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ryzh [129]
2 years ago
5

How do I Determine if ratios are equivalent

Mathematics
1 answer:
Veseljchak [2.6K]2 years ago
4 0

Step-by-step explanation:

multiply the ratio by the second ratios second number

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There are 100m of ribbon. 1 1/2m are needed to make a bouquet. How man can be made? How many m of ribbon are left over ?
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Answer:

66 can be made

1m left

Step-by-step explanation:

100 \div 1.5 = 66.667(5s.f.)

Ribbons that can be made= 66

66 \times 1.5 = 99m

length used= 99m

length remaining= 100m- 99m= 1m

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3 years ago
Use the distributive property to expand this expression: 4(2z + 3y) =
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8z+12y

Step-by-step explanation:

hope this helps

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Which of the following is not a zero of the function shown below?
erica [24]

Answer:

B

Step-by-step explanation:

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3 years ago
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Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1
Shkiper50 [21]

Answer:

\displaystyle  -\frac{1}{2} \leq x < 1

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0

Simplifying by x-1 and taking x=1 out of the possible solutions:

\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

(6x^3+14x^2+10x+12)

is always positive and doesn't affect the result. It can be neglected. The expression

(x-\frac{1}{2})^2

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0

For the expression to be negative, both signs must be opposite, that is

(x+\frac{1}{2})\geq 0, (x-1)

Or

(x+\frac{1}{2})\leq 0, (x-1)>0

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

\displaystyle  -\frac{1}{2} \leq x < 1

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

\boxed{\displaystyle  -\frac{1}{2} \leq x < 1 }

7 0
2 years ago
What would prove that the two triangles are<br> congruent?
aniked [119]
The lines on the sides!
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3 years ago
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