All categories

3 years ago

7. Factor Completely (1 Point) x^2– 25

Mathematics

1 answer:

mixas84 [53]3 years ago

7 0

Let's factor x2−25

x2−25

The middle number is 0 and the last number is -25.

Factoring means we want something like

(x+_)(x+_)

Which numbers go in the blanks?

We need two numbers that...

Add together to get 0

Multiply together to get -25

Can you think of the two numbers?

Try 5 and -5:

5+-5 = 0

5*-5 = -25

Fill in the blanks in

(x+_)(x+_)

with 5 and -5 to get...

(x+5)(x-5)

Answer:

(x+5)(x−5)

You might be interested in

Solving Exponential Equations (lacking a common base)<br> 31^d=38

Bond [772]

Answer:

d = log 38/log31

d = 1.06

Step-by-step explanation:

31 ^ d = 38

Take log of both sides since the exponential equation have different base (always)

d log 31 = log 38

d = log 38 / log 31

d = 1.58 / 1.491

d = 1.06.

We can actually check to see if this is correct.

Substituting d = 1.06

31^ 1.06 = 38.092

(You see, that's approximately correct)

Keep learning, maths is fun!

8 0

3 years ago

I need your help to find the value of x.round to the nearest hundred just the photo

True [87]

Answer:

Step-by-step explanation:

1. Cos 52° = adj/hyp

Cos 52° = x/13

x = 13×cos 52°

x = 8.00

2. Sin70° = opp/hyp

Sin70° = 30/x

x sin70° = 30

x = 30/sin70°

x = 31.93

x ≈ 32

3. Tan∅ = opp/adj

Tan∅ = 45/51

Tan∅ = 0.8824

∅ = tan-¹(0.8824)

∅ = 41.42°

∅ ≈ 41°

3 0

3 years ago

(7c) (-1) what is the answer I have no clue

N76 [4]

Answer:

-7c

Step-by-step explanation:

8 0

3 years ago

What is the sum of an 8-term geometric series if the first term is -11, the last term is 859,375, and the common ratio is -5?

NikAS [45]

Formula of the sum of the 1st nth term in a Geometric Progression:

Sum = a₁(1-rⁿ)/(1-r), where a₁ = 1st term, r = common ratio and n= rank nth of term (r≠1)

Sum = (-11)[1-(-5⁸)] /[(1-(-5)]

Sum = (-11)(1- 390625)/(6)

SUM = 716,144

3 0

3 years ago

Which of these limits evaluate to 0?

Vikentia [17]

<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

$\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\$

----------

Part II

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\$

----------

Part III

$\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\$

7 0

3 years ago