(Please help!) Solve the following system by substitution:
Y=x^2+4x+6
Y= -x+2
Thank you!

The translated point for P(6, 0) under the translation (x - 6, y - 1) is (0, -1).

let's work through this step by step.

First, let's understand the translation given: (x - 6, y - 1). This means that for any point (x, y) in the original coordinate system, the corresponding point in the translated system will be (x - 6, y - 1).

Now, let's find the translated point for P(6, 0). Using the translation rule, we substitute x = 6 and y = 0 into the formula (x - 6, y - 1):

Translated point = (6 - 6, 0 - 1)

Translated point = (0, -1)

So, the translated point for P(6, 0) under the translation (x - 6, y - 1) is (0, -1).

To solve this problem, we first understand the translation rule given. In this case, the translation rule is (x - 6, y - 1), which means we subtract 6 from the x-coordinate and subtract 1 from the y-coordinate to find the new coordinates after translation.

Next, we substitute the coordinates of point P(6, 0) into the translation rule. For x, we have 6, and for y, we have 0. Plugging these values into the rule gives us (6 - 6, 0 - 1), which simplifies to (0, -1).

Therefore, the translated point for P(6, 0) under the translation (x - 6, y - 1) is (0, -1).

complete question

P(6,0) under the translation (x -6, y -1) ?

In Einstein's theory of general relativity, gravity is treated as a phenomenon resulting from the curvature of spacetime. This curvature is caused by the presence of mass. Generally, the more mass that is contained within a given volume of space, the greater the curvature of spacetime will be at the boundary of its volume. As objects with mass move around in spacetime, the curvature changes to reflect the changed locations of those objects. In certain circumstances, accelerating objects generate changes in this curvature, which propagate outwards at the speed of light in a wave-like manner. These propagating phenomena are known as gravitational waves.

As a gravitational wave passes an observer, that observer will find spacetime distorted by the effects of strain. Distances between objects increase and decrease rhythmically as the wave passes, at a frequency corresponding to that of the wave. This occurs despite such free objects never being subjected to an unbalanced force. The magnitude of this effect decreases proportional to the inverse distance from the source. Inspiraling binary neutron stars are predicted to be a powerful source of gravitational waves as they coalesce, due to the very large acceleration of their masses as they orbit close to one another. However, due to the astronomical distances to these sources, the effects when measured on Earth are predicted to be very small, having strains of less than 1 part in 1020. Scientists have demonstrated the existence of these waves with ever more sensitive detectors. The most sensitive detector accomplished the task possessing a sensitivity measurement of about one part in 5×1022 (as of 2012) provided by the LIGO and VIRGO observatories.[20] A space based observatory, the Laser Interferometer Space Antenna, is currently under development by ESA.

Gravitational waves can penetrate regions of space that electromagnetic waves cannot. They are able to allow the observation of the merger of black holes and possibly other exotic objects in the distant Universe. Such systems cannot be observed with more traditional means such as optical telescopes or radio telescopes, and so gravitational-wave astronomy gives new insights into the working of the Universe. In particular, gravitational waves could be of interest to cosmologists as they offer a possible way of observing the very early Universe. This is not possible with conventional astronomy, since before recombination the Universe was opaque to electromagnetic radiation. Precise measurements of gravitational waves will also allow scientists to test more thoroughly the general theory of relativity.

In principle, gravitational waves could exist at any frequency. However, very low frequency waves would be impossible to detect and there is no credible source for detectable waves of very high frequency. Stephen Hawking and Werner Israel list different frequency bands for gravitational waves that could plausibly be detected, ranging from 10−7 Hz up to 1011 Hz

Hope this helped...

Now my fingers are tired.

Answer:

[tex]\frac{1}{6}[/tex]

Step-by-step explanation:

Let d be the distance from Bill to the center of the Earth and f be the gravitational force that the Earth exerts on him. Since f is inversely proportional to [tex]d^{2}[/tex], [tex]f\cdot d^2=k[/tex] for some constant k. Since the force when Bill is on the surface of Earth is 600 Newtons, [tex]k=600\cdot4000^2=960,000,000[/tex]. Therefore, if we let x be the force that the Earth acts on Bill with when he is on the Moon, [tex]x\cdot240,\!000^2=960,\!000,\!000[/tex] [tex]$x=\boxed{\dfrac{1}{6}}$[/tex]

Final answer:

Lena's garden area is calculated to be 792 square feet. She needs approximately 2.64 bags of topsoil but will purchase 3 bags since she cannot buy a fraction of a bag. The total cost for the topsoil will be $9.

Explanation:

To calculate the total cost of the topsoil Lena needs for her garden, we first need to find out the area of her garden. We do this by multiplying the garden's length by its width. The formula for area is length × width.

The garden is 36 feet long and 22 feet wide. So the area is:

Next, we calculate how many bags of topsoil Lena will require. Each bag covers 300 square feet, so she will need:

Since Lena cannot purchase a fraction of a bag, she'll need to round up to the nearest whole number. Therefore, she will purchase 3 bags of topsoil.

Finally, we find the total cost by multiplying the number of bags by the cost per bag. The cost of one bag is $3, so the total cost is:

Therefore, Lena will spend $9 on topsoil for her garden.

Answer:

What is the answer to the question

Step-by-step explanation:

The rate in gram per cup is 4.5

A rate is a measure of the quantity or change in one variable with respect to another, often expressed as a ratio.

Commonly used to describe speed, efficiency, or frequency, rates quantify how one quantity changes in relation to another, usually over time.

The rate in this scenario is the slope of the line.

slope = ∆y/∆x

∆y = change in y

∆x = change in x

slope = 18-0)/4-0

slope = 18/4

= 4.5

Therefore, the rate is 4.5 grams per cup

Answer:jgxgiddgidx

Step-by-step explanation:Hgxhfzjgxgdjgjggjdjgjgcy

Taking into account the definition of a system of linear equations, the amount of adult tickets sold was 125.

A group of two or more first-degree equations, whereby two or more unknowns are connected, is known as a system of linear equations.

Finding the value of each unknown is necessary to solve a system of equations and verify that all of the equations are satisfied. That is to say,  it is necessary to find the values of the unknowns since, when substituted, they must result in the suggested solution in both equations.

In this case, a system of linear equations must be proposed taking into account that:

You know:

The system of equations to be solved is

s= 2×a

s + a=375

There are a number of ways to solve an equation system; nevertheless, it is decided to use the substitution method, which involves clearing one of the two variables from one of the equations in the system and replacing it with its value in the other equation.

In this case, substituting the first equation in the second you get:

2×a + a= 375

Solving:

3×a= 375

a= 375÷ 3

a= 125

Finally, the amount of adult tickets sold was 125.

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Counting forwards by tens reveals a pattern where the tens digit increases and the units digit remains zero. Our decimal system is designed around powers of ten, making multiplication and division by ten intuitive by simply moving the decimal point right for multiplication and left for division.

When you count forward by tens, you can notice a pattern where the unit digit is always zero, and the digit in the tens place increases by one each time. For example, starting from 10, the sequence is 10, 20, 30, 40, 50, and so forth. This happens because each number is equivalent to a number of ten dollar bills, for example, 2 ten dollar bills make 20, 3 ten dollar bills make 30, etc.

Our numerical system is a base-10, or decimal system, influenced by the fact that humans typically have ten fingers, making it natural for us to count in tens. Each place value in a number is ten times greater than the place to its right. Therefore, when we multiply a number by ten, we effectively shift all digits one place to the left, adding a zero at the end. This is similar to when we multiply by powers of ten; for instance, 103 (or 10·10·10) is 1,000. The number of zeros in a power of ten also tells us how many places to move the decimal point to the right when multiplying.

Conversely, when we divide by powers of ten, we move the decimal point to the left by the number of zeros in that power of ten. For example, dividing 1,000 by 102 (or 100) moves the decimal two places to the left, giving us 10. This is an excellent example of the practical use of the base-10 system in arithmetic operations.

When counting forward by tens, each number is 10 greater than the previous number.

Divide the amount of each ingredient by 4.

Problem-solving is the act of defining a problem; figuring out the purpose of the trouble; identifying, prioritizing, and selecting alternatives for an answer; and imposing an answer.

Problem-solving starts with identifying the issue. As an example, a trainer may need to parent out a way to improve a scholar's overall performance on a writing talent test. To do that, the instructor will overview the writing exams looking for regions for improvement.

100 servings of apple dumplings.

The amount of each ingredient to make just 25 servings.

Divide the amount of each ingredient by 4.

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Answer:

need points ty

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savageluzti

12/14/2017

Mathematics

High School

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A bicyclist started riding at 8:00 A.M. The diagram below shows the distance the bicyclist had traveled at different times. What was the bicyclist's average rate of speed in miles per hour?

8:00 A.M. <--- 4.5 ---> 8:18 A.M. <------ 7.5 ------> 8:48 A.M.

The average rate of speed is ___ miles per hour.

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Average rate of speed is equal to total distance divided by the total time. the total distance is equal to 4.5+7.5, or 12 miles. The total time from 8:00 to 8:48 is 48/60 of one hour. 12*60/48=15 miles per hour

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Answer : The bicyclist's average rate of speed in miles per hour is 15 mile/hr.

Step-by-step explanation:

Average rate if speed : It is defined as the ratio total distance to total time.

Formula used :

As per question,

Total distance traveled by bicyclist = 4.5 mile + 7.5 mile = 12 mile

Total time taken by bicyclist = (8:48 A.M - 8:00 A.M) = 48 min = 0.8 hr

conversion used : ( 1 hour = 60 minute)

Now put all the given values in the above formula, we get the bicyclist's average rate of speed in miles per hour.

Therefore, the bicyclist's average rate of speed in miles per hour is 15 mile/hr.

Two expression with equal sign is an equation. To solve 2(x + 7) + 3x = 12, the first step in solving this equation for x  is 2x + 14 + 3x = 12. Option A is correct.

Two expression with equal sign is called as equation.

In given equation 2(x + 7) + 3x = 12 we need to solve the first step.

2(x + 7) + 3x = 12

For the first step, we need to multiply 2 with the expression in bracket,

So it will be

2x+14+3x=12.

Therefore option (A) is correct, For solveing 2(x + 7) + 3x = 12, 2x + 14 + 3x = 12  is the first step.

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The total number of apples on the tree before Bill and Sally picked any apples was [tex]\[{120}\][/tex]

Let's start by defining the total number of apples on the tree initially as [tex]\( x \).[/tex]

First, Bill picked [tex]\( \frac{1}{2} \)[/tex] of the apples. Therefore, after Bill finished picking, the number of apples left on the tree was:

[tex]\[x - \frac{x}{2} = \frac{x}{2}\][/tex]

Next, Sally picked [tex]\( \frac{1}{3} \)[/tex] of the remaining apples. After Bill picked, there were [tex]\( \frac{x}{2} \)[/tex] apples left. Thus, Sally picked:

[tex]\[\frac{1}{3} \times \frac{x}{2} = \frac{x}{6}\][/tex]

After Sally picked her apples, the number of apples left on the tree was:

[tex]\[\frac{x}{2} - \frac{x}{6}\][/tex]

To simplify this expression, find a common denominator (which is 6):

[tex]\[\frac{x}{2} - \frac{x}{6} = \frac{3x}{6} - \frac{x}{6} = \frac{2x}{6} = \frac{x}{3}\][/tex]

We are told that after Sally picked her apples, there were 40 apples left on the tree. Therefore:

[tex]\[\frac{x}{3} = 40\][/tex]

To find x, we multiply both sides by 3:

[tex]\[x = 40 \times 3 = 120\][/tex]

The values of a and b are 1/2 and 7/4 respectively. The sum of a and b is 9/4.

To find the values of a and b, we need to solve the system of equations formed by the given lines. Let's start by substituting the given coordinates (1, 2) into the equations:

x = (1/4)y + a -> 1 = (1/4)(2) + a -> 1 = 1/2 + a -> a = 1 - 1/2 = 1/2

y = (1/4)x + b -> 2 = (1/4)(1) + b -> 2 = 1/4 + b -> b = 2 - 1/4 = 7/4

Therefore, a = 1/2 and b = 7/4. To find the sum of a and b, we add them: a + b = 1/2 + 7/4 = 2/4 + 7/4 = 9/4

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To find the new mean after increasing each score by 2, we add 2 to the original mean of 84 to get a new mean of 86

The question concerns the calculation of the new mean when each score in a set is increased by a consistent amount. Given that the mean of 100 scores is 84, if we increase each score by 2, the new mean is also increased by 2. This is because the mean is the sum of all values divided by the number of values. Adding 2 to each score adds 200 to the total sum (since there are 100 scores), but the number of values remains the same. Thus, the new mean is 84 + 2, which equals 86

The solutions of -x^2 + 4x = x - 4 are -1 and 4

The equation is given as

-x^2 + 4x = x - 4

When split, it becomes

y = -x^2 + 4x

y  = x - 4

The graph represents the above equations.

So, the solution is the point of intersection of both equations

The equations intersect at x = -1 and x = 4

Hence, the solutions of -x^2 + 4x = x - 4 are -1 and 4

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137 bushes have red roses