1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
malfutka [58]
4 years ago
14

Helpppp with homework

Mathematics
2 answers:
Shtirlitz [24]4 years ago
5 0

Answer:

Pi^3 * x^4

Step-by-step explanation:

Marysya12 [62]4 years ago
5 0
The answer is pi to the 3rd and x to the 4th
Good luck!!
You might be interested in
Solve the quadratic equation by taking square roots.
Stolb23 [73]

Answer:

see explanation

Step-by-step explanation:

1

Given

2x² - 16 = 0 ( add 16 to both sides )

2x² = 16 ( divide both sides by 2 )

x² = 8 ( take the square root of both sides )

x = ± \sqrt{8} = ± 2\sqrt{2}

------------------------------------

2

Given

- 5x² + 9 = 0 ( subtract 9 from both sides )

- 5x² = - 9 ( divide both sides by - 5 )

x² = \frac{9}{5} ( take the square root of both sides )

x = ± \sqrt{\frac{9}{5} } = ± \frac{3}{\sqrt{5} }

-----------------------------------------

3

Given

6x² - 15 = 27 ( add 15 to both sides )

6x² = 42 ( divide both sides by 6 )

x² = 7 ( take the square root of both sides )

x = ± \sqrt{7}

4 0
3 years ago
Find the missing side lengths (Use sqrt in place of a square root symbol)
Kitty [74]

Step-by-step explanation:

Because this is a right triangle and it is given that one angle is 45 than also the ather angle is 45 so this triangle has 2 equal sides.So also the ather side is 6 .

Hypotenuse can be found by pythagorean theorem

6^2+6^2=x^2

36+36=x^2

x=√72

8 0
3 years ago
Use Theorem 2.1.1 to verify the logical equivalence. Give a reason for each step. -(pv –q) v(-p^q) = ~p
sertanlavr [38]

Answer:

The statement \lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) is equivalent to \lnot p, \lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv \lnot p

Step-by-step explanation:

We need to prove that the following statement \lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) is equivalent to \lnot p with the use of Theorem 2.1.1.

So

\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv

\equiv (\lnot p \land \lnot(\lnot q))\lor(\lnot p \land \lnot q) by De Morgan's law.

\equiv (\lnot p \land q)\lor(\lnot p \land \lnot q) by the Double negative law

\equiv \lnot p \land (q \lor \lnot q) by the Distributive law

\equiv \lnot p \land t by the Negation law

\equiv \lnot p by Universal bound law

Therefore \lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv \lnot p

4 0
3 years ago
Given f(x)=-x-1, solve for x when f(x)=6.
Yanka [14]

<u>Answer:</u>

x = -7

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

<em>f(x) </em>= -x - 1

when <em>f(x) </em>= 6:

- x - 1 = 6

-x = 7                   [Add 1 to both sides]

x = -7                   [Divide both sides by -1]

3 0
2 years ago
Enter the ratio as a fraction in lowest terms<br> 112 minutes to 40 minutes
bulgar [2K]

Answer:

2/1

Step-by-step explanation:

6 0
3 years ago
Other questions:
  • Identify the infinite sequence generated by the formula an =(-1)^n * 2^n
    12·1 answer
  • What is the the volume of the composite figure to the nearest whole number? The figure is not drawn to scale. Please show all wo
    11·2 answers
  • Ryans living room is 10 feet wide, 12 feet long, and 10 feet high. If one gallon of paint covers 400 square feet of surface area
    6·1 answer
  • The label on 1-gallon can of paint states that the amount of paint in the can Is sufficient to paint at least 400 square feet (o
    11·1 answer
  • I need help BAD with this question
    12·1 answer
  • Which pair or pairs of polygons are congruent ?
    10·1 answer
  • A pet store has 7 cats. they weigh, 16, 16, 5, 10, 12, 13, 16 (in pounds)
    13·1 answer
  • Please help me with this thank you
    13·2 answers
  • Help me plssssssssss ASAP
    12·2 answers
  • Given: AB || DC, AD || BC<br> Prove:
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!