PLEASE HELP!!!<br> SHOW YOUR ANSWER AND REASONING HURRY PLEASE

otez555 [7]

Answer:

D

Step-by-step explanation:

If the lines are parallel, the slope of both of them are going to be the same. So if one line is 3, the other one will be too.

6 0

3 years ago

Read 2 more answers

The length of a rectangle is twice the width. The area is 50 square inches. Find the

Airida [17]

Answer:

The dimensions are 5 and 10 inches

Step-by-step explanation:

The area is 50 square inches and the length is twice the width. 10 is the length, which is two times 5. 10 times 5 is 50.

The length is 10 and the width is 5.

4 0

3 years ago

Which equation is true when the value of x is -12

Sophie [7]

Answer:

Which equation is true when the value of x is -12 HERE YOU GO!!

Step-by-step explanation:

tricky ... let's see ...

I notice that if we subtract xy from both sides we get

7x + xy - xy = xy - xy + 21

then

7x = 21

and

x = 21/7 = 3

so there is only one value of x that satisfies the equation

x = 3

Going back to the original equation we see that any value of y will satisfy the original equation

we can see this by rearranging things:

7x + xy - xy = 21

here, I have performed the subtraction of xy on the right side as above, but have left the left side undone

(so we don't lose the presence of y)

Note that the above can also be written as

7x + (x - x)y = 21

or

7x + 0y =21

now, since anything times zero equals zero,

y may be any number.

Let's summarize:

1) x = 3

2) y = anything

looking back at the original question;

1) the equation is true for all ordered pairs

FALSE (only one x works, not all x)

2) there are no x and y pairs for which the equation is true.

FALSE (x=3, y=anything) makes it true, i.e. (3,1)

3) For each value of x, there is one and only one value of y that makes the equation true,

FALSE for each value of x, for the one value of x, x=3, y can be any number, which is an infinite number, not one

4) for each value of y, there is one and only on value of x that makes the equation true.

TRUE!! for all the infinite values of y you may pick, there is one and only one x you may pick, x = 3

ANSWER: STATEMENT 4 is CORRECT (TRUE)

3 0

3 years ago

Given that g(x) = 2x ^ 2 - 2x + 8 , find each of the following. a) g(0) b) g(- 2) C) g(3) d) g(- x) e) g(1 - t)

raketka [301]

Answer:

$g(-2)=20$

Step-by-step explanation:

Given $g(x)=2x^2-2x+8$ , substitute what is in the parentheses for $x$ to find an output.

For $g(-2)$ , the term $-2$ is in the parentheses. Thus, substitute $x=-2$ into $2x^2-2x+8$ to find $g(-2)$ :

$g(-2)=2(-2)^2-2(-2)+8,\\g(-2)=2\cdot 4+4+8,\\g(-2)=8+4+8,\\g(-2)=\boxed{20}$

5 0

3 years ago

Read 2 more answers

How many ways can eight letters be arranged into groups of five where order matters and the first two letters are already chosen

fenix001 [56]

Answer:

120

Step-by-step explanation:

Since we're dealing with a problem where the order matters and the first two letters are already chosen we need to subtract the number of letters and the number of available slots per group.

We use the permutation formula to find the answer, but before that let's check values.

n = 8

k = 5

Now since there are two letters already chosen we have to deduct two from both the value of n and k.

n = 6

k = 3

Now we can use the permutation formula:

$_{n}P_{k}=\dfrac{n!}{(n-k)!}$

$_{6}P_{3}=\dfrac{6!}{6-3)!}$

$_{6}P_{3}=\dfrac{6!}{3!}$

$_{6}P_{3}=\dfrac{6*5*4*3*2*1}{3*2*1}$

The 3*2*1 cancels out and leaves us with:

$_{6}P_{3}=6*5*4$

$_{6}P_{3}=120$

So there are 120 possible ways to arrange eight letters into groups of five where order matters and the first two letters are already chosen.

6 0

3 years ago

Read 2 more answers

PLEASE HELP!!! SHOW YOUR ANSWER AND REASONING HURRY PLEASE