All categories

4 years ago

Solve this equation. (1/2)^x-3= 16^3x-1

Mathematics

1 answer:

Vesna [10]4 years ago

4 0

This question is unsolvable this might be a trick question but it is not solvable. Hope this helps! ;D

You might be interested in

2/5 × 6/7= simplify form

kirill115 [55]

Answer:

2 x 6 = 12

5 x 7 = 35

12/35

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Solve for x C. 120 2x 3x D. (2x+6) X 126

brilliants [131]

Answer:

x=32

Step-by-step explanation:

i took the test

4 0

3 years ago

Lynn multiplies 7/8 Times a number the product is less than 7/8 which could be Lynn's number

Brums [2.3K]

Answer: 6/8 <;3..?

Step-by-step explanation:

7 0

3 years ago

Evaluate using Definite integrals

swat32

Since $[0,4]=[0,1]\cup(1,4]$ , we can rewrite the integral as

$\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt$

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

$\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt$

Both integrals are quite immediate: you only need to use the power rule

$\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}$

to get

$\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4$

Now we only need to evaluate the antiderivatives:

$\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15$

So, the final answer is 15.

4 0

3 years ago

I'm completely lost and don't really understand this.

kompoz [17]

For the third one, on the left side do 3-51.29 and keep doing it like that and you should get all the answers you need :)

6 0

3 years ago