❗️HELP❗️

Pls help meh pls Find the area of the figure. (Sides meet at right angles.)

Vlad [161]

We find the area of the total square which is 8in • 8in=64in^2 . Now we find the area of the small quadrilateral which is 4in•4in=16 in^2.
Now we subtract the area of the small quadrilateral from the big quadrilateral’s area 64in^2 -16in^2 leaving us with the are of 48in^2 of the figure

3 0

3 years ago

Simplify the expression

finlep [7]

Answer:

$\frac{2*x - 2}{2*x} - \frac{3*x + 2}{4*x} = \frac{x - 6}{4*x}$

Step-by-step explanation:

We have the expression:

$\frac{2*x - 2}{2*x} - \frac{3*x + 2}{4*x}$

The first thing we want to do, is to have the same denominator in both equations, then we need to multiply the first term by (2/2), so the denominator becomes 4*x

We will get:

$(\frac{2}{2} )\frac{2*x - 2}{2*x} - \frac{3*x + 2}{4*x} = \frac{4*x - 4}{4*x} - \frac{3*x + 2}{4*x}$

Now we can directly add the terms to get:

$\frac{4*x - 4}{4*x} - \frac{3*x + 2}{4*x} = \frac{4*x - 4 - 3*x - 2}{4*x} = \frac{x - 6}{4*x}$

We can't simplify this anymore

3 0

3 years ago

Kyle walks 3 blocks south from his home to school, and jana walks 2 blocks north from her home to kyle's home. how far and in wh

Nostrana [21]

If Jana has to walk 2 blocks north to get to Kyle's home, then 3 blocks south to get to school, this means that she has just retraced the 2 blocks she walked to get to Kyle's home.

So the school is just 1 block south of Jana's home.

6 0

2 years ago

Identify the terms of each expression 7 + 5 p + 4r + 6 s

inn [45]

13+5p+4r you add 7+6=13 then you bring down the rest

3 0

3 years ago

Help -------------------------

34kurt

Answer:

(A) -3 ≤ x ≤ 1

Step-by-step explanation:

The given function is presented as follows;

h(x) = x² - 1

From the given function, the coefficient of the quadratic term is positive, and therefore, the function is U shaped and has a minimum value, with the slope on the interval to the left of h having a negative rate of change;

The minimum value of h(x) is found as follows;

At the minimum of h(x), h'(x) = d(h(x)/dx = d(x² - 1)/dx = 2·x = 0

∴ x = 0/2 = 0 at the minimum

Therefore, the function is symmetrical about the point where x = 0

The average rate of change over an interval is given by the change in 'y' and x-values over the end-point in the interval, which is the slope of a straight line drawn between the points

The average rate of change will be negative where the y-value of the left boundary of the interval is higher than the y-value of the right boundary of the interval, such that the line formed by joining the endpoints of the interval slope downwards from left to right

The distance from the x-value of left boundary of the interval that would have a negative slope from x = 0 will be more than the distance of the x-value of the right boundary of the interval

Therefore, the interval over which h has a negative rate of change is -3 ≤ x ≤ 1

8 0

3 years ago