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40

There isnt a way the LCD can go lower, because 5 only counts by 5's and 5 and 8 meet at 40.

Answer:

least common denominator of 5 and and 8 is  40

Answer:

D

Step-by-step explanation:

Final answer:

The like radical to the expression 3√5 is y√5, as both expressions have the square root index and the same radicand, which is 5.

Explanation:

The student is asking which radical expression is like the radical 3√5. Like radicals have the same index and radicand. The index is the degree of the root, and the radicand is the number under the radical sign. The expression 3√5 means 3 times the square root of 5, or in exponential form, 3 × 51/2. The like radical for 3√5 would also need to have a square root (index of 2) and the same radicand (5). Therefore, the like radical to 3√5 from the options provided would be y√5 because it has the same index (2) and radicand (5), only with a different coefficient (y instead of 3).

Additionally, expressing radicals as fractional exponents helps to identify like radicals. For example, using the property x² = √x we can understand that if we have the same base and exponent, we can consider the expressions to be like radicals. Hence, in this case, since both 3 and y are just coefficients, the root parts √5 are the same, making them like radicals.

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Answer:

B -3.

Step-by-step explanation:

The slope = (y2-y1)/(x2-x1)

= (-2-4)/(8-6)

= -6/2

= -3.

Step-by-step explanation:

If x is the kilograms of 30% chocolate, and y is the kilograms of 50% chocolate, then:

x + y = 200

0.30x + 0.50y = 0.46(200)

Solving the system of equations with substitution:

0.30x + 0.50(200 − x) = 0.46(200)

0.30x + 100 − 0.50x = 92

8 = 0.20x

x = 40

y = 200 − x

y = 160

The distributor needs 40 kg of 30% chocolate and 160 kg of 50% chocolate.

Final answer:

To obtain 200 kilograms of a 46% fat-content chocolate, the candy distributor needs to mix 40 kilograms of a 30% fat-content chocolate and 160 kilograms of a 50% fat-content chocolate.

Explanation:

This problem can be solved using a basic mixture problem method. Let's name the amount of the 30% fat-content chocolate as 'x' and the amount of the 50% fat-content chocolate as 'y'. The total weight of the resulting chocolate is provided in the problem, 200 kilograms, therefore we know that x + y = 200.

The total fat in the chocolates should be 46% of 200kg, or 92kg. This gives us another equation based on the fat content, 0.3x + 0.5y = 92.

Solving these two equations linearly, we find the values of x and y. The amount of 30% fat content chocolate (x) is 40 kilograms and the amount of 50% fat-content chocolate (y) is 160 kilograms.

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