An amusement park charges admission plus a fee for each ride. Admission plus four ride cost $22. Admission plus seven ride cost $31. What is the charge for admission and the cost for each ride

Answer:

[tex]3\frac{3}{4}\ \frac{yd^2}{h}[/tex]

or

[tex]3.75\ \frac{yd^2}{h}[/tex]

Step-by-step explanation:

we know that

To find out the unit rate, divide the total square yards by the total time in hours

so

[tex]\frac{(5/4)}{(1/3)}=\frac{15}{4}\ \frac{yd^2}{h}[/tex]

Convert to mixed number

[tex]\frac{15}{4}\ \frac{yd^2}{h}=\frac{12}{4}+\frac{3}{4}=3\frac{3}{4}\ \frac{yd^2}{h}[/tex]

or

[tex]3.75\ \frac{yd^2}{h}[/tex]

Answer: 6(4x + 5)

Step-by-step explanation: 4(6x + 5) 24x + 20

Answer:

ax^2+bx+c=0

Step-by-step explanation:

hope this helps

Answer:

[see picture]

Step-by-step explanation:

The formula shown in the picture is the quadratic formula.

When:

[tex]ax^2+bx+c=0[/tex]

a, b, c ≠ 0

x = unknown

Answer: -8y-27

Step-by-step explanation: Simplify the expression.

Hope this helps you out.

Answer:

-8y-27.

Step-by-step explanation:

3(-2y-9)-2y

= 3*-2y+3*-9 -2y

= -6y-27-2y

= -8y-27.

Answer:

$32.50

Step-by-step explanation:

Just divide $65 by 2 and you get $32.5(0).

Juan spent $32.50 on the 2 tires.

You are very welcome!

The cost of two tires = 65 - cots of tuned up.

And we need the cost of two tires.

Subtraction is a mathematic operation. Which is used to remove terms or objects in an expression.

Given:

Juan bought two new tires for his bike.

He also had the bike tuned up.

The total bill was $65.

To find the cost of two tires:

The cost of two tires = 65 - cots of tuned up.

Therefore, we need the value of missing information.

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The complete question:

Draw a Conclusion: Juan bought two new tires for his bike. He also had the bikec tuned up. The total bill was $65. How much did Juan spend on the 2 tires?

Answer:

The distance of base of wire to the base of tower is 13.27 ft

Step-by-step explanation:

Given as :

The measure of wire connected between base ad antenna top = 20 ft

The wire is connected to the  antenna on tower at height h = 15 ft above the ground

Let The distance of base of wire to the base of tower = x ft

Let the angle drawn on ground = Ф

Now, According to question

In figure AOB

Sin angle = [tex]\dfrac{\textrm perpendicular}{\textrm hypotenuse}[/tex]

i.e SinФ = [tex]\dfrac{\textrm AB}{\textrm BO}[/tex]

Or,  SinФ = [tex]\dfrac{\textrm h}{\textrm 20}[/tex]

Or,  SinФ = [tex]\dfrac{\textrm 15}{\textrm 20}[/tex]

Or, SinФ = [tex]\dfrac{\textrm 3}{\textrm 4}[/tex]

or, Ф = [tex]sin^{-1}(\frac{3}{4})[/tex]

∴ Ф = 48.5°

Now, Again from figure

Tan angle = [tex]\dfrac{\textrm perpendicular}{\textrm base}[/tex]

i.e TanФ = [tex]\dfrac{\textrm AB}{\textrm OA}[/tex]

Or, Tan 48.5° = [tex]\dfrac{\textrm h}{\textrm x}[/tex]

Or, 1.13 = [tex]\dfrac{\textrm 15}{\textrm x}[/tex]

∴ x = [tex]\dfrac{\textrm 15}{\textrm 1.13}[/tex]

i.e x = 13.27 feet

So, The distance of base of wire to the base of tower = x = 13.27 ft

Hence, The distance of base of wire to the base of tower is 13.27 ft Answer

Final answer:

The other end of the wire will be located approximately √175 ft away from the base of the tower.

Explanation:

To find the distance from the base of the tower to the other end of the wire, we can use the Pythagorean theorem. The wire forms the hypotenuse of a right triangle, with one leg being the height of the antenna and the other leg being the distance from the base of the tower. Let's call this distance 'x'.

Using the Pythagorean theorem, we have:

x² + 15² = 20²

Simplifying this equation, we get:

x² = 20² - 15²

x² = 400 - 225

x² = 175

Taking the square root of both sides, we find:

x = √175

So the other end of the wire will be located approximately √175 ft away from the base of the tower.

Answer:

$6

The drawings will help

Answer:

Step-by-step explanation:

construct the graph with the following points.

x=0

y= -2(0)+4

y= 4

(0,4)

x=2

y=-2(2)+4

y=0

(2,0)

join these two points with a straight line.

feel free to ask any questions

Answer:

[tex]S = P(1.08)^{t}[/tex]

Step-by-step explanation:

A $10,000 deposit at the bank will double in value in 9 years.

If the interest is r% and it is compounded each year, then we can write from the formula of compound interest that

[tex]20000 = 10000(1 + \frac{r}{100})^{9}[/tex]

⇒ [tex]2 = (1 + \frac{r}{100})^{9}[/tex]

⇒ [tex]1 + \frac{r}{100} = 1.08[/tex]

⇒ r = 8%

Therefore, the formula for the accumulated amount t years after the investment is made will be  

[tex]S = P(1 + \frac{8}{100})^{t} = P(1.08)^{t}[/tex]  

where, P is the invested principal and S is the accumulated sum. (Answer)

Answer:

The first choice.

Step-by-step explanation:

We only need to know one part of this step function to figure out the answer. The first step is y=-1 if -6<x<=-4, this knocks out the 2nd and 3rd answer choices as they don't follow this equation.

Answer:

one

Step-by-step explanation:

Answer:

All numbers are divisible by 1

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

The function of the x in the table is not having a specific pattern apart from repeating 3's.

Choice A: pattern for x +1

pattern for y +1

Choice B:pattern of x -1

pattern for y +1

Choice C: pattern for x +1

pattern for y -0

Choice D:pattern for x -0

pattern for y +1

Therfore D would be the answer.

Answer:Its C

Step-by-step explanation:

ED 2023

The correct answer is provided by the third system.

To determine which system of linear inequalities includes the point (2, 1), we need to substitute this point into each inequality and check if the inequalities hold true. Let's evaluate each system step-by-step:

System 1

Evaluating at (2, 1):

The point (2, 1) does not satisfy the first inequality, so it is not in the solution set for the first system.

System 2

Evaluating at (2, 1):

The point (2, 1) does not satisfy the second inequality, so it is not in the solution set for this system either.

System 3

Evaluating at (2, 1):

The point (2, 1) satisfies both inequalities, so it is in the solution set for the third system.

System 4

Evaluating at (2, 1):

The point (2, 1) does not satisfy the first inequality, so it is not in the solution set for the fourth system.

The system of linear inequalities that includes the point (2, 1) is the third one: y [tex]\leq[/tex] -x + 3, and y [tex]\leq[/tex] 0.5x + 2.

Answer:

Step-by-step explanation:

Simplifying

h(x) = x2 + 3x + -18

Multiply h * x

hx = x2 + 3x + -18

Reorder the terms:

hx = -18 + 3x + x2

Solving

hx = -18 + 3x + x2

Solving for variable 'h'.

Move all terms containing h to the left, all other terms to the right.

Divide each side by 'x'.

h = -18x-1 + 3 + x

Simplifying

h = -18x-1 + 3 + x

Reorder the terms:

h = 3 + -18x-1 + x

Answer:

h(x)= (x+3/2 ) ^2  −  81/4

​

Step-by-step explanation:

Its correct :)

Answer: C

Step-by-step explanation:

The molecule modeled is [tex]CH_3-CH_3-C-CH_3-CH_2-COOH[/tex] or [tex]{\text} C_6H_{12}O_6[/tex].

Thus, option (C) is correct.

A chemical formula is a concise representation of the composition of a chemical compound. It uses symbols and subscripts to denote the types and numbers of atoms present in the compound.

Chemical formulas are used to describe the ratio of elements in a compound and give valuable information about its constituents.

In the given compound,

Number of Hydrogen = 12

Number of Carbon= 6

Number of Oxygen = 2

Also, there is Carboxylic function group in chain "COOH".

Therefore, the compound name is [tex]CH_3-CH_3-C-CH_3-CH_2-COOH[/tex] or [tex]{\text} C_6H_{12}O_6[/tex].

Thus, option (C) is correct.

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The portion of cake Mike ate is 1/6.

The numbers given are expressed as fractions. Fractions are numbers that consist of numerators and denominators. For example, in the fraction 2/3, 2 is the numerator and 3 is the denominator.

The first step is to determine which portion of the cake that was stored in the freezer.

Portion kept in the freezer = 2/3 x 3/4 = 1/2

The second step is to determine the amount of cake Mike ate.

2/3 - 1/2

[tex]\frac{4 - 3}{6}[/tex] = [tex]\frac{1}{6}[/tex]

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The portion of the cake that Mike ate is [tex]\frac{1}{6}[/tex].

The given parameters:

The portion of the cake that Mike wrapped and stored in the freezer is calculated as follows;

[tex]= \frac{3}{4} \times \frac{2}{3} \\\\= \frac{1}{2}[/tex]

The portion of the cake that Mike ate is calculated as follows;

[tex]remaining = original - portion \ stored\\\\remaining = \frac{2}{3} - \frac{1}{2} = \frac{4 - 3}{6} = \frac{1}{6}[/tex]

Thus, the portion of the cake that Mike ate is [tex]\frac{1}{6}[/tex].

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Answer:

y=-19.2

Step-by-step explanation:

6.4(-3)=-19.2

To decide whether to write a as a fraction or b as a decimal in the expression ax+bx, consider the context of the problem. If dealing with measurements or real-life quantities, write a as a decimal. If dealing with parts of a whole or ratios, write b as a fraction.

In the expression ax+bx, a is a decimal and b is a fraction. To decide whether to write a as a fraction or b as a decimal, you can consider the context of the problem or the specific instructions. If the problem involves measurements or real-life quantities, it is more appropriate to write a as a decimal. On the other hand, if the problem deals with parts of a whole or ratios, it is more appropriate to write b as a fraction.

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Whether you write 'a' as a fraction or 'b' as a decimal in the expression ax+bx depends on the context of the problem you're solving. 'a' can be converted to a fraction by expressing the decimal in fractional terms, whereas 'b' can be turned into a decimal by dividing the numerator by the denominator.

In the expression ax+bx, where 'a' is a decimal and 'b' is a fraction, deciding whether to write 'a' as a fraction or 'b' as a decimal depends largely on the problem you're solving and what format is most convenient or understandable. To convert 'a' from a decimal to a fraction, you can write the decimal in fractional form. If the denominator is not 100, convert it to a fraction with a denominator of 100, which then allows you to present it as a percentage. On the other hand, to convert 'b' from a fraction to a decimal, you can simply divide the numerator of the fraction by the denominator.

A practical example could be if a=0.5 and b=3/4 in the expression. Here, you can write 'a' as 1/2 (a fraction) and 'b' as 0.75 (a decimal). Therefore, the expression becomes 1/2x + 0.75x.

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Answer:

To persuade, inform and entertain.

Step-by-step explanation:

You can remember it as PIE, if that helps!

Answer:

The Area of square pyramid is 36 + 36[tex]\sqrt{5}[/tex] .

Step-by-step explanation:

Given as :

For A square pyramid

The measure of each base side = b = 6 feet

The height of the pyramid = h = 6 feet

Let The Total area of pyramid = A square feet

Now, According to question

The area of square pyramid

Area = b² + 2 × b [tex]\sqrt{\frac{b^{2} }{4}+h^{2}  }[/tex]

Where b is the square base edge

And h is the pyramid height

I.e A = 6² + 2 × 6 [tex]\sqrt{\frac{6^{2} }{4}+6^{2}  }[/tex]

Or, A = 36 + 12 × [tex]\sqrt{\frac{36 }{4}+36  }[/tex]

Or, A = 36 + 12 × [tex]\sqrt{9+36}[/tex]

Or, A = 36 + 12 × [tex]\sqrt{45}[/tex]

Or, A = 36 + 12 × 3[tex]\sqrt{5}[/tex]

Or, A = 36 + 36[tex]\sqrt{5}[/tex]

So,The Area of square pyramid = A = 36 + 36[tex]\sqrt{5}[/tex]

Hence, The Area of square pyramid is 36 + 36[tex]\sqrt{5}[/tex] . Answer

According to Newton's Third Law of Motion, the force exerted by the door on the person who kicks it, is equal and opposite to the force the person exerts on the door, i.e., 48 Newtons.

The question asks about the force exerted by a door on a person if the person kicks the door with a certain force. This scenario illustrates Newton's Third Law of Motion, which states: 'For every action, there is an equal and opposite reaction.' So, if the person exerts a force of 48 Newtons on the door, then the door will exert an equal and opposite force of 48 Newtons on the person. Therefore, the answer is A. 48 N.

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[tex]\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 4^{-\frac{11}{3}}\div 4^{-\frac{2}{3}}\implies \cfrac{4^{-\frac{11}{3}}}{4^{-\frac{2}{3}}}\implies 4^{-\frac{11}{3}}\cdot 4^{\frac{2}{3}}\implies 4^{-\frac{11}{3}+\frac{2}{3}} \\\\\\ 4^{-\frac{9}{3}}\implies 4^{-3}\implies \cfrac{1}{4^3}\implies \cfrac{1}{64}[/tex]

Answer:

C. 2/3

Step-by-step explanation:

12 female / 18 male

Dividing by 6 we get:

2 female / 3 male

or 2/3

Answer:

C.2/3

Step-by-step explanation:

Answer:

$17.35

Step-by-step explanation:

86.74*.2

Answer:

$5100 would be $76.50

$4876 would be $73.14

$5215 would be $78.23

$6225 would be $93.38

$5235 would be $78.53

Step-by-step explanation:

To find commissions take your sale price and multiply it by your percentage and divide by 100

Answer: $399.77

Step-by-step explanation:

There are no illustrations, so it is impossible to answer this question. I apologise.

Answer:

Step-by-step explanation:

Remember:

If something is:

-equal to a number

-equal to or less then a number

-euqal to or greater than a number

Then you you would use a closed mark with an arrow.

Note: Don't use the arrow for the "equal to" part

If something is greater/less than a number:

Then you would use an open mark with an arrow

So since you're using the inequality q < 69.5,

You would graph it with an open mark with a left arrow, and the open mark must be on the number 69.5

The graph the solution to the given inequality on the number line plotted below.

Inequalities are the mathematical expressions in which both sides are not equal. In inequality, unlike in equations, we compare two values. The equal sign in between is replaced by less than (or less than or equal to), greater than (or greater than or equal to), or not equal to sign.

The given inequality is q < 69.5​.

The result can be shown in multiple forms.

Inequality Form:q<69.5

Interval Notation:(−∞,69.5)

Therefore, the graph the solution to the given inequality on the number line plotted below.

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the two solutions are approximately [tex]\( x_1 \approx 2.95 \) and \( x_2 \approx -0.042 \).[/tex]

To find the value of x that satisfies the equation [tex]\( 32x^2 - 93x - 4 = 0 \)[/tex], we can use the quadratic formula:

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

Where:

- a = 32

- b = -93

- c = -4

Now, let's plug these values into the quadratic formula:

[tex]\[ x = \frac{{-(-93) \pm \sqrt{{(-93)^2 - 4 \cdot 32 \cdot (-4)}}}}{{2 \cdot 32}} \]\[ x = \frac{{93 \pm \sqrt{{8649 + 512}}}}{{64}} \]\[ x = \frac{{93 \pm \sqrt{{9161}}}}{{64}} \]\[ x = \frac{{93 \pm 95.7}}{{64}} \][/tex]

Now, we have two possible solutions:

[tex]\[ x_1 = \frac{{93 + 95.7}}{{64}} \]\[ x_2 = \frac{{93 - 95.7}}{{64}} \]\[ x_1 \approx \frac{{188.7}}{{64}} \]\[ x_2 \approx \frac{{-2.7}}{{64}} \]\[ x_1 \approx 2.95 \]\[ x_2 \approx -0.042 \][/tex]

Therefore, the two solutions are approximately [tex]\( x_1 \approx 2.95 \) and \( x_2 \approx -0.042 \).[/tex]

The probable question maybe:

What are the solutions for x in the equation [tex]\( 32x^2 - 93x - 4 = 0 \)[/tex]?