The larger of two numbers is four more than the smaller number. If the sum of the numbers is 74, find the numbers

Answer:

The answer is 6a+2b

Step-by-step explanation:

hope it's helpful :)

Answer:

21 inches is right i just took the e2020 assignment

Step-by-step explanation:

Answer:  The required solutions of the given equation are

[tex]x=\pm\sqrt7,~\pm\sqrt2i.[/tex]

Step-by-step explanation:  We are given to find the solutions of the following bi-quadratic equation :

[tex]x^4-5x^2-14=0~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

Let us consider that

[tex]x^2=y~~~~~\Rightarrow x^4=y^2.[/tex]

So, equation (i) becomes

[tex]y^2-5y-14=0,[/tex]

which is a quadratic equation in the variable y. The above equation in variable y can be solved using factorization as follows :

[tex]y^2-5y-14=0\\\\\Rightarrow y^2-7y+2y-14=0\\\\\Rightarrow y(y-7)+2(y-7)=0\\\\\Rightarrow (y-7)(y+2)=0\\\\\Rightarrow y-7=0,~~~y+2=0\\\\\Rightarrow y=7,-2.[/tex]

Therefore, we get

[tex]x^2=7\\\\\Rightarrow x=\pm\sqrt7[/tex]

and

[tex]x^2=-2\\\\\Rightarrow x=\pm\sqrt{-2}\\\\\Rightarrow x=\pm\sqrt2i,~~~~~~~~~~~~~[\textup{where }i=\sqrt{-1}][/tex]

Thus, the required solutions of the given equation are

[tex]x=\pm\sqrt7,~\pm\sqrt2i.[/tex]

Answer:

[tex]y-(-1)=-\frac{2}{7}(x-(-4))[/tex]

Step-by-step explanation:

We know that the line passes through (-4,-1) and (3,-3), where [tex](x_{1} ,y_{1} )[/tex] is [tex](-4,-1)[/tex] and [tex](x_{2} ,y_{2} )[/tex] is [tex](3,-3)[/tex].

First, we need to find the slope with the formula

[tex]m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

Replacing given points

[tex]m=\frac{-3-(-1)}{3-(-4)}=\frac{-3+1}{3+4}=\frac{-2}{7}[/tex]

Then, we use the slope, one point and the slope-point form

[tex]y-y_{1} =m(x-x_{1} )\\y-(-1)=-\frac{2}{7}(x-(-4))\\ y+1=-\frac{2}{7}x+\frac{8}{7} \\y=-\frac{2}{7}x+\frac{8}{7}-1\\y=-\frac{2}{7}x+\frac{8-7}{7} \\y=-\frac{2}{7}x+\frac{1}{7}[/tex]

The slope-intercept form is

[tex]y=-\frac{2}{7}x+\frac{1}{7}[/tex]

The point-slope form is

[tex]y-(-1)=-\frac{2}{7}(x-(-4))[/tex]

△JKL is congruent to △MNP because it can be mapped to △MNP through a reflection across the y-axis and a translation 2 units up.

To determine if △JKL is congruent to △MNP, we need to see if a series of rigid motions (rotations, reflections, translations) can map △JKL exactly onto △MNP. Looking at the positions of the vertices for both triangles, we can observe that if we carry out a reflection of △JKL across the y-axis (changing the sign of the x-coordinates) and then translate the reflected △JKL 2 units upward (+2 to the y-coordinates), we end up getting the vertices for △MNP. Therefore, △JKL is congruent to △MNP because △JKL can be mapped to △MNP by a reflection across the y-axis and a translation 2 units up. This option is not given explicitly in the original options, but this explanation provides the correct transformational sequence.

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Answer:

0.498 on edg

Step-by-step explanation:

Proving that PR = QS

Therefore, PR = QS

To learn more, visit the link below:

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x^2 +x^2 = C^2

x^2 +x^2 = 15^2

x^2 + x^2 = 225

225/2 = 112.5 = x^2

sqrt(112.5) = 10.6066

 round answer as needed

 note the answer can also be written as 15sqrt(2)/2

Answer:

Her centripetal acceleration during the turn at each end of the track is [tex]0.432\, \frac{m}{s^{2}}[/tex]

Step-by-step explanation:

Total distance covered in one round , D= 400 m

Time taken to cover one round , T = 1 min 40 s = 100 sec

Speed of runner , [tex]V=\frac{D}{T}=\frac{400}{100}\, \frac{m}{s}= 4.0\, \frac{m}{s}[/tex]

Now centripetal acceleration is given by

[tex]a_c=\frac{V^{2}}{R}[/tex]

where [tex]V= 4.0\frac{m}{s}[/tex]

[tex]Radius, R= \frac{Diameter}{2}=\frac{74}{2}m=37\, m[/tex]

[tex]\therefore a_c=\frac{4^{2}}{37}\, \frac{m}{s^{2}}=0.432\, \frac{m}{s^{2}}[/tex]

Thus her centripetal acceleration during the turn at each end of the track is [tex]0.432\, \frac{m}{s^{2}}[/tex]

The centripetal acceleration of runner at each end of track is [tex]\boxed{0.432{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}[/tex] .

Explanation:

It is given that the diameter of oval track that consists of two half circles at each end is [tex]74{\text{ m}}[/tex] .

The distance runner runs once around the track is [tex]400{\text{ m}}[/tex] . The runner covers one round of track in [tex]1{\text{ m}}[/tex]  and [tex]40{\text{ s}}[/tex] .

Our aim is obtain the value of centripetal acceleration for runner at each track.

The expression for centripetal acceleration is shown below.

[tex]{a_c}=\frac{{{v^2}}}{R}[/tex]                                                                          …… (1)

Here, [tex]v[/tex]  is the speed of the runner and [tex]R[/tex]  is the radius of track.

The speed of the runner is the ratio of distance she takes to cover the track to the time taken by her.

The time in seconds is calculated as,

[tex]\begin{aligned}t&=60{\text{ s}}+40{\text{ s}}\\&=100{\text{ s}}\\\end{aligned}[/tex]

The distance covered is [tex]400{\text{ m}}[/tex] , so the speed calculated is shown below.

[tex]\begin{aligned}{\text{speed}}&=\frac{{{\text{distance}}}}{{{\text{time}}}}\\&=\frac{{400{\text{ m}}}}{{100{\text{ s}}}}\\&=4{\text{ }}{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}[/tex]

The radius of track calculated is shown below.

[tex]\begin{aligned}R&=\frac{d}{2}\\&=\frac{{74{\text{ m}}}}{2}\\&=37{\text{ m}}\\\end{aligned}[/tex]

Substitute [tex]4[/tex]  for [tex]v[/tex]  and [tex]37[/tex]  for [tex]R[/tex]  in equation (1) to obtain the centripetal acceleration [tex]{a_c}[/tex] .

[tex]\begin{aligned}a_{c}&=\frac{(4\text{ m/s})^{2}}{37\text{ m}}\\&=\frac{16}{37}\text{ m/s}^{2}\\&=0.432\text{ m/s}^{2}\end{aligned}[/tex]

Therefore, the centripetal acceleration [tex]{a_c}[/tex]  is [tex]0.432{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}[/tex] .

Thus, the centripetal acceleration of runner at each end of track is [tex]\boxed{0.432{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}[/tex] .

Learn More:

1. Centripetal Force https://brainly.com/question/6372960

2. Velocity and Momentum https://brainly.com/question/11896510

3. Centripetal force on orbit https://brainly.com/question/7420923

Answer Details:

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords:

Track, running, diameter, circle, runner, distance, moves, constant, speed, centripetal, acceleration, force, motion, uniform, radius, velocity and oval.

Final answer:

The probability of obtaining exactly 2 heads when a fair coin is tossed 4 times is 3/8.

Explanation:

To find the probability of obtaining exactly 2 heads when a fair coin is tossed 4 times, we need to determine the number of ways to get 2 heads out of 4 tosses and divide it by the total number of possible outcomes.

Step 1: Determine the number of ways to get exactly 2 heads.

There are four coin tosses, and we want 2 of them to be heads. We can choose the positions for the 2 heads out of the 4 tosses in 4C2 = 6 ways.

Step 2: Determine the total number of possible outcomes.

Each coin toss has two possible outcomes: heads or tails. Since there are 4 tosses, the total number of outcomes is 24 = 16.

Step 3: Calculate the probability.

The probability of obtaining exactly 2 heads is 6/16 = 3/8.

multiply $9 by 80%

80% = 0.80

9 * 0.8 = 7.20

 subtract that from original price

9 - 7.20 = 1.80

 Sale price is $1.80