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
gtnhenbr [62]
3 years ago
11

Question 1

Mathematics
1 answer:
irina1246 [14]3 years ago
5 0

Answer:

Question A

Step-by-step explanation:

You might be interested in
A washer and a dryer cost $928 combined. The washer costs $78 more than the dryer. What is the cost of the dryer?
SCORPION-xisa [38]

Answer:

Step-by-step explanation:

$425

4 0
3 years ago
This question has two parts.
Evgesh-ka [11]

Answer:

C

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
I know you want to answer this question.
Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

\frac{1}{2} ^{x-4} - 3 = 4^{x-3} - 2

First, convert 4^{x-3} to base 2:

4^{x-3} = (2^{2})^{x-3}

\frac{1}{2} ^{x-4} - 3 = (2^{2})^{x-3} - 2

Next, convert \frac{1}{2} ^{x-4} to base 2:

\frac{1}{2} ^{x-4} = (2^{-1})^{x-4}

(2^{-1})^{x-4} - 3 =  (2^{2})^{x-3} - 2

Apply exponent rule: (a^{b})^{c} = a^{bc}:

(2^{-1})^{x-4} = 2^{-1*(x-4)}

2^{-1*(x-4)} - 3 = (2^{2})^{x-3} - 2

Apply exponent rule: (a^{b})^{c} = a^{bc}:

(2^{2})^{x-3} = 2^{2(x-3)}

2^{-1*(x-4)} - 3 = 2^{2(x-3)} - 2

Apply exponent rule: a^{b+c} = a^{b}a^{c}:

2^{-1(x-4)} = 2^{-1x} * 2^{4}, 2^{2(x-3)} = 2^{2x} * 2^{-6}

2^{-1 * x} * 2^{4} - 3 = 2^{2x} * 2^{-6} - 2

Apply exponent rule: (a^{b})^{c} = a^{bc}:

2^{-1x} = (2^{x})^{-1}, 2^{2x} = (2^{x})^{2}

(2^{x})^{-1} * 2^{4} - 3 = (2^{x})^{2} * 2^{-6} - 2

Rewrite the equation with 2^{x} = u:

(u)^{-1} * 2^{4} - 3 = (u)^{2} * 2^{-6} - 2

Solve u^{-1} * 2^{4} - 3 = u^{2} * 2^{-6} - 2:

u^{-1} * 2^{4} - 3 = u^{2} * 2^{-6} - 2

Refine:

\frac{16}{u} - 3 = \frac{1}{64}u^{2} - 2

Add 3 to both sides:

\frac{16}{u} - 3 + 3 = \frac{1}{64}u^{2} - 2 + 3

Simplify:

\frac{16}{u} = \frac{1}{64}u^{2} + 1

Multiply by the Least Common Multiplier (64u):

\frac{16}{u} * 64u = \frac{1}{64}u^{2} + 1 * 64u

Simplify:

\frac{16}{u} * 64u = \frac{1}{64}u^{2} + 1 * 64u

Simplify \frac{16}{u} * 64u:

1024

Simplify \frac{1}{64}u^{2} * 64u:

u^{3}

Substitute:

1024 = u^{3} + 64u

Solve for u:

u = 8

Substitute back u = 2^{x}:

8 = 2^{x}

Solve for x:

x = 3

4 0
4 years ago
Y=(6+3X)^2/3 Find the derivative of the function using the chain rule
fomenos
\bf y=(6+3x)^{\frac{2}{3}}\implies \cfrac{dy}{dx}=\stackrel{chain~rule}{\cfrac{2}{3}(6+3x)^{\frac{2}{3}-1}\cdot (0+3)}
\\\\\\
\cfrac{dy}{dx}=\cfrac{2}{3}(6+3x)^{-\frac{1}{3}}\cdot 3\implies \cfrac{dy}{dx}=2(6+3x)^{-\frac{1}{3}}\implies \cfrac{dy}{dx}=\cfrac{2}{\sqrt[3]{6+3x}}
5 0
4 years ago
Vicky is solving an equation where both sides are linear expressions. If the y-intercepts of the graphs are the same but the slo
Minchanka [31]

If the y-intercepts are the same, then the two linear have the same y-intercept as a solution. That is one common solution for both expressions. The two expressions intersect at that point. Since the slopes are different, there are no more common solutions since if two different lines intersect, they intersect at exactly one point and no more.

Answer: 1 intersection

5 0
4 years ago
Read 2 more answers
Other questions:
  • I need help with this please
    13·1 answer
  • Helloo please help! (You don’t need to give a explanation for the answer)
    14·1 answer
  • Josh thinks he can find examples of prime numbers that he can write as a product of two points
    9·1 answer
  • I will give brainliest to whoever gets it right
    8·1 answer
  • Find the area of the shaded part of circle Y.
    10·1 answer
  • Question 1
    7·1 answer
  • Please helppppppppppppppppppppppppppppp
    11·1 answer
  • Which inequality is a true statement select each correct answer
    9·1 answer
  • What is the difference between a uniform and a non-uniform probability model?
    13·1 answer
  • Solve For X, using the given triangle.
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!