The cost of producing higlighter​s with the company logo printed on them consists of a onetime setup fee of $160.00 plus $0.65 for each higlighter produced. This cost can be calculated using the formula C=160.00+0.65p, where p represents the number of higlighters produced and C is the cost. Use the formula to calculate the cost of producing 2400 higlighters. (Do not inlcude a comma in your answer)

Answer: C

Step-by-step explanation:

lets think about it

you can totally buy 0 ink cartridges. you can just be like "nope ok im gone, no ink today, the prices are so high.. maybe another time. maybe ill use the school printer or smth"

so that ELIMINATES B.

and now A and D are strange because how exactly are we going to buy half of an ink cartridge? or a fraction of an ink cartridge? that simply isnt feasible

so this eliminates A and D

now C is the best because you can buy 0, 1, 2, 3, 4 ink cartridges. but not 5 because that is $110 and Martin dont got that money, am i right fellas?

Answer:

The Answer Is C

Step-by-step explanation:

Step-by-step explanation:

a= 3×(-2)+3=-6+3=3

b=3×(-1)+3=-3+3=0

c=3×1+3=3+3=6

d=2×3+3=6+3=9

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

a=3, b=0, c=6, d=9

Good work!

Answer:

25%

Step-by-step explanation:

25% of 12 is 3

25% of 8 is 2

5 people out of 20 people

5/20 = 25%

25% percent of the entire class is going on the field trip

percentage, a relative value indicating hundredth parts of any quantity.

Given,

There are 12 girls and 8 boys in Ms. Crabapple's class.

The total number of students

12+8=20

25% of the girls and 25% of the boys are going on a field trip to a nuclear power plant,

25% of the girls =0.25×12=3

25% of the boys =0.25×8=2

So the total number of students went to field trip

3+2=5

So out of 20 student, 5 students went to the field trip.

Which means 25% of the entire class is going on the field trip

Hence, 25% percent of the entire class is going on the field trip

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Answer:

Step-by-step explanation:

what number is its own oppositie?

The number 0 is the only number that is its own opposite.

[tex]W = 23.7°[/tex]

Step-by-step explanation:

The formula given is the statement of Snell's law of refraction. We are given the following:

[tex]I_a = 1.0003[/tex]

[tex]I_w = 1.3[/tex]

[tex]A = 31.5°[/tex]

So we need to find the angle of refraction W. Starting with Snell's law,

[tex]\dfrac{I_w}{I_a} = \dfrac{\sin A}{\sin W} \Rightarrow \sin W = \left(\dfrac{I_a}{I_w}\right)\sin A[/tex]

Plugging in the given values, we get

[tex]\sin W = \left(\dfrac{1.0003}{1.3}\right)\sin{31.5°} = 0.402[/tex]

Solving for the angle W, we then get

[tex]W = \sin^{-1}(0.402) = 23.7°[/tex]

Answer:

147°

Step-by-step explanation:

Solving for x

To make a straight line the angles have to have a sum of 190. So 180°-33° would equal 147°

Answer:

x = 147°

Step-by-step explanation:

A line is 180°

180 - 33 = 147

step:

20% = 20/100

of = ×

20/100 × $1575.00

answer : $315

Step:

20% = 20/100

of = ×

20/100 × $1575.00

Answer : $315

Answer: x= 1/196

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex] = > b - \frac{7}{4} = - \frac{2}{3} [/tex]

[tex] = > b = - \frac{2}{3} + \frac{7}{4} [/tex]

[tex] = > b = \frac{ - 8 + 21}{12} [/tex]

[tex] = > b = \frac{13}{12} \: or \: 1 \frac{1}{12} [/tex]

40 grams of $ 1 per gram and 160 grams of $ 0.80 per gram should be used.

Since a grower wishes to mix a one peso per gram chemical with 80 cent per gram chemical to produce mixture of 200 grams chemical which he can sell for 84 cent a gram, to determine how many grams of each price of chemical should he use you must perform the following calculation:

Therefore, 40 grams of $ 1 per gram and 160 grams of $ 0.80 per gram should be used.

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Answer:

85%

Step-by-step explanation:

Answer:

the ansewer is 56

Step-by-step explanation:

a/(a+b) = 37/40

(a+b)/a = 40/37

a/a + b/a = 40/37

1 + b/a = 40/37

b/a = 40/37 - 1

= 3/37

1 - b/a = 34/37

(a-b)/a = 34/37

a/(a-b) = 37/34

Good Luck

Answer:

174 cm

Step-by-step explanation:

1.83 m to cm is 183.

So, 183 cm - 9 cm = 174cm

Hope you understand ^^

Answer:

174 cm

Step-by-step explanation:

1 meter is equal to 100 centimeter so 1.83m times 100 is equal to 183cm. so 183 cm-9cm=174 cm

Answer:

0.2254

Step-by-step explanation:

multiply using long multuplication

Answer:

it would be the first one!

Answer:

z ≥ -10, which is the first answer.

Step-by-step explanation:

As you can see, on the graph there is a point filled in with a line pointing to the right. That would mean the sign would be ≥, as Z is GREATER than or EQUAL TO.

Answer:

The answer is 12/7.

Step-by-step explanation:

To get the answer, simply subtract -1 to 5/7

[tex]\frac{5}{7} - (-1)\\[/tex]

Find the least common denominator.

[tex]\frac{5}{7} - (-1) = \frac{5}{7} - \frac{(-7)}{7}[/tex]

= 12/7

To check, add -1 to 12/7.

[tex]\frac{12}{7} + (-1)[/tex]

Find the least common denominator.

[tex]\frac{12}{7} + (-1) = \frac{12}{7} + \frac{(-7)}{7}[/tex]

= 5/7

Answer:

x = 10

Step-by-step explanation:

the unmarked angle in the triangle is the supplement to 6x - 7

180 - (6x - 7)

180 - 6x + 7

187 - 6x

The sum of the angles in a triangle is 180

therefor

(2x + 7) + 26 + (187 - 6x) = 180

Combine like terms

-4x + 220 = 180

Subtract 220 from both sides

-4x = -40

divide both sides by -4

x = 10

Answer:

[tex](x-1)^2(x+3)[/tex]

Step-by-step explanation:

For factorization the given expression you can use the Ruffini's rule.

First find the divisors of the independent term (these are the possible rational roots), in the given polynomial the independent term is [tex]3[/tex].

[tex]3 = \{\pm1,\pm3\}[/tex]

Now replace each number in the polynomial

[tex]P(1) = 1^3 + 1^2 - 5(1) + 3 = 0[/tex]

If the result is equal to 0 that's meaning that is a possible root of the polynomial. Then for know if is a root you have to divide the polynomial by [tex](x - \text{possible root})[/tex].

[tex](x^3 + x^2 -5x+3 )/ (x-1)[/tex]

In the first row are the coefficients of the given polynomial.

In the second row are the product between the coefficients and the independent term of the [tex]x-1[/tex].

The third row are the coefficients of the quotient polynomial (except the last that is the remainder).

[tex]\qquad 1\qquad 1\qquad-5\qquad \quad3\\1\qquad\qquad1\qquad \quad 2\qquad-3\\-------------\\ 0 \ \ \quad1 \qquad 2 \qquad -3 \qquad \quad 0[/tex]

Because the last term is the remainder and it's 0 you can factorizate [tex](x-1)[/tex] and the quotient polynomial is equal to

[tex]x^2 + 2x -3[/tex]

So our current expression is [tex](x-1)(x^2 + 2x -3)[/tex], however you can factorizate the quotient polynomial in:

[tex]x^2 +sx +rx +rs = x^2 + 3x -x + 3(-1) = (x + 3)(x -1)[/tex]

So the last expression is:

[tex](x-1)(x+3)(x-1) = (x-1)^2(x+3)[/tex]

Answer:

Step-by-step explanation:

The statement #5 and #7 give an outcome of the statement #8:

This is called transitive property of congruence:

Correct choice is C

Answer:

C. Transitive property of congruence

Step-by-step explanation:

Transitive property of congruence means, if one pair of lines or angles or triangles are congruent to a third line or angle or triangle, then the first line or angle or triangle is congruent to the third line or angle or triangle.

Step-by-step explanation: (5√5)−2x+1=

1

5

(125x)−3

0.008x+1=

1

5

(125x)−3

Step 1: Flip the equation.

1

5

(125x)−3=0.008x+1

Step 2: Add 3 to both sides.

1

5

(125x)−3+3=0.008x+1+3

1

5

(125x)=0.008x+4

Step 3: Divide both sides by 1/5.

1

5

(125x)

1

5

=

0.008x+4

1

5

125x=0.04x+20

The  value of x is 46.

Exponents and powers can be used to represent extremely big or extremely small numbers in a more straightforward fashion.

The number of times a number is multiplied by itself is defined by the exponent. The power is an expression that displays the same number or factor being multiplied repeatedly.

Given:

[tex](5\sqrt{5} )^{-2x+1 }= 1/5 * 125^{x-3}[/tex]

ow, using exponents and powers

[tex](5\sqrt{5} )^{-2x+1 }= 1/5 * 125^{x-3}[/tex]

[tex](5\sqrt{5} )^{-2x+1 }= 1/5 * 5 ^{3(x-3)\\[/tex]

[tex](5\sqrt{5} )^{-2x+1 }= 5^{3x - 9 - 1}[/tex]

[tex](5\sqrt{5} )^{-2x+1 }= 5^{3x -10}[/tex]

[tex]5^{3/2(-2x+ 1)}= 5^{3x - 10}[/tex]

[tex]5^{-x+ 3/2}= 5^{3x -10}[/tex]

Now, Comparing

-x+ 3/2 = 3x- 10

-x - 3x = -10 - 3/2

-4x = -23/2

x= 46

Hence, the value of x is 46.

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Answer:

In which town does the total number have meaning?

Answer: Town B

For Town A, it does not, cannot, does not.

For Town B, does, can, does.

9514 1404 393

Answer:

  T = I/(PR)

Step-by-step explanation:

Solve for T the same way you solve for the variable in any proportion: divide both sides of the equation by its coefficient.

  I = PRT

  I/(PR) = PRT/(PR)

  I/(PR) = T

_____

Additional comment

Please note that parentheses are required around the factors of the denominator when you write this in plain text.

  I/PR ≠ I/(PR)

  I/PR = (I/P)R . . . . . the order of operations requires the division occur before the multiplication

If you write this by hand, or in typeset form, the horizontal fraction bar serves to group the factors of the denominator:

  [tex]T=\dfrac{I}{PR}[/tex]

Answer:

Step-by-step explanation:

Total Number of possible outcomes = 52

Getting a jack from a deck of 52 cards for a certain event (A) = 4

The Probability (A) = 4/ 52

= 1 / 13 = 0.07

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Answer:

w=[tex]\frac{V}{lh}[/tex]

Step-by-step explanation:

just divide lh from both sides

14(6x-2)=x-2

84x-28=x-2

84x-x= 28-2

83x = 26

x = 26/83

there are no corect answer below

ok done. Thank to me :>

The second derivative of the composition between two functions is described by the following expression:

[tex]\frac{d^{2}}{dx^{2}} (f\,\circ\,g\,(x)) = \frac{d^{2}f}{dg^{2}}\cdot \frac{dg}{dx} + \frac{df}{dg} \cdot \frac{d^{2}g}{dx^{2}}[/tex]

Mathematically speaking, a composition between two functions is defined by the following operation:

[tex]f\,\circ\,g\,(x) =f(g(x))[/tex] (1)

By Chain Rule we get the first and second derivatives of the composition:

First derivative

[tex]\frac{d}{dx} (f\,\circ\,g\,(x)) = \frac{df}{dg}\cdot \frac{dg}{dx}[/tex] (2)

Second derivative

[tex]\frac{d^{2}}{dx^{2}} (f\,\circ\,g\,(x)) = \frac{d^{2}f}{dg^{2}}\cdot \frac{dg}{dx} + \frac{df}{dg} \cdot \frac{d^{2}g}{dx^{2}}[/tex] (3)

The second derivative of the composition between two functions is described by the following expression:

[tex]\frac{d^{2}}{dx^{2}} (f\,\circ\,g\,(x)) = \frac{d^{2}f}{dg^{2}}\cdot \frac{dg}{dx} + \frac{df}{dg} \cdot \frac{d^{2}g}{dx^{2}}[/tex]

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