Question 1:

Where [tex]2 \times 4^{1-x} = 8^{-x}[/tex], find [tex]x[/tex].

A. -3
B. -2
C. 2

Question 2:

Three points [tex]P, Q[/tex], and [tex]R[/tex] are in the same plane in a field. Given [tex]PQ = 25 \, \text{cm}[/tex], [tex]PR = 24 \, \text{cm}[/tex], and [tex]\angle QRP = 90^{\circ}[/tex], calculate [tex]QR[/tex].

A. 7 cm
B. 8 cm
C. 9 cm
D. 10 cm
E. 12 cm



Answer :

Sure, let's address each question step-by-step:

### Question 1:
Solve [tex]\(2 \times 4^{1-x} = 8^{-x}\)[/tex].

To solve the equation, we can rewrite both sides of the equation using powers of 2.

1. [tex]\(4\)[/tex] is [tex]\(2^2\)[/tex], so [tex]\(4^{1-x} = (2^2)^{1-x} = 2^{2(1-x)} = 2^{2 - 2x}\)[/tex].
2. [tex]\(8\)[/tex] is [tex]\(2^3\)[/tex], so [tex]\(8^{-x} = (2^3)^{-x} = 2^{3(-x)} = 2^{-3x}\)[/tex].

Substituting these back into the equation:

[tex]\[2 \cdot 2^{2 - 2x} = 2^{-3x}\][/tex]

Combine the terms on the left-hand side:

[tex]\[2^1 \cdot 2^{2 - 2x} = 2^{-3x}\][/tex]

[tex]\[2^{1 + 2 - 2x} = 2^{-3x}\][/tex]

[tex]\[2^{3 - 2x} = 2^{-3x}\][/tex]

Since the bases are the same, we can equate the exponents:

[tex]\[3 - 2x = -3x\][/tex]

Solve for [tex]\(x\)[/tex]:

[tex]\[3 - 2x = -3x\][/tex]

Add [tex]\(2x\)[/tex] to both sides:

[tex]\[3 = -x\][/tex]

Multiply both sides by [tex]\(-1\)[/tex]:

[tex]\[x = -3\][/tex]

So, the answer to the first question is:
[tex]\[ \boxed{-3} \][/tex]

### Question 2:
Points [tex]\(P, Q\)[/tex], and [tex]\(R\)[/tex] are in the same plane in a field, [tex]\(PQ = 25 \, \text{cm}\)[/tex], [tex]\(PR = 24 \, \text{cm}\)[/tex], and [tex]\(\triangle QRP = 90^{\circ}\)[/tex]. Calculate the length of [tex]\(QR\)[/tex].

Given:
- [tex]\(PQ = 25 \, \text{cm}\)[/tex]
- [tex]\(PR = 24 \, \text{cm}\)[/tex]
- [tex]\(\angle QRP = 90^{\circ}\)[/tex]

We can use the Pythagorean theorem to find [tex]\(QR\)[/tex].

In [tex]\(\triangle QRP\)[/tex] which is a right-angled triangle:
[tex]\[ PQ^2 = PR^2 + QR^2 \][/tex]

Substitute the known values:
[tex]\[ 25^2 = 24^2 + QR^2 \][/tex]

Calculate the squares:
[tex]\[ 625 = 576 + QR^2 \][/tex]

Rearrange to solve for [tex]\(QR^2\)[/tex]:
[tex]\[ QR^2 = 625 - 576 \][/tex]
[tex]\[ QR^2 = 49 \][/tex]

Take the square root of both sides:
[tex]\[ QR = \sqrt{49} \][/tex]
[tex]\[ QR = 7 \, \text{cm} \][/tex]

So, the answer to the second question is:
[tex]\[ \boxed{7 \, \text{cm}} \][/tex]

Both questions have been solved step-by-step as requested.