Omg please help!!!!! And fast if you can

Tom [10]

First of all gall, the end is NOT an exponent, its the unit of measurement for finding the area (thats square units)

Heres the answer:
31.5 square units.
Explanation:
The formula for the area of a triangle is bh/2
The formula for the area of a triangle is bh
(Use this formula and try to find the real explanation)

3 0

2 years ago

Read 2 more answers

If WXYZ is a square, which statements must be true? Check all that apply.

OLga [1]

Answer:

A might be right

hope this is helpful

8 0

2 years ago

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$

Since 9 is greater than 0, therefore, function is increasing on interval $(-\infty, -1) \text{ or } x$ .

Now we will check 0 for 2nd interval.

$f'(0)=3(0)^{2}-3=0-3=-3$

Since -3 is less than 0, therefore, function is decreasing on interval $(-1,1) \text{ or } -1$ .

We will check 2 from our 3rd interval.

$f'(2)=3(2)^{2}-3=3(4)-3=12-3=9$

Since 9 is greater than 0, therefore, function is increasing on interval $(1,\infty) \text{ or } x>1$ .

6 0

3 years ago

In ΔIJK, the measure of ∠K=90°, KJ = 65, IK = 72, and JI = 97. What is the value of the cosine of ∠J to the nearest hundredth?

kow [346]

Answer:

$\cos(J) = 0.67$

Step-by-step explanation:

Given

$\angle K = 90^o$

$KJ = 65$

$IK = 72$

$JI = 97$

Required

$\cos(J)$

The question is illustrated with the attached image.

From the image, we have:

$\cos(J) = \frac{KJ}{JI}$

This gives:

$\cos(J) = \frac{65}{97}$

$\cos(J) = 0.67010309278$

$\cos(J) = 0.67$ --- approximated

7 0

3 years ago

Check whether x² +4x +4 is prefect spuare expression or not . if so factorise the expression<br>

irinina [24]

Answer:

It is a perfect square expression.

It factors to (x+2)^2

===========================================================

Explanation:

The general format for a perfect square is

(a+b)^2 = a^2 + 2ab + b^2

Which can be seen through the use of the FOIL rule.

Compare x^2+4x+4 with a^2+2ab+b^2, and we have these three equivalences:

x^2 = a^2 .... first terms
4x = 2ab .... middle terms
4 = b^2 .... last terms

Since x^2 = a^2, we can apply the square root to both sides to get a = x. Similarly, 4 = b^2 leads to b = 2. We could get b = -2, but that would mean 2ab = 2x*(-2) = -4x instead of 4x. So we'll stick with b = 2 instead.

Because a = x and b = 2, we then can say:

(a+b)^2 = a^2 + 2ab + b^2

(x+2)^2 = x^2 + 2*x*2 + 2^2

(x+2)^2 = x^2 + 4x + 4

Showing that x^2+4x+4 factors to (x+2)^2.

7 0

2 years ago

Omg please help!!!!! And fast if you can​

Omg please help!!!!! And fast if you can