Question

1. what is the length of the hypotenuse of the right triangle below? side lengths are measured in centimeters.

a. $sqrt{2 + 8}$ cm
b. $8 - 2$ cm
c. $sqrt{2^{2}+8^{2}}$ cm
d. $2^{2}+8^{2}$ cm
e. i have not learned this yet.

1. write the expression using a single exponent. $2^{m}cdot2^{x}$

a. $2^{m + x}$
b. $4^{m + x}$
c. $4^{mx}$
d. $2^{mx}$
e. i have not learned this yet.

1. solve. $\begin{cases}y=-x + 8\\x + y=7end{cases}$

a. $(3,4)$
b. $(2,6)$
c. $(-\frac{1}{2},\frac{17}{2})$
d. no solutions
e. i have not learned this yet.

Explanation:

Step1: Apply Pythagorean theorem

For a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(c=\sqrt{a^{2}+b^{2}}\). Here \(a = 2\) and \(b = 8\), so \(c=\sqrt{2^{2}+8^{2}}\).

Step2: Use exponent rule for multiplication

The rule for multiplying two exponential terms with the same base \(a^m\cdot a^n=a^{m + n}\). For \(2^m\cdot2^x\), with \(a = 2\), \(m=m\) and \(n=x\), we get \(2^{m + x}\).

Step3: Solve the system of equations

Substitute \(y=-x + 8\) into \(x + y=7\). We have \(x+(-x + 8)=7\), which simplifies to \(8=7\), a contradiction. So there are no solutions.

Answer:

1. C. \(\sqrt{2^{2}+8^{2}}\text{ cm}\)
2. A. \(2^{m + x}\)
3. D. No solutions