A system of equations has 1 solution. If 4x - y = 5 is one of the equations, which could be the other equation?

Let a and b be non-zero integers

r-ruslan [8.4K]

Answer:

The correct answer is d

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

20 POINTS PLEASE HELP

Rom4ik [11]

What are the measurements of the container

8 0

3 years ago

Find the derivative.

krek1111 [17]

Answer:

$\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding/Factoring

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = \frac{\sqrt{x}}{e^x}$

Step 2: Differentiate

Derivative Rule [Quotient Rule]: $\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}$
Basic Power Rule: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}$
Exponential Differentiation: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}$
Simplify: $\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}$
Rewrite: $\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}$
Factor: $\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

7 0

2 years ago

Applying properties of Exponents in exercise,use the properties of exponents to simplify the expression.See example 1.

evablogger [386]

Answer: a) 15625, b) 0.2, c) 625, d) 0.008.

Step-by-step explanation:

Since we have given that

a) $(5^2)(5^3)$

As we know that

$a^m\times a^n=a^{m+n}$

So, it becomes,

$5^2\times 5^3=5^{2+3}=5^6=15625$

(b) $(5^2)(5^{-3})$

So, it becomes,

$5^2\times 5^{-3}=5^{2-3}=5^{-1}=\dfrac{1}{5}=0.2$

(c) $(5^2)^2$

As we know that

$(a^m)^n=a^{mn}$

So, it becomes,

$5^{2\times 2}=5^4=625$

(d) $5^{-3}$

$a^{-m}=\dfrac{1}{a^m}\\\\5^{-3}=\dfrac{1}{5^3}=\dfrac{1}{125}=0.008$

Hence, a) 15625, b) 0.2, c) 625, d) 0.008.

4 0

3 years ago

In a certain sequence of numbers, each term after the first is found by doubling and then adding $3$ to the previous term. If th

eimsori [14]

Given $a_7=125$ you can find a_6:

$a_6= \frac{125-3}{2} =61$
$a_5= \frac{61-3}{2} =29 \\ a_4= \frac{29-3}{2}=13 \\ a_3= \frac{13-3}{2} =5 \\ a_2= \frac{5-3}{2} =1 \\ a_1= \frac{1-3}{2}=-1$
Answer: -1

6 0

2 years ago