in triangle abc, angle b is a right angle. the length of side ab is 10√37 and the length of side bc is 24√37. what is the length of side ac? a) 14√37 b) 26√37 c) 34√37 d) √34·37 the function f is defined by f(x)=a√(x + b), where a and b are constants. in the xy - plane, the graph of y = f(x) passes through the point (-24,0), and f(24) b d) a

Question

in triangle abc, angle b is a right angle. the length of side ab is 10√37 and the length of side bc is 24√37. what is the length of side ac?
a) 14√37
b) 26√37
c) 34√37
d) √34·37

the function f is defined by f(x)=a√(x + b), where a and b are constants. in the xy - plane, the graph of y = f(x) passes through the point (-24,0), and f(24)<0. which of the following must be true?
a) f(0)=24
b) f(0)= - 24
c) a > b
d) a < b

f(x)=(1.84)^(x/4)
the function f is defined by the given equation. the equation can be rewritten as f(x)=(1 + p/100)^x, where p is a constant. which of the following is closest to the value of p?
a) 16
b) 21
c) 46
d) 96

in the xy - plane, a circle has center c with coordinates (h,k). points a and b lie on the circle. point a has coordinates (h + 1,k + √102), and ∠acb is a right angle. what is the length of ab?
a) √206
b) 2√102
c) 103√2
d) 103√3

Accepted Answer

### First problem (triangle $ABC$):
# Explanation:
## Step1: Apply Pythagorean theorem
In a right - triangle $ABC$ with $\angle B = 90^{\circ}$, by the Pythagorean theorem $AC^{2}=AB^{2}+BC^{2}$. Given $AB = 10\sqrt{37}$ and $BC = 24\sqrt{37}$.
$$
\begin{align*}
AB^{2}&=(10\sqrt{37})^{2}=10^{2}\times37 = 3700\
BC^{2}&=(24\sqrt{37})^{2}=24^{2}\times37=576\times37 = 21312
\end{align*}
$$
## Step2: Calculate $AC^{2}$
$$
AC^{2}=3700 + 21312=25012
$$
## Step3: Simplify $AC$
$$
AC=\sqrt{25012}=\sqrt{37\times676}=\sqrt{37\times26^{2}} = 26\sqrt{37}
$$

# Answer:
B. $26\sqrt{37}$

### Second problem ($f(x)=a\sqrt{x + b}$):
# Explanation:
## Step1: Use the point $(-24,0)$
Since the graph of $y = f(x)$ passes through $(-24,0)$, then $0=a\sqrt{-24 + b}$. Since $a
eq0$ (otherwise $f(x)$ would be a constant function and $f(24)$ would not be well - defined in the non - trivial sense), we have $b = 24$. So $f(x)=a\sqrt{x + 24}$.
## Step2: Analyze $f(24)$
$f(24)=a\sqrt{24 + 24}=a\sqrt{48}$. Given $f(24)<0$, then $a<0$.
## Step3: Calculate $f(0)$
$f(0)=a\sqrt{0 + 24}$. Since $a<0$, $f(0)
eq24$. Now, compare $a$ and $b$. Since $a<0$ and $b = 24>0$, then $a < b$.

# Answer:
D. $a < b$

### Third problem ($f(x)=(1.84)^{\frac{x}{4}}$):
# Explanation:
## Step1: Rewrite the function
We know that $y = a^{kx}=(1 +\frac{p}{100})^{x}$, where $a=(1.84)^{\frac{1}{4}}$. Then $(1 +\frac{p}{100})=(1.84)^{\frac{1}{4}}$. Raise both sides to the power of $4$: $(1+\frac{p}{100})^{4}=1.84$.
## Step2: Use the binomial approximation $(1 + x)^{n}\approx1+nx$ (for $|x|\ll1$, here we can also solve $(1+\frac{p}{100})^{4}=1.84$ directly. $(1+\frac{p}{100})=\sqrt[4]{1.84}\approx1.16$.
## Step3: Solve for $p$
$1+\frac{p}{100}=1.16$, then $\frac{p}{100}=0.16$, $p = 16$

# Answer:
A. $16$

### Fourth problem (circle with center $C(h,k)$ and points $A(h + 1,k+\sqrt{102})$ and $\angle ACB = 90^{\circ}$):
# Explanation:
## Step1: Use the distance formula for $r = AC$
The distance formula $d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}$. If $C(h,k)$ and $A(h + 1,k+\sqrt{102})$, then $r^{2}=AC^{2}=1^{2}+(\sqrt{102})^{2}=1 + 102=103$.
## Step2: Use the property of a right - angled triangle in a circle ($AB$ is the hypotenuse of right - triangle $ACB$ with $AC = BC=r$)
By the Pythagorean theorem $AB^{2}=AC^{2}+BC^{2}$. Since $AC = BC=r$, $AB^{2}=2r^{2}$. Substitute $r^{2}=103$, we get $AB^{2}=2\times103$, $AB=\sqrt{206}$

# Answer:
A. $\sqrt{206}$

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QUESTION IMAGE

Question

Explanation:

First problem (triangle \(ABC\)):

Step1: Apply Pythagorean theorem

Step2: Calculate \(AC^{2}\)

Step3: Simplify \(AC\)

Step1: Use the point \((-24,0)\)

Step2: Analyze \(f(24)\)

Step3: Calculate \(f(0)\)

Step1: Rewrite the function

Step2: Use the binomial approximation \((1 + x)^{n}\approx1+nx\) (for \(|x|\ll1\), here we can also solve \((1+\frac{p}{100})^{4}=1.84\) directly. \((1+\frac{p}{100})=\sqrt[4]{1.84}\approx1.16\).

Step3: Solve for \(p\)

Answer:

Second problem (\(f(x)=a\sqrt{x + b}\)):