Find the length of the missing side.
14.5 m
7.3 m
X

Answer: 7500

Step-by-step explanation:

multiply 150 by 50

Answer:

$60

Step-by-step explanation:

$240 = 100%

Discount = 25%

100% - 25% = 75%

75% as a decimal is 0.75

Multiply:

240(0.75) = 180

You could also do 3/4 of 240 because 0.75 is 3/4

To find out how much she saved, let's subtract the sale price from the original price:

240 - 180 = 60

Ella's mom saved $60 from the discount

If my answer is incorrect, pls correct me!

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

-Chetan K

Answer:

B Center: (-4, 8) Radius = 4

Step-by-step explanation:

Answer:

$52.00

Step-by-step explanation:

35% of $80

(35÷100) ×80

0.35× 80

$28.00

$80 -$28

$52.00

Answer:

[tex]h(x) = (\frac{f}{g})(x)[/tex]

Step-by-step explanation:

Given

[tex]f(x) = 3x^3 + 9x^2 -12x[/tex]

[tex]g(x) = x - 1[/tex]

[tex]h(x) = 3x^2 + 12x[/tex]

Required

What defines h(x)

Looking at the degree of f(x), g(x) and h(x), we have:

[tex]h(x) = (\frac{f}{g})(x)[/tex]

See proof

[tex]h(x) = (\frac{f}{g})(x)[/tex]

This gives:

[tex]h(x) = \frac{f(x)}{g(x)}[/tex]

[tex]h(x) = \frac{3x^3 + 9x^2 -12x}{x - 1}[/tex]

Factorize

[tex]h(x) = \frac{(3x^2 + 12x)(x - 1)}{x - 1}\\[/tex]

[tex]h(x) = 3x^2 + 12x[/tex]

Answer: (b)

Step-by-step explanation:

Given

The function is given as

[tex]f(x)=\dfrac{x^2-12x+20}{x-2}[/tex]

Solving the function

[tex]f(x)=\dfrac{x^2-2x-10x+20}{x-2}\\\\f(x)=\dfrac{(x-2)(x-10)}{(x-2)}\\\\f(x)=x-10[/tex]

for [tex]x=2[/tex]

[tex]f(2)=2-10\\f(2)=-8[/tex]

The function is continuous at [tex]x=2[/tex] because [tex]\lim_{x \to 2} f(x)[/tex] exists.

If the limit exists at a point, then the function is continuous.  

Answer:

on edge its fs not b or c

Step-by-step explanation:

Answer:

The graph has a domain of {x|0 < x < infinite} and approaches 0 as x decreases

The graph has a domain of {y| -infinite < y < infinite} and decreases as x approaches 0.

Step-by-step explanation:

I was confused at first because I was like "There is 2 answers" and I reread the question again and found my mistake.

And sorry for the wait because "I am in class"

Answer: Hello the scatter plot related to your question is missing attached below is the scatter plot

answer :  The model fails to capture the relationship between the variables accurately, and there may exist nonlinear relationship ( C )

Step-by-step explanation:

The conclusion that can be drawn based upon the scatter chart is that The model fails to capture the relationship between the variables accurately, and there may exist nonlinear relationship

A scatter plot helps in observing the relationship within different numeric variables but the scatter plot attached fails in the showing the actual relationship

Slope : -2

y - intercept : 3

Equation : y = -2x+3

Answer:

Step-by-step explanation:

Slope: -0.5

Y-Intercept: 3

y = -0.5x+3

Answer:

if you ant to find the area of the traporziord prism you have to multiply the height and with to find the area

Answer:

C) 56

Step-by-step explanation:

DEG and FEG are on a straight line forming 180 degree.

Set your formula up as

180 = (3x+31)+(2x-6)

180 = 3x +2x +31 - 6

180 - 31 + 6 = 5x

155 = 5x

155/5 = x

31 = x

Now substitute 31 in place of x

FEG = (2*31-6)

FEG = 56

Answer:

4/10 is the same as 40/100

there fore 4/10+40/100=40/100+40/100=

80/100

hope it helps :)

Answer:

80/100

4/5    as reduced fraction

Step-by-step explanation:

4/10 + 40/100

use common denominator of 100

40/100 + 40/100 = 80/100

As a reduced fraction

4/5

9514 1404 393

Answer:

  C.  7

Step-by-step explanation:

To find the value of 2Ω3 using the definition of AΩB, we are replacing A with 2, and B with 3. This seems to give us ...

  2Ω3 = 2² +3² -2·3 = 4 + 9 - 6 = 7

The value of 2Ω3 is 7.

Answer:

Hence, the azimuth, bearing, and length of the line connecting station Shore to station Rock are [tex]270^{\circ},90^{\circ}[/tex] and [tex]650.94[/tex] feet.

Given :

The shore station point [tex](X_1,Y_1)[/tex] in feet,

        [tex](X_1,Y_1)=(2058.97, 6980.06)[/tex]

The Rock station point [tex](X_2,Y_2)[/tex] in feet

       [tex](X_2,Y_2)=(1408.03,6980.06)[/tex]

From the figure

     [tex]x=2058.97-1408.03=650.94[/tex] feet

    [tex]y=6980.06-6980.06=0\\[/tex] feet

Length of the line [tex]L=\sqrt{x^2+y^2}[/tex]

                           [tex]\Rightarrow L=\sqrt{(650.94)^2+0}=650.94[/tex]

                            [tex]\Rightarrow L=650.94[/tex] feet

[tex]\because[/tex] [tex]\tan \theta=\frac{y}{x}=\frac{0}{x}=0[/tex]

[tex]\Rightarrow \theta=\tan^{-1}(0)=0[/tex]

Azimuth of line[tex]=270^{\circ}+\theta[/tex]

                       [tex]=270^{\circ}+0[/tex]

                      [tex]=270^{\circ}[/tex]

[tex]\therefore[/tex] Bearing [tex]=360^{\circ}-\text{Azimuth}[/tex]

                [tex]=360^{\circ}-270^{\circ}[/tex]

                [tex]=90^{\circ}[/tex]

9514 1404 393

Answer:

  $2560.00

Step-by-step explanation:

The account balance is given by the formula ...

  A = Pe^(rt) . . . . principal P invested at rate r for t years

Filling in your numbers, we have ...

  $3287.11 = Pe^(0.05·5)

  P = $3287.11e^(-0.25) = $2560.00

The initial amount was $2560.00.

Answer:

d. 63%

Step-by-step explanation:

percent, will be white?

A41% b50% c59% d63%

The probability of white = P (w) = 41/65= 0.63

The  probability of yellow = P (y)= 24/65= 0.369=0.37

The probability of choosing white is 0.63 . When rounded to  nearest percent gives

0.63*100/100

=0.63*100 percent

= 63 percent

= 63%

the probability of getting   the next marble white is the same as the probability of getting a white.

Answer:

A.) The aircraft rose quickly into the air at takeoff, and then continued at a constant altitude.:)

Answer:

Omg, thank u so much I am on this question rn on Edulatic and I have 100 Q's and I'm only on Q 45, This Question rly helped me a lot bc it came with the answer. God Bless you!!

Answer:

look at the picture i have sent

Answer:

The cities are 35 cm apart in map.

The scale on a map is 5 cm : 8 km.

Step-by-step explanation:

mrk me brainliest please

Answer:

About 292.72

Step-by-step explanation:

We can imagine a circle circumscribed around the triangle, then solve it as we normally do using the apothem. By dividing 360 by the number of sides (3) we know that the central angle constructed by the apothem and radiuses would be 120. Divided by 2 (because we are using a right triangle to solve it) is 60. Now with this you can either use trig ratios (sin, cos, or tan) or use special right triangles (30-60-90 or 45-45-90). I used special rights. With this I know my apothem is [tex]\frac{13\sqrt{3} }{3}[/tex]. Perimeter is 26 times 3 which is 78. The formula for area of regular polygons is 1/2(perimeter times apothem). When we plug our values in the answer is about 292.72. You are lucky that I just had my test on this, I hope it helps.

Here is my work since this is very visual for me:

Answer:

the answer is 6.

Step-by-step explanation:

Answer:

6.

Step-by-step explanation:

3+3=6 = 2×3=6

you can do draw 3 circles 2 times and add it all together.

Answer:

1/20

Step-by-step explanation:

According to G0ogle 5 percent equals 1/20 in lowest terms.

Answer:

5% equals 5/100 which is 1/20 in lowest terms.

Step-by-step explanation:

5% is basically equivalent to 5/100. 5/100 in lowest terms is 1/20 since you divide the numerator and denominator by 5.

I hope this helps, have a nice day.

Answer:14

Step-by-step explanation:

Answer:

1/12

Step-by-step explanation:

To find experimental probability you have to solve the Number of times an event occurs / Total number of trials. So since the 5 only appears once and there are 12 numbers, it would be 1/12. Did this help?

The experimental probability of rolling 5 will be 0.18.

Its fundamental concept is that someone will nearly surely occur. The proportion of positive events in comparison to the total of occurrences. Then the probability is given as,

P = (Favorable event) / (Total event)

The table is given below:

Number      Total

    1                12

   2                16

   3                21

   4                23

   5                18

   6                10

The experimental probability of rolling a 5 is calculated as,

P = 18 / (12 + 16 + 21 + 23 + 18 + 10)

P = 18 / 100

P = 0.18

Thus, the probability is 0.18.

More about the probability link is given below.

https://brainly.com/question/795909

#SPJ2

Answer:

The probability that they will meet is 0.4375

Step-by-step explanation:

Let A and B be two friends who choose to meet  the same one-hour period.Since there are 60 minutes in an hour therefore n= 60

A can wait for 15 minutes

So the probability of A to meet is given by P(A)=  (15/60)= 1/4

Similarly

B can wait for 15 minutes

So the probability of B to meet is given by P(B)=  (15/60)= 1/4

The  probability  that A cannot meet is given by 1- P(A)=  1-1/4= 3/4

Similarly

The  probability  that B cannot meet is given by 1- P(B)=  1-1/4= 3/4

And

The  probability  that both  A  and B cannot meet is given by = 3/4*3/4= 9/16

So the probability that both will meet = 1- 9/16= 7/16= 0.4375

Answer:

its a rectangle

Step-by-step explanation:

Answer:

[tex]\frac{3}{2}[/tex]

Step-by-step explanation:

3 [tex]\frac{3}{4}[/tex] x [tex]\frac{2}{5}[/tex]

[tex]\frac{15}{4}[/tex] x [tex]\frac{2}{5}[/tex]

[tex]\frac{30}{20}[/tex] = [tex]\frac{3}{2}[/tex]

or another way

[tex]\frac{15}{4}[/tex] x [tex]\frac{2}{5}[/tex]

[tex]\frac{3}{2}[/tex] x [tex]\frac{1}{1}[/tex] = [tex]\frac{3}{2}[/tex]

Answer:

[tex]\frac{8x^{18} }{y^{2} }[/tex]

Step-by-step explanation:

Answer:

Hence, the [tex]y(t)=e^{-2 t}\left(\frac{75}{74} \cos 6 t+\frac{75}{148} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t))\end{array}[/tex] and approximately value of [tex]y(t)[/tex] is [tex]-0.844[/tex].

Given :

 [tex]my''+cy'+ky=F(t), y(0)=0, y'(0)=0,[/tex]

Where  [tex]m=2[/tex] kilograms

           [tex]c=8[/tex] kilograms per second

           [tex]k=80[/tex] Newtons per meter

          [tex]F(t)=20\sin (6t)[/tex] Newtons

Explanation :

(1)

Solve the initial value problem. [tex]y(t)[/tex]

   [tex]my''+cy'+ky=F(t), y(0)=0, y'(0)=0,[/tex]

[tex]\Rightarrow 2y''+8y'+80y=20\sin (6t)[/tex]

[tex]\Rightarrow y''+4y'+40y=10\sin (6t)[/tex]

Auxilary equations :[tex]F(t)=0[/tex]

[tex]\Rightarrow r^2+4r+40=0[/tex]

[tex]\Rightarrow r=\frac{-4\pm\sqrt{4^2-4\times 1\times 40}}{2\times 1}[/tex]

[tex]\Rightarrow r=\frac{-4\pm\sqrt{16-160}}{2}[/tex]

[tex]\Rightarrow r=\frac{-4\pm\sqrt{-144}}{2}[/tex]

[tex]\Rightarrow r=\frac{-4\pm12i}{2}[/tex]

[tex]\Rightarrow r=-2\pm6i[/tex]

The complementary solution is [tex]y_c=e^{-2t}\left(c_1\cos 6t+c_2\sin 6t\right)[/tex]

The particular Integral, [tex]y_p=\frac{1}{f(D)}F(t)[/tex]

[tex]y_{y} &=\frac{1}{D^{2}+4 D+40} 25 \sin (6 t) \\\\ y_{y} &=\frac{25}{-6^{2}+4 D+40} \sin (6 t) \quad\left(D^{2} \text { is replaced with }-6^{2}=-36\right) \\\\y_{y} &=\frac{25}{4 D+4} \sin (6 t) \\\\y_{y} &=\frac{25}{4(D+1)} \sin (6 t) \\\\y_{y} &=\frac{25(D-1)}{4(D+1)(D-1)} \sin (6 t) \\\\y_{y} &=\frac{25(D-1)}{4\left(D^{2}-1\right)} \sin (6 t) \\\\y_{y} &=\frac{25(D-1)}{4(-36-1)} \sin (6 t) \\\\y_{y} &=-\frac{25}{148}(D-1) \sin (6 t) \\y_{y} &=-\frac{25}{148}\left(\frac{d}{d t} \sin (6 t)-\sin (6[/tex]

Hence the general solution is :[tex]y=y_c+y_p=e^{-2t}(c_1\cos 6t+c_2\sin 6t)-\frac{25}{148}(6\cos 6t-\sin 6t)[/tex]

Now we use given initial condition.

[tex]y(t) &=e^{-2 t}\left(c_{1} \cos 6 t+c_{2} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\\y(0) &=e^{-\alpha 0)}\left(c_{1} \cos (0)+c_{2} \sin (0)\right)-\frac{25}{148}(6 \cos (0)-\sin (0)) \\\\0 &=\left(c_{1}\right)-\frac{25}{148}(6) \\\\c_{1} &=\frac{75}{74} \\\\y(t) &=e^{-2 t}\left(c_{1} \cos 6 t+c_{2} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\[/tex]

[tex]y^{\prime}(t)=-2 e^{-2 t}\left(c_{1} \cos 6 t+c_{2} \sin 6 t\right)+e^{-2 t}\left(-6 c_{1} \sin 6 t+6 c_{2} \cos 6 t\right)-\frac{25}{148}(-36 \sin (6 t)-6 \cos (6 t)) \\\\y^{\prime}(0)=-2 e^{0}\left(c_{1} \cos 0+c_{2} \sin 0\right)+e^{0}\left(-6 c_{1} \sin 0+6 c_{2} \cos 0\right)-\frac{25}{148}(-36 \sin 0-6 \cos 0) \\\\0=-2\left(c_{1}\right)+\left(6 c_{2}\right)-\frac{25}{148}(-6) \\\\0=-2 c_{1}+6 c_{2}+\frac{75}{74} \\\\0=-2\left(\frac{75}{74}\right)+6 c_{2}+\frac{75}{74} \\\\[/tex][tex]\begin{array}{l}0=-\frac{150}{74}+6 c_{2}+\frac{75}{74} \\\\\frac{150}{74}-\frac{75}{74}=6 c_{2}\end{array}[/tex]

[tex]\begin{array}{l}c_{2}=\frac{25}{148}\\\\\text { Substitute } c_{1} \text { and } c_{2} \text { in } y(t) \text { . Then }\\\\y(t)=e^{-2 t}\left(\frac{75}{74} \cos 6 t+\frac{75}{148} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t))\end{array}[/tex]

(2)

[tex]y(t)=e^{-2 t}\left(\frac{75}{74} \cos 6 t+\frac{75}{148} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\y(t)=\left(\frac{75}{74} e^{-2 t} \cos 6 t+\frac{75}{148} e^{-2 t} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\y(t)=\frac{75}{74}\left(e^{-2 t}-1\right) \cos 6 t+\frac{25}{148}\left(3 e^{-2 t}+1\right) \sin 6 t \\\\|y(t)| \leq \frac{75}{74} e^{-2 t}-1|\cos 6 t|+\frac{25}{148}\left|3 e^{-2 t}+1\right||\sin 6 t| \\\\[/tex]

[tex]|y(t)| \leq \frac{75}{74}\left|e^{-2 t}-1\right|+\frac{25}{148}\left|3 e^{-2 t}+1\right| \\\\\lim _{t \rightarrow \infty} y(t) \leq \lim _{t \rightarrow \infty}\left\{\frac{75}{74}\left|e^{-2 t}-1\right|+\frac{25}{148}\left|3 e^{-2 t}+1\right|\right\} \\\\\lim _{t \rightarrow \infty} y(t)=\left\{\frac{75}{74}\left|\lim _{t \rightarrow \infty}\left(e^{-2 t}-1\right)\right|+\frac{25}{148}\left|\lim _{t \rightarrow \infty}\left(3 e^{-2 t}+1\right)\right|\right\} \\[/tex]

[tex]\lim _{t \rightarrow \infty} y(t)=\left\{\frac{75}{74}(-1)+\frac{25}{148}(1)\right\}=-\frac{75}{74}+\frac{25}{148}=-\frac{-150+25}{148}=-\frac{125}{148} \approx-0.844[/tex]

Using the sine rule,

[tex] \frac{a}{sin(a)} = \frac{b}{sin(b)} = \frac{c}{sin(c)} [/tex]

Here we are going to use the values of A and C,

[tex] \frac{12}{sin(a)} = \frac{14}{sin(90)} \\ \frac{12}{sin(a)} = \frac{14}{1} \\ sin(a) = 12 \div 14 \\ sin(a) = 0.8571[/tex]

So sin(A) = 12/14 = 6/7 = 0.8571, but since the question says in its simplest radical form, I think the closest answer to it should be

[tex] \frac{ \sqrt{3} }{2} [/tex]