Ask question

Ask question

All categories

WINSTONCH [101]

3 years ago

13

Two lines pass through the origin. The lines have slopes that are opposite. Compare and contrast the lines. What is the slope of

the x axis? Explain

1 answer:

Ira Lisetskai [31]3 years ago

6 0

The compare is hat they both have the same y-intercept, the contrast is that tone line is positive and the other line is negative

You might be interested in

Is 11/16 bigger than 3/4

Goryan [66]

No 3/4 is bigger

if you multiply the top and bottom of 3/4 by 4, you get 12/16.

11/16 <12/16

8 0

3 years ago

Please awnser correct awnser! < 3

Gwar [14]

Answer:

y = (1/4)x + 0

or

y = (1/4)x

the (1/4) is supposed to be a fraction of 1 in the numerator and 4 in the denominator

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

What is the sign of f on the interval -2

TEA [102]

Answer:

$f$ is sometimes positive and sometimes negative.

Step-by-step explanation:

$f(x)=(x-3)(x+2)(x+4)(x+4)(x-1)(2x-9)$

Take $x=-1\in(-2,\frac{9}{2})\ as-2$

$f(-1)=(-1-3)(-1+2)(-1+4)(-1-1)(-2-9)\\\\=(-4)(1)(3)(-2)(-11)\\\\=-264\\\\f(-1)$

Take $x=2\in(-2,\frac{9}{2})\ as\ -2$

$f(2)=(2,-3)(2+2)(2+4)(2-1)(2\times2-9)\\\\=(-1)(4)(6)(1)(-5)\\\\=120\\\\f(2)>0$

Hence $f(x)$ for $x=-1$ and $f(x)>0$ for $x=2$

So $f$ is sometimes positive and sometimes negative in the interval.

3 0

3 years ago

Which expression is equivalent to the area of square A, in square inches?

Setler [38]

Answer:

length of 1 side of A, using the Pyth. Thm. and the dimensions of the other two squares: (side of A)^2 = (10 in)^2 + (24 in)^2. Then:

(side of A)^2 = 100+ 576 in^2 = 676 in^2.

Here I have not bothered to solve for the length of the side of A, since we want the area of square A. But if you do want the side length, find it: sqrt(676) = 26 in. Then the area of A is (26 in)^2 = 676 in^2.

Then the area of square A is (26 in)^2 =

Read more on Brainly.com - brainly.com/question/10676137#readmore

Step-by-step explanation:

8 0

3 years ago

Given the arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103,...

omeli [17]

We are given

arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103

so, first term is 124

u(1)= 124

now, we can find common difference

$d=u(2)-u(1)$

$d=117-124$

$d=-7$

now, we can find kth term

$u(k)=u(1)+(k-1)d$

now, we can plug values

and we get

$u(k)=124+(k-1)*-7$

$u(k)=124-7k+7$

$u(k)=131-7k$

u(k) must be negative

so,

$u(k)=131-7k$

$131-7k$

now, we can solve for k

$7k>131$

$k>18.714$

so, it's closest integer value is

$k=19$ ..............Answer

3 0

3 years ago