Select one of the factors of 3x^2 + 10x+ 3

Evaluate this expression. 5^-3

sweet-ann [11.9K]

Answer:

$\Huge \boxed{\frac{1}{125}}$

Step-by-step explanation:

<h2>Order of operations</h2>

PEMDAS

Parenthesis, exponent, multiply, divide, add, and subtract from left to right.

Do exponent.

$\displaystyle \frac{1}{5^3}$

$\displaystyle 5^3=5\times5\times5=125$

$\displaystyle \frac{1}{125}=125$

$\large \boxed{\frac{1}{125}}$ , which is our answer.

Hope this helps!

4 0

2 years ago

The principal P is borrowed and the loan's future value A at time t is given. Determine the loan's simple interest rater.

Anni [7]

Interest rate = 7%

I= PRT where I is the interest earned, p is the principal, r is rate as a decimal, and t is time in years.

A - P will give us the interest earned.

12,150 - 9,000 = 3,150

3150 = (9000)(r)(5)
3150 = 45,000r
r = 0.07
r = 7%

4 0

2 years ago

What is the y value for the line that passes through (−2,3) and (0,8)

sp2606 [1]

Answer:

y=5x/2 + 9

Step-by-step explanation:

Check the above picture for workings

3 0

2 years ago

A company is designing a new cylindrical water bottle. The volume of the bottle will be 211 cm^3. The height of the water bottle

Zinaida [17]

Answer:

The correct answer is 2.912 cm.

Step-by-step explanation:

A company is designing a new cylindrical water bottle.

Volume of a cylinder is given by π × $r^{2}$ × h, where h is the height of the cylinder and r is the radius of the cylinder.

The volume of each bottle will be 211 $cm^{3}$ .

The height (h) of the water bottle be 7.9 cm.

Let the radius of the bottle be r cm.

∴ π × $r^{2}$ × h = 211 ; (π = 3.15)

⇒ $r^{2}$ × 24.885 = 211

⇒ r = 2.912

The radius of the water bottle is 2.912 cm.

4 0

2 years ago

Find the indicated limit, if it exists.

kondor19780726 [428]

Answer:

d) The limit does not exist

General Formulas and Concepts:

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$
Left-Side Limit: $\displaystyle \lim_{x \to c^-} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

Step-by-step explanation:

*Note:

In order for a limit to exist, the right-side and left-side limits must equal each other.

Step 1: Define

Identify

$\displaystyle f(x) = \left\{\begin{array}{ccc}5 - x,\ x < 5\\8,\ x = 5\\x + 3,\ x > 5\end{array}$

Step 2: Find Right-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^+} 5 - x$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} 5 - x = 5 - 5 = 0$

Step 3: Find Left-Side Limit

Substitute in function [Limit]: $\displaystyle \lim_{x \to 5^-} x + 3$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} x + 3 = 5 + 3 = 8$

∴ Since $\displaystyle \lim_{x \to 5^+} f(x) \neq \lim_{x \to 5^-} f(x)$ , then $\displaystyle \lim_{x \to 5} f(x) = DNE$

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

Unit: Limits

5 0

2 years ago