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2 years ago

The quotient of the sum of 2t and 2 and twice the cube of s

1 answer:

8 0

The equation according to the given verbal sentence is $\frac{2 t+2}{2 s^{3}}$ (after solving this $\frac{t+1}{s^{3}}$ ).

Step-by-step explanation:

Rewrite the given verbal sentence and so it will be easier to translate.

Given: The quotient of the sum of 2 t and 2 and twice the cube of s

sum of 2 t and 2 can be written as

2 t + 2

twice the cube of s

$2 s^{3}$

Now, combine the above according to the given verbal. The quotient of the sum of 2 t and 2 and twice the cube of s means,

2 t + 2 divided by $2 s^{3}$

Now, the equation would be $\frac{2 t+2}{2 s^{3}}$

When solving this equation, we get

$\frac{2(t+1)}{2 s^{3}}=\frac{t+1}{s^{3}}$

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What is the y-intercept of the line shown below? A:3/4 B:2 C:3 D:4

olya-2409 [2.1K]

The y-intercept is the y value where the blue line crosses the Y axis which is the vertical black line.

The line crosses at the number 4, so the y-intercept is 4

Answer: D. 4

5 0

2 years ago

The formula f(x + 1) = Two-thirds(f(x)) defines a geometric sequence where f(1) = 18. Which explicit formula can be used to mode

astraxan [27]

Answer:

The explicit formula is Tn = 18[(2/3)^(n-1)]

Where n is the term we are looking for

Step-by-step explanation:

Here, we want to get an explicit formula to model the equation

Now, F(2) = 2/3 * f1 = 2/3 * 18 = 12

F(3) = 2/3 * f(2) = 2/3 * 12 = 8

F(4) = 2/3 * F(3) = 2/3 * 8 = 16/3

F(5) = 2/3 * 16/3 = 32/9

Thus, seeing how the equations are progressing, we can definitely see a pattern.

That is Tn = (2/3)^(n-1)(18)

7 0

3 years ago

Solve. -7x=x^2+12 help

Nat2105 [25]

Hey There!

The answer to your problem is $x=-3$ or $x=-4$ both answers are correct!

Step 1: Subtract x²+12 from both sides.

−7x−(x²+12)=x²+12−(x²+12)

−x²−7x−12=0

Step 2: Factor left side of equation.

$(-x-3)(x+4)=0$

Step 3: Set factors equal to 0.

$-x-3=0$ or $x+4=0$

Get your answer:

$x=-3$ or $x=-4$

3 0

2 years ago

Prove the following by induction. In each case, n is apositive integer. 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.

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

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

Evaluate x+3y when x=2, y-3

AlekseyPX

Answer:

-7

Step-by-step explanation:

x+3y

Let x =2 and y = -3

2 +3(-3)

2 -9

-7

6 0

3 years ago

Read 2 more answers