Step-by-step explanation:

The function [tex]y=asin(bx-c)+d[/tex] has an amplitude of [tex]|a|[/tex]. Therefore the amplitude of the function [tex]y=5sin(2x-1)+3[/tex] is [tex]|a|=|5|=5[/tex].

Answer:

[tex]\theta_1 \ n\ \theta_2 = 120, 240[/tex]

Step-by-step explanation:

The question is incomplete, as the angles of rotation are not stated.

However, I will list the angles less than 360 degrees that will carry the hexagon and the nonagon onto itself

We have:

[tex]Nonagon = 9\ sides[/tex]

[tex]Hexagon = 6\ sides[/tex]

Divide 360 degrees by the number of sides in each angle, then find the multiples.

Nonagon

[tex]\theta = \frac{360}{9} =40[/tex]

List the multiples of 40

[tex]\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320[/tex]

Hexagon

[tex]\theta = \frac{360}{6} =60[/tex]

List the multiples of 60

[tex]\theta_2 = 60, 120, 180, 240, 300[/tex]

List out the common angles

[tex]\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320[/tex]

[tex]\theta_2 = 60, 120, 180, 240, 300[/tex]

[tex]\theta_1 \ n\ \theta_2 = 120, 240[/tex]

This means that, only a rotation of [tex]120, 240[/tex] will lift both shapes onto themselves, when applied to both shapes.

The other angles will only work on one of the shapes, but not both at the same time.