Having trouble can someone help me solve this

Answer:

x ≤ 135/2

Step-by-step explanation:

x is the cost per night. You have to pay for 2 nights, and you do not want to spend more than $135 total. Another way to say that, is you want to spend less than or equal to $135.

x = cost per night

2 = number of nights

2x = total cost

Total cost must be less than or equal to $135.

2x ≤ 135

Divide each side by 2.

x ≤ 135/2

Answer:A or C

Step-by-step explanation: i guessed plus 19 hours ago

Answer:

[tex]\huge\boxed{\sf \{(-2,1)\}}[/tex]

Step-by-step explanation:

Given equations:

y = x + 3

y = -x - 1

By comparing the above equations, we get:

x + 3 = -x - 1

Add x to both sides

Subtract 3 to both sides

2x = -1 - 3

2x = -4

Divide 2 to both sides

x = -4/2

Substitute x = -2 in first equation.

y = x + 3

y = -2 + 3

So,

[tex]\rule[225]{225}{2}[/tex]

Answer:

[tex]x = - 2 \\ y = 1[/tex]

Step-by-step explanation:

The above are Simultaneous equations.

Simultaneous equations are two or more algebraic equations which share same variables like X,Y or X,Y and Z.

They are called simultaneous equations because they can be solved the same time.

From the question,

[tex]y = x + 3 - - - - - - (1) \\ y = - x - 1 - - - - - - (2) \\ Substitute \: equation \: (2)into \: (1) \\ - x - 1 = x + 3 \\ Collect \: liketerms \\ - x - x = 3 + 1 \\ - 2x = 4 \\ Divide \: bothsides \: by \: - 2 \\ \frac{ - 2x}{ - 2} = \frac{4}{ - 2} \\ x = - 2 \\ Substitute \: the \: value \: of \: x \: into \: equation \: (1) \\ y = x + 3 \\ y = - 2 + 3 \\ y = 1 \\ Therefore \: x = - 2 \: \: and \: y = 1[/tex]

The equation of the line in explicit form is y = - 2 · x + 5, whose implicit form is 2 · x + y = 5. (Correct choice: D)

In this question we must find the equation of the line perpendicular to the segment BC and that goes through the point A(x, y) = (1, 3). First, the slope of the line segment BC is:

m = [1 - (- 3)]/[5 - (- 3)]

m = 4/8

m = 1/2

And the slope of the line is:

m' = - 1/(1/2) = - 2

The intercept of the line is found by using the explicit form of equation of the line, m' = - 2 and (x, y) = (1, 3):

3 = - 2 · 1 + b

b = 5

The equation of the line in explicit form is y = - 2 · x + 5, whose implicit form is 2 · x + y = 5. (Correct choice: D)

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The proportion cranberry juice should be mixed with 84 ounces of apple juice 1:5 that is 1 parts of cranberry juice and 5 parts of lemon juice.

According to the question,

A punch recipe calls for mixing 2 parts of cranberry juice with 5 parts of apple juice.

84/2 = 42

5*42=210

Hence, the proportion cranberry juice should be mixed with 84 ounces of apple juice 1:5 that is 1 part of cranberry juice and 5 parts of lemon juice.

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The coordinates of the vertex that A maps to after Daniel's reflections are (3, 4) and the coordinates of the vertex that A maps to after Zachary's reflections are (3, 2)

From the given figure, the coordinate of the vertex A is represented as:

A = (-5, 2)

The coordinates of the vertex that A maps to after Daniel's reflections

The rule of reflection across the line x = -1 is

(x, y) ⇒ (-x - 2, y)

So, we have:

A' = (5 - 2, 2)

Evaluate the difference

A' = (3, 2)

The rule of reflection across the line y = 2 is

(x, y) ⇒ (x, -y + 4)

So, we have:

A'' = (3, -2 + 4)

Evaluate the difference

A'' = (3, 4)

Hence, the coordinates of the vertex that A maps to after Daniel's reflections are (3, 4)

The coordinates of the vertex that A maps to after Zachary's reflections

The rule of reflection across the line y = 2 is

(x, y) ⇒ (x, -y + 4)

So, we have:

A' = (-5, -2 + 4)

Evaluate the difference

A' = (-5, 2)

The rule of reflection across the line x = -1 is

(x, y) ⇒ (-x - 2, y)

So, we have:

A'' = (5 - 2, 2)

Evaluate the difference

A'' = (3, 2)

Hence, the coordinates of the vertex that A maps to after Zachary's reflections are (3, 2)

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The correct option is (d).

The other beaker contains 17 cc (cubic centimetres) of the liquid.

The volume of a cube with dimensions of 1 cm ×1 cm× 1 cm is equal to one cubic centimetre (or cubic centimetre in US English) (SI unit symbol: cm³; non-SI abbreviations: cc and ccm). A millilitre's volume is equal to one cubic centimetre.

According to the question,

The total liquid is 32 in cubic centimetres which is stored in the two beakers.

In the first beaker the liquid stored is 15 cubic centimetres.

The second beaker will contain the liquid = total liquid stored in both beakers - liquid stored in the first beaker.

Liquid in second beaker = 32 - 15

                                        = 17

Therefore, the liquid stored in the second beaker is 17 cc (cubic centimetres).

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[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Here we go ~

[tex]\qquad \sf  \dashrightarrow \: (1.6 \times 10 {}^{5} ) \sdot(2 .3 \times {10}^{1} )[/tex]

[tex]\qquad \sf  \dashrightarrow \: (1.6 \times 2.3) \sdot(10 {}^{5} \times 10 {}^{1} )[/tex]

[tex]\qquad \sf  \dashrightarrow \: 3.68 \times 10 {}^{6} [/tex]

Answer:

  D.  completing the square

Step-by-step explanation:

The process shown rearranges the standard-form quadratic equation to vertex form. It involves factoring out the leading coefficient and distributing the constant so that the variable terms can be written as a perfect square.

A perfect square trinomial is the square of a binomial. It has the form ...

  (x +b)² = x² +2bx +b²

Given two variable terms, x² and 2bx, the perfect square trinomial can be formed by adding b², which is the square of half the x-term coefficient.

When b² is added to the sum (x² +2bx), the square is complete. The sum (x² +2bx +b²) can be written as the square (x +b)².

This process of adding b² to the sum (x² +2bx) will change the polynomial unless a similar quantity is subtracted. That is why the 3rd line of the problem statement show 4 being added and subtracted inside parentheses:

  y = 5(x² +4x +4 -4) -17

When we're applying this transformation, we do not want to change the equation. We simply want to rearrange it.

In the next step, the subtracted term is brought outside the parentheses. This lets us write the quantity in parentheses as a perfect square.

  y = 5(x² +4x +4) +5(-4) -17

  y = 5(x² +4x +4) -20 -17

This process of adding and subtracting a suitable quantity and rearranging the equation so the variable terms make a perfect square is called ...

  "completing the square."

__

Additional comment

The equation in this form is said to be in "vertex form."

  y = a(x -h)² +k

In this form, the constant 'a' is a vertical scale factor; the ordered pair (h, k) identifies the vertex of the quadratic on a graph.

The domain of the exponential function given is the set of all real numbers while the range of the exponential function is the set of all real numbers greater than zero.

As with other exponential functions, it follows that the domain of the exponential function given is the set of all real numbers while the range of the exponential function is the set of all real numbers greater than zero.

Additionally, by observation, the function has a positive variable correlation, hence;

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Answer:

If you divide it out and it comes out a whole number unless the original set of numbers is a fraction

Considering the area of the rectangle, we have that the length is of 8 units and the width is of 10 units.

The area of a rectangle of length l and width w is given as follows:

A = lw.

In this problem, the area is of 80 square units, while the length and the width are consecutive even integers, hence:

Then:

l(l + 2) = 80

l² + 2l - 80 = 0

(l + 10)(l - 8) = 0.

We need the positive measure, hence:

l -  8= 0 -> l = 8 units.

w = l + 2 = 10 units.

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Answer:

100.5 inches

Step-by-step explanation:

If each course has 3/8 inch of concrete below it  (including the bottom one)

  8 in * 12   + 3/8 in * 12 = 100 1/2 inches

Answer:

y = -3x + 9

Step-by-step explanation:

The slope is

[tex] \frac{6 - 0}{1 - 3} = - 3[/tex]

Substituting into point-slope form, we get

y = -3(x-3)

y = -3x + 9

Based on the length and width of the rectangular lawn, and the percentage the man has mowed, the strip width would be 40.5 m.

first, find the area of the lawn:

= 100 x 200

= 20,000 ft²

the area mowed is:

= 40.5% x 20,000

= 8,100 ft²

the width is therefore:

= 8,100 / 200

= 40.5 m

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Answer:

  $70,300

Step-by-step explanation:

The amount Kaydence saved is the value of an arithmetic series with first term 100 and common difference 100. The number of terms in the sum is 37. The formula for the sum can be used.

The formula for the sum of n terms of an arithmetic series is ...

  Sn = (2a1 +d(n -1))(n/2) . . . . . . first term a1, common difference d

The number of terms being added is ...

  2050 -2014 +1 = 37 . . . . . . years Kaydence made deposits

The formula with a1=100, d=100, and n=37 is ...

  [tex]S_{37}=(2\cdot100+100(37-1))\dfrac{37}{2}=\dfrac{(200+3600)\cdot37}{2}\\\\S_{37}=70,\!300[/tex]

Kaydence would have saved $70,300 by the end of Jan 1, 2050.

The statements that are true about the domain and range of the functions f(x), g(x), and h(x) include:

It should be noted that a function simply means a relation between a set of inputs to a set of possible outputs.

In this case, f(x) has a domain of all real numbers as well as g(x) which also has a domain of all real numbers.

In conclusion, the correct options are A and D.

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Answer:

A and D

Step-by-step explanation:

Edge 2022

OPTION A is the correct answer

The equation of a parabola whose vertex is (0, 0) and focus is (1 / 8, 0) is equal to x = 2 · y².

Herein we have the case of a parabola whose axis of symmetry is parallel to the x-axis. The standard form of the equation of this parabola is shown below:

(x - h) = [1 / (4 · p)] · (y - k)²     (1)

Where:

The distance from the vertex to the focus is 1 / 8. If we know that the location of the vertex is (0, 0), then the standard form of the equation of the parabola is:

x = 2 · y²     (1)

The equation of a parabola whose vertex is (0, 0) and focus is (1 / 8, 0) is equal to x = 2 · y².

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Since the geometric series has 6 terms, increases by a factor of 4, and has a sum of 1365, the value of the first term is 1.

The sum of a geometric series is given by

Sₙ = a(rⁿ - 1)/(r - 1) with r > 1 where

Now, since our Geometric Series has 6 terms, n = 6. Also, it increases by a factor of 4, so, r = 4 and has a sum of 1365, so Sₙ = 1356. So,we have that

Since we require the first term, a , making a subject of the formula, we have

a = Sₙ(r - 1)/(rⁿ - 1)

Substituting the values of the variables into the equation, we have

a = Sₙ(r - 1)/(rⁿ - 1)

a = S₆(r - 1)/(r⁶ - 1)

a = 1365(4 - 1)/(4⁶ - 1)

a = 1365(3)/(4096 - 1)

a = 4095/4095

a = 1

So, the value of the first term is 1.

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The general form of the equation for the given circle centered at O(0, 0) is x^2 + y^2 = 41

The given parameters are

Center = (0, 0)

Point = (4, 5)

Rewrite the given parameters are

(a, b)= (0, 0)

(x, y) = (4, 5)

The general form of the equation for the given circle is represented as:

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

Substitute the known values in the above equation to calculate the radius r

(4 - 0)^2 + (5 - 0)^2 = r^2

Evaluate the difference

4^2 + 5^2 = r^2

Evaluate the exponent

16 + 25 = r^2

Evaluate the sum

41= r^2

Rewrite as

r^2 = 41

Substitute r^2 = 41 in (x - a)^2 + (y - b)^2 = r^2

(x - a)^2 + (y - b)^2 = 41

Substitute (a, b)= (0, 0) in (x - a)^2 + (y - b)^2 = 41

(x - 0)^2 + (y - 0)^2 = 41

Evaluate the difference

x^2 + y^2 = 41

Hence, the general form of the equation for the given circle centered at O(0, 0) is x^2 + y^2 = 41

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Answer:

Step-by-step explanation:

Elasticity Demand

(Q1 - Q2)/(Q1 + Q2)

(P1 - P2)/(P1 + P2)

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[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{Equation a.}[/tex]

[tex]\rm{2x^2 - 162 = 0}[/tex]

[tex]\huge\textbf{Solving for:}[/tex]

[tex]\rm{2x^2 - 162 = 0}[/tex]

[tex]\huge\textbf{Add \boxed{\bf 162} to both sides:}[/tex]

[tex]\rm{2x^2 - 162 + 162 = 0 + 162}[/tex]

[tex]\huge\textbf{Simplify it:}[/tex]

[tex]\rm{2x^2 = 0 + 162}[/tex]

[tex]\rm{2x^2 = 162}[/tex]

[tex]\huge\textbf{Divide \boxed{\bf 2} to both sides:}[/tex]

[tex]\rm{\dfrac{2x^2}{2} = \dfrac{162}{2}}[/tex]

[tex]\huge\textbf{Simplify it:}[/tex]

[tex]\rm{x^2 = \dfrac{182}{2}}[/tex]

[tex]\rm{x^2 = 81}[/tex]

[tex]\huge\textbf{Take the square root of \boxed{\bf 81}}[/tex]

[tex]\rm{x = \pm \sqrt{81}}[/tex]

[tex]\rm{x = 9\ or\ x = -9}[/tex]

[tex]\huge\textbf{Therefore, your answer should be:}[/tex]

[tex]\huge\boxed{\textsf{x = } \frak{9\ or } \ \textsf{x = }\frak{-9}}\huge\checkmark[/tex]

[tex]\huge\textbf{Equation b.}[/tex]

[tex]\rm{-\dfrac{1}{2}(x - 3)^2 = -2}[/tex]

[tex]\huge\textbf{Solving for:}[/tex]

[tex]\rm{-\dfrac{1}{2}(x - 3)^2 = -2}[/tex]

[tex]\huge\textbf{Simplify both sides of your equation:}[/tex]

[tex]\rm{-\dfrac{1}{2}x^2 + 3x - \dfrac{9}{2} = -2}[/tex]

[tex]\huge\textbf{Subtract \boxed{\bf -2} to both sides:}[/tex]

[tex]\rm{-\dfrac{1}{2}x^2 + 3x - \dfrac{9}{2} - (-2) = -2 - (-2)}[/tex]

[tex]\huge\textbf{Simplify it:}[/tex]

[tex]\rm{- \dfrac{1}{2}x^2 + 3x - \dfrac{5}{2} = 0}[/tex]

[tex]\huge\textbf{Now, we can convert the equation to:}[/tex]

[tex]\rm{-0.5x^2 + 3x - 2.5 = 0}[/tex]

[tex]\large\text{When we changed the equation entirely, we made it easier to solve}\uparrow[/tex]

[tex]\huge\textbf{Use the quadratic formula to solve.}[/tex]

[tex]\large\textsf{The quadratic formula:}[/tex]

[tex]\mathsf{x= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}}[/tex]

[tex]\huge\textbf{Here are your labels:}[/tex]

[tex]\text{a = }\rm{-0.5}\\\\\text{b = }\rm{ 3}\\\\\rm{c = }\rm{\ -2.5}[/tex]

[tex]\huge\textbf{Your new equation:}[/tex]

[tex]\rm{x = \dfrac{-(3) \pm \sqrt{3^2 - 4(-0.5)(-2.5)}}{2(-0.5)}}[/tex]

[tex]\huge\textbf{Simplify it:}[/tex]

[tex]\rm{x = \dfrac{-3 \pm \sqrt{4}}{-1}}[/tex]

[tex]\rm{x = 1\ or \ x = 5}[/tex]

[tex]\huge\textbf{Therefore, your answer should be: }[/tex]

[tex]\huge\boxed{\textsf{x = }\frak{1}\ \textsf{or x = }\frak{5}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

[tex]a)\ x_1=-9,\ x_2=9\\\\b)\ x_1=1,\ x_2=5[/tex]

Given equations:

[tex]a)\ 2x^2 - 162 = 0\\\\b) -\dfrac{1}{2}(x-3)^2=-2[/tex]

A) 2x² - 162 = 0

Step 1: Divide both sides by 2.

Step 2: Take the square root of both sides (using both the positive and negative roots).

[tex]\\\implies \sqrt{x^2}=\sqrt{81}\\\\\implies x=\pm\ 9[/tex]

Step 3: Separate into two cases.

[tex]\implies x_1 = -9,\ x_2 =-9[/tex]

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

B) -1/2(x - 3)² = -2

Step 1: Multiply both sides by -2.

[tex]\\\implies -2\left(-\dfrac{1}{2}(x-3)^2\right)=-2(-2)\\\\\implies (x-3)^2=4[/tex]

Step 2: Take the square root of both sides (using both the positive and negative roots).

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

Step 3: Separate into two cases and solve each one.[tex]1)\ x-3=-2\implies x=-2+3\implies \boxed{x=1}\\\\2)\ x-3=2\implies x=2+3\implies \boxed{x=5}[/tex]

Using it's concepts, the domain and the range of the graph are given as follows:

In this graph, have that the function is defined for all values of x except x = -1, and assumes all real values, hence the domain and the range of the graph are given as follows:

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The volume of the oblique cylinder is 6867.18 cubic units

The given parameters are:

Height = 27

Radius = 9

The volume is calculated as:

[tex]V = \pi r^2 h[/tex]

So, we have:

[tex]V = 3.14 * 9^2 * 27[/tex]

Evaluate

V = 6867.18

Hence, the volume of the oblique cylinder is 6867.18 cubic units

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Subsets with bit strings:

To find the subsets with bit strings:

We need to express a) b) and c) into bit strings.

Firstly, a number of elements in the universal set represent the number of bits in the bit string.

Secondly, 1 = yes element is present in both universal sets as well as in subset.

0 = No, the element is not present in the subset but present in the universal set.

Hence, we have:

a) Subset {3,4,5} = 0011100000  (As there are 3 1's which means only 3,4,5 are present in both universal set and subset.

Similarly,

b) Subset {1, 3, 6, 10} = 1010010001

c) Subset {2, 3, 4, 7, 8, 9} = 0111001110

Therefore, Subsets with bit strings:

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Complete question:

Suppose that the universal set is U={1,2,3,4,5,6,7,8,9,10}. Express each of the following subsets with bit strings (of length 10) where the ith bit (from left to right) is 1 if i is in the subset and zero otherwise.

a) {3, 4, 5}

b) {1, 3, 6, 10}

c) {2, 3, 4, 7, 8, 9}.

Answer:

[tex]\left( x+\frac{1}{2} \right)^{2} + \frac{3}{4}[/tex]

Step-by-step explanation:

NOTE :

[tex]x^2+ax=\left( x+\frac{a}{2} \right)^{2} -\left( \frac{a}{2} \right)^{2}[/tex]

………………………………………

f(x) = x² + x +1

     [tex]=x^{2}+2\times \frac{x}{2} +1[/tex]

     [tex]=\left( x+\frac{1}{2} \right)^{2} -\left( \frac{1}{2} \right)^{2} +1[/tex]

     [tex]=\left( x+\frac{1}{2} \right)^{2} - \frac{1}{4}+1[/tex]

     [tex]=\left( x+\frac{1}{2} \right)^{2} + \frac{3}{4}[/tex]