Help me please its my last question I need help pleaseee

A woman starts out walking due north for 2 miles, and then she turns due west and continues for another 2 miles. How far from he

Mrac [35]

For this case we can model the problem as a rectangle triangle.
We have:
Length of the sides of the triangle.
We want to find:
Length of the hypotenuse.
Using the Pythagorean theorem we have:
$d ^ 2 = 2 ^ 2 + 2 ^ 2
$
$d = \sqrt{2 ^ 2 + 2 ^ 2} 

d = \sqrt{4 + 4} 

d = \sqrt{8} 

d = 2\sqrt{2} 
$
Answer:
She is:
$d = 2\sqrt{2}$
miles far from her starting point

4 0

3 years ago

Help thank you appreciated will give brainliest

ivann1987 [24]

I think that it the secound one then the fourth one then the sixth one then the first one then the fifth one then third one the eighth one then the ninth one

6 0

3 years ago

In a positive correlation, __________. A. the independent variable is always negative B. an increase in the independent variable

lyudmila [28]

The answer would be, "B".

8 0

3 years ago

Read 2 more answers

What is the other square root of 119 + 120i?

MakcuM [25]

Answer:

$\sqrt{119+120 i}=\pm (12.24+i 4.90)$

Step-by-step explanation:

Given

z = 119 + 120 i

Let $\sqrt{119+120 i}=p+iq$

Squaring both sides

$119+120 i=p^2-q^2+2ipq$

Comparing real and imaginary part

Re(LHS)=Re(RHS)

$119=p^2-q^2$ ...........................(1)

comparing Im(LHS)=Im(RHS)

120=2pq

$q=\frac{60}{p}$

Substitute q in 1

$119=p^2-(\frac{60}{p})^2$

$p^4-119p^2-(68)^2=0$

Let $x=p^2$

$x^2-119x-4624=0$

$x=\dfrac{119\pm \sqrt{119^2+4\times 4624}}{2}$

$x=\frac{119\pm 180.71}{2}$

we take only Positive value because $p^2=x$

x=149.85

$p^2=149.85$

thus $p=\pm 12.24$

$q=\pm 4.90$

thus,

$\sqrt{119+120 i}=\pm (12.24+i 4.90)$

8 0

3 years ago

1. Which of the following is equivalent to x5y2/xy2 when x ≠ 0 and y ≠ 0?

Aleonysh [2.5K]

Q1. The answer is D. x4

Let's first rewrite the expression:
x⁵y²/xy² = x⁵/x * y²/y²

Using the rule xᵃ/xᵇ = x(ᵃ⁻ᵇ), we can write the expression as following:
x⁵y²/xy² = x⁵/x * y²/y² = x⁵⁻¹ * y²⁻² = x⁴ * y⁰ = x⁴ * 1 = x⁴

Thus, the correct answer is D.

Q2. The answer is A. 5(5^3/2/5)^2

125 in the form of exponent is 5³.
125 = 5³
Now, let's calculate all choices.

The rules we will use are:
xᵃ * xᵇ = x(ᵃ⁺ᵇ)
xᵃ/xᵇ = x(ᵃ⁻ᵇ)
(xᵃ)ᵇ = xᵃ*ᵇ

A. 5(5³/2/5)² = 5 * (5³ * 5/2)²
= 5 * (5³⁺¹/2)²
= 5 * (5⁴/2)²
= 5 * (5⁴)²/(2)²
= 5 * 5⁴*²/4
= 5 * 5⁸ / 4
= 5¹⁺⁸ / 4
= 5⁹/4
≠ 5³ ≠ 125

B. (5³/5⁴)⁻³ = (5³⁻⁴)⁻³
= (5⁻¹)⁻³
= 5⁽⁻¹⁾*⁽⁻³⁾
= 5³
= 125

C. 5⁻²/5⁻⁵ = 5⁽⁻²⁾⁻⁽⁻⁵)
= 5⁽⁻²⁾⁺⁵
= 5³
= 125

D. 5(5⁵/5³) = 5 * 5⁵⁻³
= 5 * 5²
= 5¹⁺²
= 5³
= 125

Therefore, the only expression that is not equal to 125 is A.

Q3. The answer is 63x5

Let's check all choices
The rules we will use are:
xᵃ * xᵇ = x(ᵃ⁺ᵇ)
xᵃ/xᵇ = x(ᵃ⁻ᵇ)
(xᵃ)ᵇ = xᵃ*ᵇ

A. 6³x
6³x/6x⁵ = 6³/6 * x/x⁵
= 6³⁻¹ * x¹⁻⁵
= 6²x⁻⁴
= 36x⁻⁴
≠ 36

B. 6³x⁵
6³x⁵/6x⁵ = 6³/6 * x⁵/x⁵
= 6³⁻¹ * x⁵⁻⁵
= 6² * x⁰
 = 36 * 1
= 36

C. 6x⁵
6x⁵/6x⁵ = 1
≠ 36

D. 6⁷x⁵
6⁷x⁵/6x⁵ = 6⁷/6 * x⁵/x⁵
= 6⁷⁻¹ * x⁵⁻⁵
= 6⁶ * x⁰
= 46656 * 1
≠ 36

Therefore, the correct choice is B.

Q4. The answer is

We will use the rule: xᵃ/xᵇ = x(ᵃ⁻ᵇ)

5.4 x 10¹²/1.2 x 10³ = 5.4 / 1.2 x 10¹²/10³
= 4.5 x 10¹²⁻³
= 4.5 x 10⁹

Q5. The answer is B. To subtract powers with the same base, divide the exponents

Some of the rules regarding operations with exponents are:
xᵃ/xᵇ = x(ᵃ⁻ᵇ) - choice A
xᵃ * xᵇ = x(ᵃ⁺ᵇ) - choice C
(xᵃ)ᵇ = xᵃ*ᵇ - choice D

Through the process of elimination, choice B is not true.

8 0

3 years ago