PLZZZ HELP MEEE

emmainna [20.7K]

To solve this equation we must first plug in 5 for x.

$\frac{1}{3}$ (5 - 4x)
↓
$\frac{1}{3}$ (5 - 4 · 5)

The next step would be to multiply 4 by 5 to get rid of the parenthesis within our main parenthesis.

$\frac{1}{3}$ (5 - 4 · 5)

$\frac{1}{3}$ (5 - 20)

The next step would be to subtract both of the numbers within the parenthesis.

$\frac{1}{3}$ (-15)

Now we must multiply.

- $\frac{1}{3}$ · 15 = - $\frac{1×15}{3}$

- $\frac{15}{3}$

Now we simplify our answer to get our final answer.

-5

8 0

3 years ago

The law of exponent used in is _____

Lady bird [3.3K]

A is the answer I just did that on my usa test prep

5 0

2 years ago

Read 2 more answers

A jar contains only pennies, nickels, dimes, and quarters. There are 29 pennies, 13 dimes, and 21 quarters. The rest of the

aleksklad [387]

Answer:

First Question: 63 coins are not nickels.

Second Question: 63+n=91

Step-by-step explanation:

For your first question: 29+21=50 and 50+13=63. 63 coins are not nickels.

For your second question: You can do 29+21+13+n=91, I just simplified and said 63+n=91. You can choose.

Hope that this helps! Good luck with what you're working on!

6 0

2 years ago

Read 2 more answers

Without solving determine the number of real solutions for each quadratic equation

seropon [69]

Answer:

b^2-4b+3=0

b²-3x-b+3=0

b(b-3)-1(b-3)=0

(b-3)(b-1)=0

either

b=3 or b=1

.

2n^2 + 7 = -4n + 5

2n²+4n+7-5=0

2n²+4n+2=0

2(n²+2n+1)=0

(n+1)²=0/2

:.n=-1

.

x - 3x^2 = 5+ 2x - x^2

0=5+ 2x - x^2-x +3x^2

0=5+x+2x²

2x²+x+5=0

comparing above equation with ax²+bx +c we get

a=2

b=1

c=5

x={-b±√(b²-4ac)}/2a ={-1±√(1²-4×2×5)}/2×1

={-1±√-39}/2

3 0

2 years ago

Find the critical numbers of the function f(x) = x6(x − 1)5 what does the first derivative test tell you that the second derivat

patriot [66]

The given function is f(x) = x⁶(x-1)⁵

The first derivative is
f'(x) = 6x⁵(x-1)⁵ + 5x⁶(x-1)⁴
= x⁵(x-1)⁴(6x - 6 + 5x)
= x⁵(x-1)⁴(11x - 6)
The critical values are the zeros of f'(x). They are
x = 0, 6/11, and 1.
The critical values indicate that turning points exist for f(x) at the critical points. However, we do not know the nature of the turning points.

Write the first derivative in the form f'(x) = (x-1)⁴(11x⁶ - 6x⁵).
The second derivative is
f''(x) = 4(x-1)³(11x⁶ - 6x⁵) + (x-1)⁴(66x⁵ - 30x⁴)
= (x-1)³(44x⁶ - 24x⁵ + 66x⁶ - 30x⁵ - 66x⁵ + 30x⁴)
= (x-1)³(110x⁶ - 120x⁵ + 30x⁴)
= 10x⁴(x-1)³(11x² - 12x + 3)
The sign of f''(x) at the critical values tell us the nature of the turning point.

f''(0) = 0, therefore a point of inflection exists at x = 0.
f''(6/11) > 0, therefore a local minumum exists at x = 6/11.
f''(1) = 0, therefore a point of inflection exists at x=1.

The graphs shown below confirm these results.

8 0

2 years ago