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3 years ago

Two quantities, x and y, are directly proportional. If x is halved, what happens to y?

2 answers:

5 0

Y should also be halved.

For example, if x=4 and y=2, x=2y.
If x is halved for x=2, you get 2=2y, or y=1, which is still one half of x, so the proportion remains the same.

raketka [301]3 years ago

5 0

Answer:

y is also halved

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $y/x=k$ or $y=kx$

If x increases then y increases

If x decreases then y decreases

therefore

If x is halved then y is also halved

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A translation moves point V(–2, 3) to V prime (negative 2, 7). Which are true statements about the translation? The point moves

Travka [436]

Answer:

The transformation rule is (x, y) → (x + 0, y + 4).

There occurs a transformation of vertical translation.

Step-by-step explanation:

A translation moves point V(–2,3) to V'(-2,7).

We have to choose the true statement about the translation from the given options.

The transformation rule is (x, y) → (x + 0, y + 4).

There occurs a transformation of vertical translation. (Answer)

{Since the x-value does not change and the y-value changes by 4 unit to the upward direction}

5 0

3 years ago

Read 2 more answers

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

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