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3 years ago

Can someone please help me with 4 and 5 ???

Mathematics

1 answer:

Montano1993 [528]3 years ago

5 0

Answer:

4. D

A: -25.5=a

B: b=-4

C: c=12

D: d=8

Step-by-step explanation:

4.Between a and d they are the 2 bigger values but D was the greatest out of all of them. - divided by a - will result as a positive and 7/9=0.77 and 19/12 *I looked for the least common factor and multiplied numerator and denominator to 12 depending on the denominators value. Ex: since one of my denominators was 4, I multiply the whole fraction by 3 to get 9/12 and the other was 5/6 times 2 is 10/12 and just add 10/12 by 9/12 which is 19/12.

A: All I did was multiply 8.5 and -3 and get -25.5=a.

B: I add 7 on both sides of the equation and -7 and 7 get canceled off and -11+7=-4. b=4.

C: I multiplied - to -3 and got 3, now I can subtract -3 on both sides. 15-3=12 so c=12.

D. I had to divide by 4 on both sides to get d by itself. 32/4=8 so final answer would be d=8.

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Answer to a unit test in math grade 6

Lelu [443]

Answer:

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Step-by-step explanation:

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2 years ago

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What is equivalent to -1/4y -2 1/4y + 1/2 (4-2y) a) -3y + 2 b) -3 1/2y + 2 c) -4y + 4

Amanda [17]

Answer:

B. -3 1/2y + 2

Step-by-step explanation:

Our expression is: $\frac{-1}{4} y-2\frac{1}{4} y+\frac{1}{2} (4-2y)$ .

Let's first distribute out that parentheses. Remember that distribution is simply taking the sum of the product of the outside term with each of the inside terms. Here, the outside term is 1/2 and the inside terms are 4 and -2y:

$\frac{1}{2} (4-2y)=\frac{1}{2} *4+\frac{1}{2} *(-2y)=2-y$

Now, we have:

$\frac{-1}{4} y-2\frac{1}{4} y+2-y$

We want to combine like terms, which means combining all the terms with y in them:

$\frac{-1}{4} y-2\frac{1}{4} y-y+2=\frac{-1}{4} y-\frac{9}{4} y-\frac{4}{4} y+2=\frac{-1-9-4}{4} y+2=\frac{-14}{4} y+2=\frac{-7}{2} y+2$

Remember that -7/2 can be written as the mixed number -3 1/2, so our final answer is:

-3 1/2y + 2

The answer is thus B.

<em>~ an aesthetics lover</em>

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3 years ago

Your task is to build a road joining a ranch to a highway that enables drivers to reach the city in the shortest time. The perpe

lubasha [3.4K]

Answer:

(a)In the attachment

(b)The road of length 35.79 km should be built such that it joins the highway at 19.52km from the perpendicular point P.

Step-by-step explanation:

(a)In the attachment

(b)The distance that enables the driver to reach the city in the shortest time is denoted by the Straight Line RM (from the Ranch to Point M)

First, let us determine length of line RM.

Using Pythagoras theorem

$|RM|^{2}=30^2+x^2\\|RM|=\sqrt{30^2+x^2}$

The Speed limit on the Road is 60 km/h and 110 km/h on the highway.

Time Taken = Distance/Time

Time taken on the road $=\frac{\sqrt{30^2+x^2}}{60}$

Time taken on the highway $=\frac{50-x}{110}$

Total time taken to travel, T $=\frac{\sqrt{30^2+x^2}}{60}+\frac{50-x}{110}$

Minimum time taken occurs when the derivative of T equals 0.

$T^{'}=\frac{x}{60\sqrt{30^2+x^2}}-\frac{1}{110}\\\frac{x}{60\sqrt{30^2+x^2}}-\frac{1}{110}=0\\\frac{x}{60\sqrt{30^2+x^2}}=\frac{1}{110}\\110x=60\sqrt{30^2+x^2}\\$

Square both sides

$12100x^2=3600(30^2+x^2)\\12100x^2=3240000+3600x^2\\12100x^2-3600x^2=3240000\\8500x^2=3240000\\x^2=\frac{3240000}{8500} =381.18\\x=\sqrt{381.18} =19.52$