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3 years ago

3x + y = 4 x + 2y = -2

Mathematics

1 answer:

dsp733 years ago

7 0

3x+y=4 (y=-3x+4)

Plug it in for y x+2(-3x+4)=-2
x=2

Plug in for x 2+2y=-2
y=-2

Solution: (2,-2)

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4d + 5 >= 13 or 7d - 2 <12

vivado [14]

I'm guessing that the >= is a greater than or equal to sign, so the answer should be d>=2 and d<2, please tell me if I am wrong and I will do my best to correct it!

6 0

3 years ago

Find 62(9) . Express your answer as a fraction in simplest form

Bas_tet [7]

Answer:

558/1

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

What are the coordinates of the turning point for the function f(x) = (x - 1)3 - 3? A. (-1, -3)

Naddik [55]

ANSWER

The turning point is

$(1,-3).$

EXPLANATION

The function given to us is

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

At turning point,
$f '(x) = 0$

So we need to differentiate the given function and equate it to zero.

We using the chain rule of differentiation, we obtain,
$f '(x) =3 (x - 1) ^{2}$

We equate this to zero to obtain,

$3 (x - 1) ^{2} = 0$

We divide through by 3.

$(x - 1) ^{2} = 0$

We solve for x to get,

$x - 1 = 0$

$x = 1$

We substitute this x-value in to the function to obtain the corresponding y-value of the turning point.

$f(1) = (1- 1) ^{3} - 3$

$f(1) = 0 - 3$

$f(1) = - 3$

Therefore the turning point is

$(1,-3)$

C is the correct answer.

6 0

3 years ago

Read 2 more answers

Write the expression that shows 3 time the sixth power of 10

77julia77 [94]

3 x 10^6

Shouldn't be too hard for ya ;)

8 0

3 years ago

What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (5 points)

zzz [600]

Answer:

B) 4√2

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Parametric Differentiation

Integration

Integrals
Definite Integrals
Integration Constant C

Arc Length Formula [Parametric]: $\displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.$

Interval [0, π]

Step 2: Find Arc Length

[Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]: $\displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.$
Substitute in variables [Arc Length Formula - Parametric]: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx$
[Integrand] Simplify: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx$
[Integral] Evaluate: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}$

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametric Integration

Book: College Calculus 10e

4 0

3 years ago