Photo attached. Someone please help
Thanks

The gross price for the 210 pieces that Lance completed last week is $468.6.

Percentage is a measurement to find value of given number out of hundred.

Given that,

Amount received by Lance Greene for one piece = $2.20.

Also, Lance gets 10% bonus per piece after 180 pieces.

Given that, last week Lance completed 210 pieces .

Price for 210 piece = 210 x 2.20  = 462

Also, bonus gets by Lance on 30 pieces = 30 x 2.2 x 10/100 = 6.6

Total price = 462 + 6.6 = 468.6

The required gross price is $468.6.

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Answer:

Step-by-step explanation:

distance = [tex]\sqrt{(-5/2+1/2)^{2} +(9/4+3/4)^{2} }=\sqrt{13}[/tex]=3.61

The design of a simulation that would allow Ian to determine how likely it would

be to guess all answers correctly on this quiz can be Illustrated through a coin.

Ian can design a simulation to determine the likelihood of guessing all answers correctly on the quiz by using a fair coin. He can flip the coin four times, with each flip representing a guess on the quiz. If the coin lands heads up, he marks it as a true answer, and if it lands tails up, he marks it as a false answer.

He can repeat this process a large number of times (e.g. 1000) and count the number of times that the sequence T, T, F, T is generated. The proportion of times that this sequence is generated out of the total number of simulations will be an estimate of the probability of guessing all answers correctly on the quiz by chance.

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Answer:

1/24

Step-by-step explanation:

Solve for y =4

48x=2

Answer:

[tex]A.\ \tan(x) \to 2.\ \sin(x) \sec(x)[/tex]

[tex]B.\ \cos(x) \to 5. \sec(x) - \sec(x)\sin^2(x)[/tex]

[tex]C.\ \sec(x)csc(x) \to 3. \tan(x) + \cot(x)[/tex]

[tex]D. \frac{1 - (cos(x))^2}{cos(x)} \to 1. \sin(x) \tan(x)[/tex]

[tex]E.\ 2\sec(x) \to\ 4.\ \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}[/tex]

Step-by-step explanation:

Given

[tex]A.\ \tan(x)[/tex]

[tex]B.\ \cos(x)[/tex]

[tex]C.\ \sec(x)csc(x)[/tex]

[tex]D.\ \frac{1 - (cos(x))^2}{cos(x)}[/tex]

[tex]E.\ 2\sec(x)[/tex]

Required

Match the above with the appropriate identity from

[tex]1.\ \sin(x) \tan(x)[/tex]

[tex]2.\ \sin(x) \sec(x)[/tex]

[tex]3.\ \tan(x) + \cot(x)[/tex]

[tex]4.\ \frac{cos(x)}{1 - sin(x)} + \frac{1 - \sin(x)}{cos(x)}[/tex]

[tex]5.\ \sec(x) - \sec(x)(\sin(x))^2[/tex]

Solving (A):

[tex]A.\ \tan(x)[/tex]

In trigonometry,

[tex]\frac{sin(x)}{\cos(x)} = \tan(x)[/tex]

So, we have:

[tex]\tan(x) = \frac{\sin(x)}{\cos(x)}[/tex]

Split

[tex]\tan(x) = \sin(x) * \frac{1}{\cos(x)}[/tex]

In trigonometry

[tex]\frac{1}{\cos(x)} =sec(x)[/tex]

So, we have:

[tex]\tan(x) = \sin(x) * \sec(x)[/tex]

[tex]\tan(x) = \sin(x) \sec(x)[/tex] --- proved

Solving (b):

[tex]B.\ \cos(x)[/tex]

Multiply by [tex]\frac{\cos(x)}{\cos(x)}[/tex] --- an equivalent of 1

So, we have:

[tex]\cos(x) = \cos(x) * \frac{\cos(x)}{\cos(x)}[/tex]

[tex]\cos(x) = \frac{\cos^2(x)}{\cos(x)}[/tex]

In trigonometry:

[tex]\cos^2(x) = 1 - \sin^2(x)[/tex]

So, we have:

[tex]\cos(x) = \frac{1 - \sin^2(x)}{\cos(x)}[/tex]

Split

[tex]\cos(x) = \frac{1}{\cos(x)} - \frac{\sin^2(x)}{\cos(x)}[/tex]

Rewrite as:

[tex]\cos(x) = \frac{1}{\cos(x)} - \frac{1}{\cos(x)}*\sin^2(x)[/tex]

Express [tex]\frac{1}{\cos(x)}\ as\ \sec(x)[/tex]

[tex]\cos(x) = \sec(x) - \sec(x) * \sin^2(x)[/tex]

[tex]\cos(x) = \sec(x) - \sec(x)\sin^2(x)[/tex] --- proved

Solving (C):

[tex]C.\ \sec(x)csc(x)[/tex]

In trigonometry

[tex]\sec(x)= \frac{1}{\cos(x)}[/tex]

and

[tex]\csc(x)= \frac{1}{\sin(x)}[/tex]

So, we have:

[tex]\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)}[/tex]

Multiply by [tex]\frac{\cos(x)}{\cos(x)}[/tex] --- an equivalent of 1

[tex]\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)} * \frac{\cos(x)}{\cos(x)}[/tex]

[tex]\sec(x)csc(x) = \frac{1}{\cos^2(x)}*\frac{\cos(x)}{\sin(x)}[/tex]

Express [tex]\frac{1}{\cos^2(x)}\ as\ \sec^2(x)[/tex] and [tex]\frac{\cos(x)}{\sin(x)}\ as\ \frac{1}{\tan(x)}[/tex]

[tex]\sec(x)csc(x) = \sec^2(x)*\frac{1}{\tan(x)}[/tex]

[tex]\sec(x)csc(x) = \frac{\sec^2(x)}{\tan(x)}[/tex]

In trigonometry:

[tex]tan^2(x) + 1 =\sec^2(x)[/tex]

So, we have:

[tex]\sec(x)csc(x) = \frac{\tan^2(x) + 1}{\tan(x)}[/tex]

Split

[tex]\sec(x)csc(x) = \frac{\tan^2(x)}{\tan(x)} + \frac{1}{\tan(x)}[/tex]

Simplify

[tex]\sec(x)csc(x) = \tan(x) + \cot(x)[/tex]  proved

Solving (D)

[tex]D.\ \frac{1 - (cos(x))^2}{cos(x)}[/tex]

Open bracket

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

[tex]1 - \cos^2(x) = \sin^2(x)[/tex]

So, we have:

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \frac{sin^2(x)}{cos(x)}[/tex]

Split

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \frac{sin(x)}{cos(x)}[/tex]

[tex]\frac{sin(x)}{\cos(x)} = \tan(x)[/tex]

So, we have:

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \tan(x)[/tex]

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) \tan(x)[/tex] --- proved

Solving (E):

[tex]E.\ 2\sec(x)[/tex]

In trigonometry

[tex]\sec(x)= \frac{1}{\cos(x)}[/tex]

So, we have:

[tex]2\sec(x) = 2 * \frac{1}{\cos(x)}[/tex]

[tex]2\sec(x) = \frac{2}{\cos(x)}[/tex]

Multiply by [tex]\frac{1 - \sin(x)}{1 - \sin(x)}[/tex] --- an equivalent of 1

[tex]2\sec(x) = \frac{2}{\cos(x)} * \frac{1 - \sin(x)}{1 - \sin(x)}[/tex]

[tex]2\sec(x) = \frac{2(1 - \sin(x))}{(1 - \sin(x))\cos(x)}[/tex]

Open bracket

[tex]2\sec(x) = \frac{2 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]

Express 2 as 1 + 1

[tex]2\sec(x) = \frac{1+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]

Express 1 as [tex]\sin^2(x) + \cos^2(x)[/tex]

[tex]2\sec(x) = \frac{\sin^2(x) + \cos^2(x)+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]

Rewrite as:

[tex]2\sec(x) = \frac{\cos^2(x)+1 - 2\sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}[/tex]

Expand

[tex]2\sec(x) = \frac{\cos^2(x)+1 - \sin(x)- \sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}[/tex]

Factorize

[tex]2\sec(x) = \frac{\cos^2(x)+1(1 - \sin(x))- \sin(x)(1-\sin(x))}{(1 - \sin(x))\cos(x)}[/tex]

Factor out 1 - sin(x)

[tex]2\sec(x) = \frac{\cos^2(x)+(1- \sin(x))(1-\sin(x))}{(1 - \sin(x))\cos(x)}[/tex]

Express as squares

[tex]2\sec(x) = \frac{\cos^2(x)+(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}[/tex]

Split

[tex]2\sec(x) = \frac{\cos^2(x)}{(1 - \sin(x))\cos(x)} +\frac{(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}[/tex]

Cancel out like factors

[tex]2\sec(x) = \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}[/tex] --- proved

Answer:

10 females, 15 males

Step-by-step explanation:

Let the number of females in the class be x. As the number of males is 5 more than the number of females, which is x, the number of males would be shown as x+5.

The number of females plus the number of males equal 25, so we get this equation:

x + (x+5) = 25

x + x + 5 = 25

2x + 5 = 25

2x = 20

x = 10.

So, there are 10 females, and since there are 5 more males than females, there are 10 + 5 = 15 males.

Answer:

it will take Johnny 7 hours

Step-by-step explanation:

It was easy, i divided 105 by 15 and got 7

Have a wonderful day :)

Answer:

7 hours

Step-by-step explanation:

15 * x = 105

---          ---

15           7

x=7

Answer:

$432,189

Step-by-step explanation:

$1,000,000 - $567,811 = $432,189

Answer:

Hello There!!

Step-by-step explanation:

The answer is 15 as 3a=3×a so 3×5=15.

hope this helps,have a great day!!

~Pinky~

Answer:

Assuming data is from population, 0.1

Step-by-step explanation:

Answer:

The length of the diagonal is 12.53

Step-by-step explanation:

Since we have a rectangle, the diagonal is the hypotenuse of a right triangle. So we can use Pythagorean Theorem.

a^2 + b^2 = c^2

We know the two legs are 6 and 11, so we can find the hypotenuse (which is the diagonal) see image.

See image.

Answer:

When solving a compound inequality, you can express the solution in two different ways: using a union or intersection.

Using a union:

A union represents the solutions that belong to either one inequality or the other, or both. You can express a union using the symbol "or," or by writing the solutions for each inequality separately and combining them using a "union" symbol (∪). For example, if you have the compound inequality "x > 2 or x < -3," you can express the solution as "x ∈ (-3, 2) ∪ (-∞, -3) ∪ (2, ∞)."

Using an intersection:

An intersection represents the solutions that belong to both inequalities. You can express an intersection using the symbol "and," or by writing the solutions for each inequality separately and combining them using an "intersection" symbol (∩). For example, if you have the compound inequality "x > 2 and x < -3," you can express the solution as "x ∈ (-3, 2) ∩ (-∞, -3) ∩ (2, ∞)."

Which method you use to express the solution will depend on the specific inequalities and the type of solutions you are looking for. It is important to carefully consider the inequality symbols (>, <, ≥, ≤) and the use of "and" or "or" in order to determine the correct solution.

When solving a compound inequality, the solution can be expressed in one of two ways: as an interval or as a union of intervals.

Inequality is defined as mathematical statements that have a minimum of two terms containing variables or numbers that are not equal.

An interval is a set of all real numbers between two given numbers, and is written in the form (a, b) or [a, b], where a and b are real numbers, and a < b. If a and b are included in the interval, then the interval is written as [a, b]; if a and b are not included in the interval, then the interval is written as (a, b).

A union of intervals is a combination of two or more intervals, and is written in the form (a, b) ∪ (c, d) or [a, b] ∪ [c, d], where a, b, c, and d are real numbers, and a < b and c < d.

When solving a compound inequality, it is important to first simplify the inequality by combining like terms, isolating the variable on one side of the inequality, and combining any separate inequalities into a single inequality. Then, the solution can be found by graphing the inequality on a number line, and shading the portion of the number line that represents the solution.

Once the solution is found the solution can be expressed either as an interval or as a union of intervals depending on the inequality. For example, if the inequality is x>1 and x<3, then the solution would be expressed as (1,3) and this is an example of an interval. If it's x>1 and x≥3 then the solution would be expressed as (1,3] U [3,∞) and this is an example of union of intervals.

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Answer: A. Quadrilateral

Step-by-step explanation:

The name for a polygon with 4 sides is a quadrilateral. Quadrilateral is a two-dimensional shape with four straight sides and four angles. The most common type of quadrilateral is the rectangle, which has four right angles. Other types of quadrilaterals include squares, parallelograms, trapezoids, and rhombuses.

Answer:

quadrilateral!!

Step-by-step explanation:

look at the beginning of the word, quad. Quad means 4 and is referring to a 4 sided figure.

Answer:

[tex]y=3x-7[/tex]

Step-by-step explanation:

Parallel lines, by definition, never intersect. They have the same slope but different y-intercepts (otherwise they would be the same line) on a graph.

Slope-Intercept Form:

Slope-Intercept form is expressed as: [tex]y=mx+b[/tex], where

This form is super useful for linear equations as it gives us the two key features of a linear equation. It's how each of the options are formatted, so we know we'll have to use this form.

The line is parallel to [tex]y=3x-4[/tex], meaning it has a slope of 3, but also a y-intercept other than -4 (otherwise they would be the same equation).

So we can generally form an equation: [tex]y=3x+b\text{, }b\ne-4[/tex], so we can plug any value for "b" here (except -4) and have a parallel line

Since we not only want to find a parallel line, but also one that passes through a specific point, we can use our general parallel equation: [tex]y=3x+b\text{, }b\ne-4[/tex], and plug in known values. We of course already have the slope plugged in, but now we have a (x, y) coordinate, which we can plug in for x and y in the equation.

Original Equation:

[tex]y=3x+b[/tex]

Point given: (3, 2), substitute in x=3 and y=2:

[tex]2=3(3)+b[/tex]

Simplify:

[tex]2=9+b[/tex]

Subtract 9 from both sides:

[tex]-7=b[/tex]

Now we can take this value. and plug it back into the general equation:

[tex]y=3x+(-7)\implies y=3x-7[/tex]

Now we have our answer!

Answer:

8 × 10 + 5 × 1. is the correct answer .STAY BRAINLY

Proportionately, the quantity of ore required to obtain 30 kilograms of silver is 2,000 kilograms.

Proportion refers to two or more ratios equated to each other.

We can use decimals, fractions, or percentages to show the proportional value of one quantity or variable to another.

The quantity of ore that can produce 30 kilograms of silver, based on the given proportion, can be obtained by equating the ratio to the required silver quantity.

The percentage of ore that is pure silver = 1¹/₂% = (0.015)

The quantity of silver needed = 30 kilograms

The quantity of ore that can produce 30 kilograms of silver = 2,000 kilograms (30/0.015)

1¹/₂% = 30 kg

100% = 2,000 kg (30/1¹/₂%)

Thus, given the ratio or proportion of pure silver to ore, the quantity of ore needed to obtain 30 kilograms of silver is 2,000 kilograms.

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now, we're assuming this is some rectangular area, and thus is simply the product of both quantities, let's change all mixed fractions to improper fractions firstly.

[tex]\stackrel{mixed}{12\frac{1}{2}}\implies \cfrac{12\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{25}{2}} ~\hfill \stackrel{mixed}{11\frac{3}{4}} \implies \cfrac{11\cdot 4+3}{4} \implies \stackrel{improper}{\cfrac{47}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{25}{2}\cdot \cfrac{47}{4}\implies \cfrac{1175}{8}\implies 146\frac{7}{8}~ft^2[/tex]

Volume of trianglular prism:

= ½ × b × h × l

= ½ × 8 × 4 × 7

= 4 × 28

= 112 cm³

Answer:

112 sq cm

Step-by-step explanation:

8x4=32

32/2=16

16x7=112

Answer:

8, 45, 18, 8.75 pretty sure

Step-by-step explanation:

cross multiplying: when a/b=x/c you type in a*c/b to get x in calculator

Answer:

it would be 10392

Step-by-step explanation:

10392÷8=1299

Answer:

10392

Step-by-step explanation:

10391 ÷ 8 = 1298.875

Not divisible by 8

10392 ÷ 8 = 1299

Divisible by 8

The claim is not correct as both the triangles do not meet the SAS triangle congruence criterion.

If any two sides and angle included between the sides of one triangle are equivalent to the corresponding two sides and the angle between the sides of the second triangle, then the two triangles are said to be congruent by SAS rule.

As from the definition above we can understand that for the triangles to be congruent the sides and the angle included between them should be same.

In the given triangles

The two sides of both the triangle has same length but the angle 30 degree is not the angle included for both the traingles

Therfore we can say that the triangles are not congruent with SAS congruency.

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Answer:

8 = x²

Step-by-step explanation:

Logarithmic form: logₓ(8) = 2

Exponential form: 8 = x²

Answer:

(1/6) of the students did not pass.

Ans: (1/6)30 = 5 of the students did not pass

To solve this question, we need to first figure out, in fractional form, the number of students who did not pass. The problem tells us that 5/6 of the students passed. Therefore, we know that 1/6 of the students did not pass, because 5/6 + 1/6 = 1. So, all we need to do is multiply 1/6 by the total number of students:

1/6 * 36 = 6

So, 6 students did not pass.

Answer:

x=17 degrees

Step-by-step explanation:

We know that a right angle is equal to 90 degrees. Therefore, two angles between a right angle have to equal 90. Therefore, 73+x=90. Isolate x and get x=90-73. x=17. Therefore, x=17 degrees.

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The given polynomial is a factor of P(x) because P(-5)=0.

The Remainder Theorem states that if a synthetic division of a polynomial by x = a results in a zero remainder, then x = an is a zero of the polynomial (thanks to the Remainder Theorem), and x an is also a factor of the polynomial (courtesy of the Factor Theorem).

The Factor Theorem states that x - c is only a factor of P(x) when P(c) = 0.

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The mapping does not represents a function

A function is a relationship between inputs where each input is related to exactly one output. Every function has a domain and codomain or range. A function is generally denoted by f(x) where x is the input

In this case, for the mapping represent a function, every element in set A must be mapped to exactly one element in set B.

Since there is more than one arrow from set A (5) pointing to different values in set B (-2 and 9), then it is not a function. Also, the two different value (5 and 7) in set A has the same value (-2) in set B.

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Answer:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c[/tex]

Step-by-step explanation:

Given

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]

Required

Integrate

We have:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]

Let

[tex]u = x^5[/tex]

Differentiate

[tex]\frac{du}{dx} = 5x^4[/tex]

Make dx the subject

[tex]dx = \frac{du}{5x^4}[/tex]

So, we have:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, \frac{du}{5x^4}[/tex]

[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{10}}} \, du[/tex]

Express x^(10) as x^(5*2)

[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{5*2}}} \, du[/tex]

Rewrite as:

[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{5)^2}}} \, du[/tex]

Recall that: [tex]u = x^5[/tex]

[tex]\frac{1}{5} \int\ {\frac{1}{1 + u^2}}} \, du[/tex]

Integrate

[tex]\frac{1}{5} * \arctan(u) + c[/tex]

Substitute: [tex]u = x^5[/tex]

[tex]\frac{1}{5} * \arctan(x^5) + c[/tex]

Hence:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c[/tex]

Answer:

$36

Step-by-step explanation:

we need to find the cost of 12 items, and the function of determining the cost is C(x)=2x+12, where x is the number of items (the input value of the function)

we know that there are 12 items, so in this function, x=12

substitute x as 12 in the function; the x in C(x) also gets substituted as 12, as 12 is the input value and the value of C(x) is the output

C(12)=2(12)+12

multiply

C(12)=24+12

add

C(12)=36

that means, for 12 items, the cost will be $36

The value of x = -7 and y = 9 from the system of equations .

A finite collection of equations for which common solutions are sought is referred to as a set of simultaneous equations, often known as a system of equations or an equation system.

simultaneous equations, system of equations Two or more equations in algebra must be solved jointly (i.e., the solution must satisfy all the equations in the system). The number of equations must match the number of unknowns for a system to have a singular solution.

To create an equation in one variable using the elimination approach, you may either add or subtract the equations. To remove a variable, add the equations when the coefficients of one variable are in opposition, and subtract the equations when the coefficients of one variable are in equality.

The equations are :

-x - y = -2          (i)

and

5x + 7y = 28        (ii)

multipling eqation (i) with 5 and then adding both equations we get,

-5x - 5y +5x + 7y = -10 + 28

⇒ 2y = 18

⇒ y = 9

Again, -x -9 = -2

or, x = -7

The value of x = -7 and y = 9 from the system of equations

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