18. For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line
2. Is each pair of lines parallel, perpendicular, or neither?
18. Line 1: Passes through (0, 6) and (3, -24 Line 2: Passes through (-1, 19) and (8, -71)

Based on the figures above, the numbers of children that attended the park that day is 10,000.

Fist we have to use the determinant to solve for c and as such:

[tex]\frac{det.\frac{30000}{500000} \frac{1}{20} }{det. \frac{1}{10} \frac{1}{20} }[/tex]

=[tex]\frac{30000 x 20 - 500000 x 1}{1 x 20 - 1 x 10}[/tex]

= 100000/10

=10,000

Therefore, Based on the figures above, the numbers of children that attended the park that day is 10,000.

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Step-by-step explanation:

[tex] = 3 {x}^{2} + 10x + 3[/tex]

[tex] = 3 {x}^{2} + 9x + x + 3[/tex]

[tex] = 3x(x + 3) + 1.(x + 3)[/tex]

[tex] = (x + 3)(3x + 1)[/tex]

[tex] = (3x + 1)(x + 3)[/tex]

The answer is C.

Answer:

9.57 moles

Step-by-step explanation:

There is 1.1  mole per liter....

1.1 M/ l * 8.7 L = 9.57 moles

Answer:

Step-by-step explanation:

do a squared plus b squared = c squared like

2 and 5 thingy squared by 2 + 6 squared = c squared

unsquare c and you get your answer

The equivalent expression of x^3 (x^2 + 5x + 7) is x^5 + 4x^4 + x^4 + 8x^3 - x^3

The expression is given as:

x^3 (x^2 + 5x + 7)

Expand the expression

x^3 * x^2 + x^3 * 5x + x^3 * 7

Evaluate the product

x^5 + 5x^4 + 7x^3

Expand the expression

x^5 + 4x^4 + x^4 + 8x^3 - x^3

Hence, the equivalent expression of x^3 (x^2 + 5x + 7) is x^5 + 4x^4 + x^4 + 8x^3 - x^3

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The addition of the polynomials 14x² - 8x + 3 and - 6x² + 7x - 11 is 8x² - x - 8

(14x^2 - 8x + 3) + (-6x^2 + 7x - 11)

= 14x² - 8x + 3 - 6x² + 7x - 11

= 14x² - 6x² - 8x + 7x + 3 - 11

= 8x² - x - 8

Therefore, the addition of the polynomials 14x² - 8x + 3 and - 6x² + 7x - 11 is 8x² - x - 8

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Answer:

.5

Step-by-step explanation:

Point = P (1, 0, -2)

Origin= 0 (0, 0, 0)

The distance formula in 3D = √(n2 - n1)² + (y2 - y1)² + (Z2 - Z1)²

                                             = √(1 - 0)² + (0 - 0)² + (-2 - 0)²

                                             = √1 + 4

                                             = √5

As a special case of the distance formulation, think we need to recognize the distance of a point (x,y) to the beginning. in step with the gap formula, this is √(x−0)2+(y−0)2=√x2+y2. A point (x,y) is at a distance r from the starting place if and simplest if √x2+y2=r, or, if we square each aspect: x2+y2=r2.

Conclusion: The method to find the distance among the two factors is commonly given with the aid of d=√((x2 – x1)² + (y2 – y1)²). This formula is used to find the gap among any two factors on a coordinate aircraft or x-y plane.

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Answer:

0.00694444444444444444444444444444

Rounded to the nearest thousandths: 0.007

Step-by-step explanation:

Use calculator.

Hope this helps!

If not, I am sorry.

The value of x is 184.

What is the value of x?

Given Information

The measure of angle 2 = 92°

The measure of angle 4 = (x/2)°

From the figure, it can be seen that angle 2 and 4 make vertically opposite angles.

Vertically Opposite Angles

The pair of angles that are formed by two intersecting lines and are opposite to each other are defined as vertically opposite angles. The values of vertically opposite angles are always equal.

Here, in this case, ∠1 and ∠3 make a pair of vertically opposite angles. ∠2 and ∠4 form another pair of vertically opposite angles. Hence,

∠4 = ∠2

(x/2) = 92

x = 92 × 2

x = 184

Thus, the value of x is 184.

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Answer:

184

Step-by-step explanation:

edge

Answer:

I think that the awnser to 59 is 24x⁴-15x²+27x

Answer:

152 m²

Step-by-step explanation:

Area of flower garden

Assuming the flower garden is rectangular and its dimensions are measured in meters:

[tex]\begin{aligned}\textsf{Area of a rectangle} & = \sf width \times length\\\implies \textsf{Area of flower garden} & = \sf 14 \times 20\\& = \sf 280\:\:m^2 \end{aligned}[/tex]

Total area of path and garden

If there is a 2 m wide path all around the garden, the dimensions of the total area will be the dimensions of the garden extended by 4 m:

[tex]\begin{aligned}\textsf{Area of a rectangle} & = \sf width \times length\\\implies \textsf{Total area} & = \sf 18 \times 24\\& = \sf 432\:\:m^2 \end{aligned}[/tex]

Area of path

To calculate the area of the path, subtract the area of the garden from the total area:

[tex]\begin{aligned}\implies \textsf{Area of path} & = \sf Total\:area-Area\:of\:garden\\& = \sf 432-280\\& = \sf 152\:\:m^2 \end{aligned}[/tex]

The equation of the circle is (x - 1)^2 +(y + 9)^2 = 144

The given parameters are:

Center, (a,b) = (1, -9)

Radius, r = 12

The circle equation is calculated as:

(x - a)^2 + (y- b)^2 = r^2

This gives

(x - 1)^2 +(y + 9)^2 = 12^2

Evaluate the exponent

(x - 1)^2 +(y + 9)^2 = 144

Hence, the equation of the circle is (x - 1)^2 +(y + 9)^2 = 144

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Answer:

x<47

Step-by-step explanation:

Take -11 to the side which 36 is. Once it crosses it stops being negative and becomes positive so the equation becomes x<36+11 which is x<47

Hence, The percentage of households in the town with one or more cars = 75.8%, a standard error to the estimate= 1.92%

Total sample households  = 500

Households in the town with one car = 172

Households in town with two or more cars  =207

Households in the town with one or more cars = 172 + 207

                                                                            = 379

The percentage of households in the town with one or more cars =

(379 / 500)*100 = 75.8%.

A standard error to the estimate  = 1.92%.

Hence, The percentage of households in the town with one or more cars = 75. , a standard error to the estimate= 1.92%

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Answer:

a: (x + 3)(x + 4)

Step-by-step explanation:

assuming x^2 + 7x + 12 is the equation you are factoring

you would first multiply a * c

a is 1

b is 7

c is 12

1 * 12 = 12

then you take b and find what adds up to b but multiplies to ac

what adds up to 7 but multiplies to 12

3 and 4

so

you split the equation up like this

x^2 + 3x + 4x + 12

then you factor

x(x + 3) + 4(x + 3)

because x + 3 and x + 3 are the same we can do this

(x + 4)(x + 3)

that is your equation

so a is your answer

it doesnt matter if you have the (x + 3) or (x + 4) in the front

-------------------------------------------------------------------------------------------------------------

Answer:  [tex]\textsf{Option A, (x + 3)(x + 4)}[/tex]

-------------------------------------------------------------------------------------------------------------

In order to factor the polynomial we need to first split the middle term so it add up to 7 but multiplies to 12.  We know that 3 and 4 can do that.  We would the factor by grouping which would give us our final answer.

[tex]x^2 + 7x + 12[/tex]

[tex]x^2 + 4x + 3x + 12[/tex]

[tex]x(x + 4) + 3(x + 4)[/tex]

[tex](x + 3)(x + 4)[/tex]

Answer: [tex]\textsf{Option A, (x + 3)(x + 4)}[/tex]

Step-by-step explanation:

So a quadratic equation in factored form is generally given in the form: [tex]a(x-b)(x+c)[/tex]. Sometimes x may have a coefficient but I'll just create a simple quadratic equation. In this form it's easy to see that the solutions to the equation are b and -c, because if you input them in as x, you'll get one of the factors equal to 0, and it'll make the y-value zero, thus it's a solution. So b and c can literally be anything. But for this example I'll chose 3 and 6 to keep it simple. Inputting these as b and c you'll get: [tex]a(x-3)(x+6)[/tex]. Now the value of a will determine the stretch/compression, but to keep it really simple I'll just say that a=1. So now all you have to do is expand the two factors. When you expand them you'll get it in the form: [tex](x-b)(x+c) = x^2+cx-bx-bc[/tex] by using the foil method. So if you expand the two factors in this example you'll get: [tex]x^2+6x-3x-18[/tex] which simplifies to [tex]x^2+3x-18[/tex]. So cool now we have our equation. Now it's time to solve it, you can either factor it (since you literally just had it in factored form), or use the quadratic formula I'll show both methods

Factoring:

   When you factor an equation in the form: [tex]ax^2+bx+c[/tex]. You usually multiply the coefficients a and c. Then you'll find factors of ac that add up to b. In this example ac is really just c because a is 1...

   So given the equation: [tex]x^2+3x-18[/tex] You want to look for factors of -18 that add up to 3. And whenever you have a negative number as AC you really want to look for factors that have a distance of b which in this case is 3. For example I would say that 9 and 3 have a distance of 6. And then after that depending on whether b is negative or positive. I can choose which factor will be negative and which will be positive. If b was 6 well then 9 would have to be positive and 3 would be negative, but if b was -6 then 9 would have to be negative and 3 would have to be positive.

 So in this example look for factors that have "distance" of 3. Well 6 and 3 have a "distance" of 6 so I know those 2 will be the factors. Since 3 is positive that means 6 will be positive, and 3 will be negative. so they add up to 3 and multiply to get -18

   AC factors that add up to 3 and multiply to -18: 6 and -3

   Now write it in factored form using the two factors:

   [tex](x-3)(x+6)[/tex]

  Now set up each factor equal to 0 and solve for x:

   [tex]x-3 = 0\\x=3[/tex]

   [tex]x+6=0\\x=-6[/tex]

   This gives you the two solutions: x=3 and x=-6

Quadratic Formula:

   The quadratic formula is usually used when a equation can't be easily factored. The quadratic formula gives both zeroes in the equation: [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]. Notice the plus or minus sign next to the square root? That's where the the two solutions will come from, or possibly one solution if it's at the vertex. or maybe even no real solutions if the quadratic never intersects the x-axis.

  Ok so let's identify  a, b, and c. We're given the equation: [tex]x^2+3x-18[/tex]. so a=1, b=3, and c=-18. Plugging these values into the equation you get:

   [tex]x=\frac{-(3)\pm\sqrt{(3)^2-4(1)(-18)}}{2(1)}\\[/tex]

   Simplifying it a bit gets you the equation:

   [tex]x=\frac{-3\pm\sqrt{81}}{2}[/tex]

   Simplifying the radical gives you:

   [tex]x=\frac{-3\pm9}{2}[/tex]

  Use the positive sign:

   [tex]x=\frac{-3+9}{2}\\x=\frac{6}{2}\\x=3[/tex]

   Use the negative sign:

   [tex]x=\frac{-3-9}{2}\\x=\frac{-12}{2}\\x=-6[/tex]

  This gives you the two solutions x=3, and x=-6
   

Answer:

70 km/hr

Step-by-step explanation:

You just have to add both of the average velocity as you are finding the combined average velocity of both vehicles

Answer:

7x-13y=-117

Step-by-step explanation:

I don't know how this question wants me to solve it, but I just used the slope intercept form. So first, start off with y = mx+b. Plug in the slope for m, so 7/13. Then plug in your points for x,y which is 0,9. You'll get 9 = b. So then you have y = 7/13x+9. So now you need to change into standard form. So, multiply by the denominator (13) and you'll get 13y = 7x+117. Then move the variable over and you'll get 13y - 7x = 117. But it doesn't fit any, so then just multiply by -1 and you'll get -13y + 7x = -117.  

Yes, it is true that Jimmy John was underfilling the lettuce bags with sample mean of 14.39 pounds and sample standard deviation of 0.539.

Given sample mean =14.39 pound, sample standard deviation=0.539 and stated weight=15 pounds.

We know that mean is the number which comes after dividing sum of numbers divided by numbers. Standard deviation shows how much dispersed data is. Firstly we observe through mean and because it is less than 15, it means Jimmy John's had underfilled in atleast one bag otherwise if he had not underfilled the mean may come greater than or equal to 15 pounds.

Now come along standard deviation, we can observe that standard deviation is also low means the points are not in such a way that one point lies at 15 and other one is 25. It also shows underfilling of lettuce bags.

Hence it is true that Jimmy John had underfilled the lettuce bags.

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Answer:

1. a+c is larger than b+d

2. No way to tell whether a+d or b+c is larger.

Step-by-step explanation:

1. Which is larger, a+c or b+d?

Let a, b, c, and d be any numbers such that [tex]a > b > c > d[/tex].

Specifically, note that [tex]a > b[/tex], and subtracting b from both sides of the inequality, observe that [tex]a-b > 0[/tex].

Similarly, [tex]c > d[/tex], and subtracting d from both sides of the inequality, observe that [tex]c-d > 0[/tex].

From this, add "a-b" (a positive number, as proven above) to both sides of the inequality.

[tex](a-b)+(c-d) > (a-b)+0[/tex]

Addition by zero (the additive identity) doesn't change anything, so the right side remains "a-b"...

[tex](a-b)+(c-d) > a-b[/tex]

... and "a-b" is positive...

[tex](a-b)+(c-d) > a-b > 0[/tex]

... so, by the transitive property of inequality...

[tex](a-b)+(c-d) > 0[/tex]

Recall that subtraction is addition by a negative number...
[tex]a+(-b)+c+(-d) > 0[/tex]

...and that addition is associative and commutative, so things can be added in any order, so the middle two terms on the left side can be rearranged...

[tex]a+c+(-b)+(-d) > 0[/tex]

Adding b + d to both sides of the inequality

[tex](a+c+(-b)+(-d))+(b+d) > 0+(b+d)[/tex]

... and simplifying

[tex]a+c > b+d[/tex]

So, a+c is larger than b+d.

2. Which is larger, a+d or b+c?

Consider the following two examples:

Example 1

Suppose a=10; b=3; c=2; d=1.

Note that [tex]a > b > c > d[/tex] ([tex]10 > 3 > 2 > 1[/tex]) and, also observe that [tex]a+d=(10)+(1)=11[/tex], and [tex]b+c=(3)+(2)=5[/tex], so a+d is larger than b+c.

Example 2

However, suppose a=10; b=9; c=8; d=1.

Note that [tex]a > b > c > d[/tex] ([tex]10 > 9 > 8 > 1[/tex]) but that [tex]a+d=(10)+(1)=11[/tex], and [tex]b+c=(9)+(8)=17[/tex], so a+d is smaller than b+c.

So, in one example, a+d is bigger, and in the other, a+d is smaller.  Therefore, there is no way to tell which of a+d or b+c is larger from only the given information.

Answer:

Step-by-step explanation:Rectangles have 4 sides--2 = lengths and 2 = widths, so perimeter = 2l + 2 w = 2(l + w).

l + w = (2x - 1) + (x + 3) = 3x + 2.  Then double with distributive property P = 2(3x + 2)

P = 6x + 4

You could also do distributive property twice:  2(2x - 1) + 2(x + 3) = 4x - 2 + 2x + 6 = 6x + 4.

Answer:

See below.  I assume that (x) = 8x2 - 7x + 3 is really (x) = 8x^2 - 7x + 3

Step-by-step explanation:

Substitute the value of x given in f(x) into the equation f(x) = 8x^2 - 7x + 3

For example, f(0) would be f(0) = 8(0)^2 - 7(0) + 3.  f(0) = 3

f(-2) would be f(-2) = 8(-2)^2 - 7(-2) + 3.  

                                =   8*4  + 14 +3

                                = 32 + 17        therefore      f(-2) = 49

x f(x)

-2 49

-1 18

0 3

1 4

2 21

Answer:

62 and 64

Step-by-step explanation:

Let x be the first number

Let y be the 2nd number

Given,

x + 2 = y (Consecutive even numbers)

x - y = -2 (rearranged) - Equation 1

x + y = 126 - Equation 2

Now we can solve for x and y to find the 2 numbers by using substitution method in solving simultaneous equations.

x = y-2 (rearranged equation 1)

Now we substitute equation 1 into equation 2.

y - 2 + y = 126

2y - 2 = 126

2y = 126 + 2

2y = 128

y = 128 / 2

= 64

Now we will substitute y into equation 1.

x - 64 = -2

x = -2 + 64

= 62

Therefore the 2 numbers are 62 and 64.

[tex]\Large\maltese\underline{\textsf{A. What is Asked}}[/tex]

Find two consecutive even numbers whose sum is 126.

[tex]\Large\maltese\underline{\textsf{B. This problem has been solved!}}[/tex]

For now, let the first number be x.

Let the second number be x+2. (consecutive even numbers are right next to each other, like 2 and 4)

These two integers add up to 126. This gives us an equation that we can solve in terms of x.

[tex]\bf{x+x+2=126}[/tex] | arrange the like terms

[tex]\bf{2x+2=126}[/tex] | subtract 2

[tex]\bf{2x=124}[/tex] | 2 was subtracted from BOTH sides

[tex]\bf{x=62}[/tex]

[tex]\cline{1-2}[/tex]

Now, the second integer is [tex]\bf{x+2}[/tex].

[tex]\bf{So\;let's\;put\;the\;first\;number\;into\;the\;formula\;x+2}[/tex].

[tex]\bf{62+2}[/tex] | add (mental arithmetic)

[tex]\bf{64}[/tex]

[tex]\cline{1-2}[/tex]

[tex]\bf{Result:}[/tex]

                  [tex]\bf{=Number\;1=62}\\\\=Number\;2=64[/tex]

[tex]\LARGE\boxed{\bf{aesthetic \not1 \theta l}}[/tex]

Answer:

The GCF is 2.

Step-by-step explanation:

In my opinion, one of the easier ways to find the greatest common factor of two numbers is to get the prime factors of both and multiply the common ones. The prime factorization of 8 is 2*2*2, and the prime factorization of 162 is 2*3*3 (you can make a factor tree for both of these numbers to verify this). The only common prime factor for both of these numbers is 2, which means the greatest common factor of 8 and 18 is 2.

Answer:

122

Step-by-step explanation:

Given information, line t is parallel to line s.

angle 1 and angle 2 are alternate angles and so their measure is equal to each other:

10x + 2 = 12x - 22 transfer like terms to the same side of the equation

2 + 22 = 12x - 10x

24 = 2x divide both sides by 2

12 = x to find the measure of angle 2 replace x with the value we found

12*12 - 22 = 122

The ordered pair that is not a solution of the system of inequalities is the one in option D (1, 7).

Here we have the system of inequalities:

y > x² - 4x-5

y < -x²  - 5x+6

If at least one of the inequalities is false for the given ordered pairs, then the ordered pair is not a solution for the system.

Then we only need to evaluate all the ordered pairs on the given inequalities and see when we have one that is false.

If you look at the pair D (1,  7) and we evaluate that on the second inequality, we get:

7 < (1)² - 5*1 + 6

7 < -2

This is false, then (1, 7) is not a solution of the sytem.

The correct option is D.

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The simplified expression of [tex](4^3)(\frac{4^4^0x^4}{3^2x^2y^2})[/tex] is [tex]\frac{4^7x^2}{3^2y^2}[/tex]

The expression is given as:

[tex](4^3)(\frac{4^4^0x^4}{3^2x^2y^2})[/tex]

Expressions raise to power 0 is 1.

So, we have:

[tex](4^3)(\frac{4^4x^4}{3^2x^2y^2})[/tex]

Apply the law of indices to expression with common base

[tex](\frac{4^{4+3}x^{4-2}}{3^2y^2})[/tex]

Evaluate the sum and difference

[tex](\frac{4^7x^2}{3^2y^2})[/tex]

Remove the bracket

[tex]\frac{4^7x^2}{3^2y^2}[/tex]

Hence, the simplified expression of [tex](4^3)(\frac{4^4^0x^4}{3^2x^2y^2})[/tex] is [tex]\frac{4^7x^2}{3^2y^2}[/tex]

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