A function is graphed on the coordinate plane. What is the value of the function when x = 2 ? Question 4 options:

Determine algebraically all points where the graphs of XY equal 10 and Y equals X +3 intersect

Gelneren [198K]

XY=10
Y=X+3
X(X+3)=10
X^2+3X=10
X^2+3X-10
(X-2)(X+5)

6 0

3 years ago

The vertices of the triangle is A(-2,4), B(1,2), C(-2,-2)

spin [16.1K]

See description in the pic.

5 0

2 years ago

The perimeter of a rectangle is 128 cm, and the ratio of its sides is 5: 3. Find the area of a rectangle

Juliette [100K]

Answer:

Area = 960 cm^2

Step-by-step explanation:

Perimeter (P) = 2 (l + w)

P = 2l + 2w

Let the length be 5x

Let the width be 3x

From the question, P = 128.

Therefore:

128 = 2l + 2w

Divide both sides by 2.

l + w = 64

Recall we said:

l = 5x

w = 3x

So then:

5x + 3x = 64

8x = 64

Divide both sides by 8

x = 64/8

x = 8.

Since x = 8

Length = 5x = 5*8 = 40cm

Width =3x = 3 * 8 = 24cm

Area of a RECTANGLE:

Area = length * width

Area = 40 * 24

Area = 960 cm^2

7 0

3 years ago

FREEEEEEE POINTSSSSSSSSFREEEEEEE POINTSSSSSSSSFREEEEEEE POINTSSSSSSSSFREEEEEEE POINTSSSSSSSSFREEEEEEE POINTSSSSSSSSFREEEEEEE POI

monitta

Answer:

OMG THANK YOU SO MUCH I NEEDED THIS SO MUCH FOR MY MATH TEST!!!!

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

o-na [289]

Answers:

(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

(d) f is concave down at $(-\infty, 2)$

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

3 0

3 years ago