enya exchanges $200 for euros (€). Suppose the conversion rate is €1 = $1.321. How many euros should Kenya receive? (1 point) €264.20 €151.40 €132.10 €66.05

Answer: $ 0.41

Step-by-step explanation:

Given : The Factors shot 5 rolls of film with 36 exposures on each roll.

i.e. One roll contains 36 exposures .

It cost $14.85 to process each roll.

Then, the cost to process each exposure will be :_

[tex]\$14.85\div36=\$0.4125\\\\\approx\$0.41\ \ \ \ [\text{ Rounded to the nearest penny}][/tex]

Therefore, the cost to process each exposure= $0.41

The approximate instantaneous rate of change in the intensity of light at exactly 3m from the flash bulb is -20/27 lux per meter.

The photographer uses a light meter to measure the intensity of light from a flash bulb. The intensity of the flash bulb, I, in lux, can be represented by the function I(d) = 10/d^2, where d is the distance from the light in meters.

To determine the approximate instantaneous rate of change in the intensity of light at exactly 3m from the flash bulb, we need to find the derivative of the intensity function with respect to d. Taking the derivative, we get dI/dd = -20/d^3.

Substituting d = 3 into the derivative, we get dI/dd = -20/3^3 = -20/27.

Therefore, the approximate instantaneous rate of change in the intensity of light is -20/27 lux per meter at exactly 3m from the flash bulb.

The approximate instantaneous rate of change in the intensity of light at exactly 3m from the flash bulb is approximately -0.741 lux/m.

To determine the instantaneous rate of change in the intensity of light at exactly 3m from the flash bulb, we need to find the derivative of the intensity function with respect to distance.

The intensity function is given as I(d) = 10/d^2, where d is the distance from the light in meters. To find the derivative, we can use the power rule for differentiation.

Taking the derivative of the intensity function gives us: dI/d(d) = -20/d^3.

At exactly 3 meters, the instantaneous rate of change in the intensity of light can be calculated by substituting d=3 into the derivative equation: dI/d(d) = -20/3^3 = -20/27. Therefore, the approximate instantaneous rate of change in the intensity of light at exactly 3m from the flash bulb is approximately -0.741 lux/m.

Answer:

Sample response: The number 4 only has three factors: 1, 2, and 4. Adam probably counted the factor 2 twice.

Step-by-step explanation:

The transformation of a function may lead to the rigid motion based on what type of transformation you applied in it. Thus, option D is correct.

We need to check the transformation that is the rigid motion for the function f(x).

Now, the different transformation is given below:

Let us take the transformations of the function from the top to the bottom.

(1) In the first transformation, we changed our function f(x) to 2f(x)+3.

When we used multiplication of division operations towards any functions then it will change the amplitude of the function, and hence it is not a rigid motion.

(2) In the second transformation, we changed our function f(x) to f(x/2)+1.

When we used multiplication of division operations towards any functions then it will change the amplitude of the function, and hence it is not a rigid motion.

(3) In the third transformation, we changed our function f(x) to 2f(x/2).

When we used multiplication of division operations towards any functions then it will change the amplitude of the function, and hence it is not a rigid motion.

(4) In the fourth transformation, we changed our function f(x) to f(x-1/2)+3.

When we used multiplication of division operations towards any functions then it will change the amplitude of the function, and when we use addition or subtraction towards any function then the amplitude remains constant. Hence, it is a rigid motion.

Thus, option D is correct.

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-6 because you keep change and flip the equation so....

8/-3 stays the same

you change the equation from a divide to a times

and u flip the -4/9 to get 9/-4

8/-3 X 9/-4

you can then just times the top and bottom to get

72/-12= -6

To solve the equation C = 5b + 200 for b in terms of C, subtract 200 from both sides of the equation and then divide both sides by 5.

To solve the equation C = 5b + 200 for b in terms of C, we need to isolate the variable b. First, subtract 200 from both sides of the equation:

C - 200 = 5b

Next, divide both sides of the equation by 5:

b = (C - 200)/5

So, the equivalent equation when solved for b in terms of C is b = (C - 200)/5.

The correct answer is:

12 rides

Explanation:

You have $35 to spend.  Taking away the admission price, $12.99, we have

35-12.99 = $22.01 left.

Each ride costs $1.75.  This means you can ride

22.01/1.75 = 12.577 or 12 rides.  (This does not round up, as you would not have enough money to ride 13 rides.)

divide total depth by time:

-56 /4 =14

 so it is diving at 14 miles per minute

 so after the first minute it was at -14 miles

The modifications made to the list of assets and liabilities that would result in an increase in net worth is Option D. Net worth is defined as the value of all the non-financial and financial assets owned by an individual or an institution.

All of the above (Putting $100 in savings, Paying $100 on credit cards and Getting paid $100)

Its the letter D ik because i just do aha

Final answer:

The product of 7 times 300 is 2100, which can be expressed as 7 times 21 hundreds when considering that 300 equals 3 hundreds.

Explanation:

The student asked to find the product of 7 times 300, which can be represented algebraically as 7 × 300. To express this in terms of hundreds, we would have 7 × 3 hundreds, since 300 is the same as 3 hundreds. Following the multiplication of 7 and 3, we get 21 hundreds, which is equal to 2100. Therefore, the completed statement would be 7 × 300 = 7 × 21 hundreds.

This multiplication utilizes the concept where multiplying by a power of ten, like hundreds (102), is equivalent to moving the decimal point to the right in this case by two places.

The number of candy bars Roland sold during the three weeks will be 152.

Algebra is the study of abstract symbols, while logic is the manipulation of all those ideas.

Roland sold 26 candy bars during the first week.

The number of treats he sold during the subsequent week was 4 under twice the quantity of pieces of candy he sold during the principal week.

Then the number of candies sold in the second week will be calculated as,

⇒ 2 x 26 - 4

⇒ 48

The number of sweet treats he sold during the third week was 6 a greater number than 1 ¹/₂ time the number of pieces of candy he sold during the subsequent week. Then the number of candies sold in third week will be given as,

⇒ 48 x (1 ¹/₂) + 6

⇒ 48 x 1.5 + 6

⇒ 72 + 6

⇒ 78

The total candies are calculated as,

⇒ 26 + 48 + 78

⇒ 152

The number of candy bars Roland sold during the three weeks will be 152.

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The equation of the line is [tex]\( y = \frac{1}{2}x - 2 \)[/tex].

To find the equation of a line with a given slope passing through a specific point, you can use the point-slope form of a linear equation: [tex]\( y - y_1 = m(x - x_1) \)[/tex], where ( m ) is the slope and [tex]\( (x_1, y_1) \)[/tex] is the given point.

Given that the slope [tex]\( m = \frac{1}{2} \)[/tex] and the point ( (2, -1) ), we substitute these values into the point-slope form:

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

This simplifies to:

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

To remove the fraction, distribute [tex]\( \frac{1}{2} \)[/tex]:

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

Now, isolate ( y ) by subtracting 1 from both sides:

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

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

So, the equation of the line is [tex]\( y = \frac{1}{2}x - 2 \)[/tex].