I need help understanding this

Mathematics

2 answers:

vovikov84 [41]3 years ago

8 0

I'm assuming that the question's really just asking for the slopes of the straight lines that are both horizontal and vertical, as you wouldn't actually need to graph the rectangle in any way, since you know that a rectangle has two sets of parallel sides with one set of the sides longer than the other. Remember to just use your knowledge of special types of slope to figure this out.

telo118 [61]3 years ago

5 0

Well, the slope for each side is simple.

All you have to do to remember slope is rise over run.

So for example:
For Vertices E and F the Slope would be 3/9
For Vertices G to H the slope would be 3/9
For vertices G to F the slope would be 6/2
For vertices H to E the slope would be 6/2

What do you notice about slopes on opposite sides?
Answer:
( H to E,and G to F) Have the same slope
(And G to H and E to F.) Have the same slope.

What do you notice about slopes of adjacent lines?
adjacent means adjoining or next to. So can you figure this question out? Using the information above?

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?/8 = 3/?<br> I am missing 2 numbers for a equivalent fraction problem it is a bit complicated.

lyudmila [28]

?/8 = 3/?
Multiplying both sides by 8, then “?” gives

8? * ?/8 = 8? * 3/?
?*? = 24
The number, then, is sqrt(24), or 2 sqrt(6), which is approximately 4.90.

4 0

4 years ago

Please answer all parts.

andrew-mc [135]

Answer:

1. Line c

2. She subtract when she suppose to add

3. 2x-12+12 = 10+12

2x = 22

x = 22/2 = 11

x is 11

4 0

3 years ago

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

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