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3 years ago

Help please with this question

Mathematics

1 answer:

Svetllana [295]3 years ago

8 0

Step-by-step explanation:

0 = x² - 6x - 27

a = 1, b = -6, c = -27 Factor using AC method.

ac = (1)(-27) = -27

Factors of -27 that add up to -6 are -9 and 3.

Therefore:

0 = (x - 9) (x + 3)

x = -3, x = 9

The x-intercepts are (-3, 0) and (9, 0).

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My brainly username is kinda hot

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Answer: Lol, it just says “Invisible”, but ngl it is pretty cool

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3 years ago

What are the coordinates of point A? A (–4, 3) B (3, –4) C (–3, –4) D(–4, –3)

fomenos

Answer:

The coordinates of point a according to graph would be (-4,-3).

x = -4 , y = -3

hope it helps!

6 0

3 years ago

Read 2 more answers

7 Solve for x. x + 1 12 15 X =

dangina [55]

X=4. First, you do 15/12 and get 1.25. You do this to see what number you have to multiply 12 to, to get 15. Since the sides a proportional it should be the same for the sides with their variables. Then you create an expression that puts the numbers the same place where 15 and 12 were to equal 1.25. You cross multiply, solve and get 4.

6 0

3 years ago

Can someone please help me solve this? thank you!:)

Nataliya [291]

First, we need to set up our two equations. For the picture of this scenario, there is one length (L) and two widths (W) because the beach removes one of the lengths. We will have a perimeter equation and an area equation.

P = L + 2W

A = L * W

Now that we have our equations, we need to plug in what we know, which is the 40m of rope.

40 = L + 2W

A = L * W

Then, we need to solve for one of the variables in the perimeter equation. I will solve for L.

L = 40 - 2W

Now, we can substitute the value for L into L in the area equation and get a quadratic equation.

A = W(40 - 2W)

A = -2W^2 - 40W

The maximum area will occur where the derivative equals 0, or at the absolute value of the x-value of the vertex of the parabola.

V = -b/2a

V = 40/2(2) = 40/4 = 10

Derivative:

-4w - 40 = 0

-4w = 40

w = |-10| = 10

To find the other dimension, use the perimeter equation.

40 = L + 2(10)

40 = L + 20

L = 20m

Therefore, the dimensions of the area are 10m by 20m.

Hope this helps!

4 0

3 years ago

Read 2 more answers

Consider the following. (See attachment)

Furkat [3]

Answer:

Area: 16

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Calculus

Derivatives

Derivative Notation

Integrals - Area under the curve

Trig Integration

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = 8sin(x) + sin(8x)$

$\displaystyle y = 0$

Bounds of Integration: 0 ≤ x ≤ π

Step 2: Find Area Pt. 1

Set up integral: $\displaystyle A = \int\limits^{\pi}_0 {[8sin(x) + sin(8x)]} \, dx$
Rewrite integral [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^{\pi}_0 {8sin(x)} \, dx + \int\limits^{\pi}_0 {sin(8x)} \, dx$
[1st Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 8\int\limits^{\pi}_0 {sin(x)} \, dx + \int\limits^{\pi}_0 {sin(8x)} \, dx$
[1st Integral] Integrate [Trig Integration]: $\displaystyle A = 8[-cos(x)] \bigg| \limits^{\pi}_0 + \int\limits^{\pi}_0 {sin(8x)} \, dx$
[1st Integral] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = 8(2) + \int\limits^{\pi}_0 {sin(8x)} \, dx$
Multiply: $\displaystyle A = 16 + \int\limits^{\pi}_0 {sin(8x)} \, dx$

Step 3: Identify Variables

Identify variables for u-substitution.

u = 8x

du = 8dx

Step 4: Find Area Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 16 + \frac{1}{8}\int\limits^{\pi}_0 {8sin(8x)} \, dx$
[Integral] U-Substitution: $\displaystyle A = 16 + \frac{1}{8}\int\limits^{8\pi}_0 {sin(u)} \, du$
[Integral] Integrate [Trig Integration]: $\displaystyle A = 16 + \frac{1}{8}[-cos(u)] \bigg| \limits^{8\pi}_0$
[Integral] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = 16 + \frac{1}{8}(0)$
Simplify: $\displaystyle A = 16$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Integration - Area under the curve

Book: College Calculus 10e

4 0

3 years ago