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2 years ago

7

What is the circumference of a circle with a diameter of 2.9 cm?

2 answers:

steposvetlana [31]2 years ago

8 0

Answer:

C≈9.11

Step-by-step explanation:

slega [8]2 years ago

7 0

Answer:

The answer is 9.111

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Describe the nature of the roots for this equation.

kirza4 [7]

Answer:

Option D. two complex roots

Step-by-step explanation:

we know that

In a quadratic equation of the form $ax^{2} +bx+c=0$ the discriminant D is equal to

$D=(b^{2}-4ac)$

in this problem we have

$3x^{2} -2x+1=0$

so

$a=3\\b=-2\\c=1$

substitute the values

$D=(-2^{2}-4(3)(1))$

$D=(4-12)$

$D=-8$

The discriminant is negative

therefore

The quadratic equation has two complex roots

5 0

2 years ago

1. Thom has a scale model of his car. The

MariettaO [177]

$x =\frac{4}{3}inch$ is the size in wheels on the scale model .

<u>Step-by-step explanation:</u>

Correct Question : Tom has a scale model of his car. The scale factor is 1 : 12. If the actual car has 16-inch wheels, what size are the wheels on the scale model?

We have , The scale factor is 1 : 12. We need to find If the actual car has 16-inch wheels, what size are the wheels on the scale model .Let's find out:

Ratio of size of wheels to actual size of wheels is 1:12 , but actual car has 16-inch wheels So ,

⇒ $\frac{1}{12} = \frac{x}{16}$ { x is size of wheel in scale model }

⇒ $\frac{1(16)}{12} = \frac{x}{16}(16)$

⇒ $x =\frac{16}{12}$

⇒ $x =\frac{4}{3}inch$

Therefore , $x =\frac{4}{3}inch$ is the size in wheels on the scale model .

5 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

2 years ago

Factor the algebraic expression below in terms of a single trigonometric function. csc 2x - 1

hjlf

(csc(x) + 1)(csc(x) -1)

5 0

2 years ago

How many times larger is 70 than 7

Dennis_Churaev [7]

Answer:

63

Step-by-step explanation:

8 0

3 years ago