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2 years ago

(Number 8 answer ?) I’m having trouble with this concept.

Mathematics

1 answer:

coldgirl [10]2 years ago

5 0

Answer: C

I put the working in the other question you put up, but to put it briefly, multiply everything by 6 because all the options provided have a -6 inside that was meant to be the -1

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Evaluate the following. 106 A. 60 B. 100,000 C. 10,000 D. 1,000,000

Ivan

Answer: the answer would be c

Step-by-step explanation:

7 0

2 years ago

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Which oddered pair is the the system of equation?

Mnenie [13.5K]

Answer:

(3, -9)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS
Equality Properties

Algebra I

Solving systems of equations using substitution/elimination
Solving systems of equations by graphing

Step-by-step explanation:

Step 1: Define systems

-5x - 3y = 12

y = x - 12

Step 2: Solve for x

Substitution

Substitute in y: -5x - 3(x - 12) = 12
Distribute -3: -5x - 3x + 36 = 12
Combine like terms: -8x + 36 = 12
Isolate x term: -8x = -24
Isolate x: x = 3

Step 3: Solve for y

Define original equation: y = x - 12
Substitute in x: y = 3 - 12
Subtract: y = -9

Step 4: Graph systems

Check the solution set.

8 0

2 years ago

Does the point (0, 37) satisfy the equation y = 79x - -37? yes no

AlexFokin [52]

Answer:

yes

Step-by-step explanation:

y = 79x - -37

37 = 79(0) - - 37

37 = 0 - - 37

37 = 37

5 0

2 years ago

What formula can we use to find the circumference of a circle?

Assoli18 [71]

Answer:

B plz mrk brainliest

Step-by-step explanation:

6 0

2 years ago

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For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?

Bond [772]

Given that $r=\theta$ , then $r'=1$

The slope of a tangent line in the polar coordinate is given by:

$m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}$

Thus, we have:

$m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}$

Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

$\sin\theta+\theta\cos\theta=0 \\ \\ \theta\cos\theta=-\sin\theta \\ \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta$

Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

θ = 0

θ = 2.02875783811043

θ = 4.91318043943488

Part B:

For vertical tangent lines, $\frac{1}{m} =0$

Thus, we have:

$\cos\theta-\theta\sin\theta=0 \\ \\ \Rightarrow\theta\sin\theta=\cos\theta \\ \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta$

Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

θ = 4.91718592528713

3 0

3 years ago