Mathematics Test 1 Sat Examination

A fish swam a distance of 5,104 yards. How far did the fish swim, in miles? (1 mile = 1,760 yards)
If 1 mile = 1760 
Then 5104 yards = 5104/1760 = 2.9 miles

The correct answer is C.7x³

12x³ − 5x³ = (12 - 5)x³ = 7x³

x + y = 18   _________Eqn1
5y = x    ____________Eqn2
Subttute Eqn 2 into 1

5y + y = 18
y = 18/6 = 3
Subtitute in the value of y into Eqn 2
5y = x
5(3) = x
x = 15

Correct answer: A) x > 0, y > 0

The point given is (8, 2).

x = 8, which is greater than 0.
y = 2, which is also greater than 0.

So the point satisfies both:

x>0
y>0

Given the function:

f(x) = 7x + 1

Substitute x = 7:

f(7) = 7(7) + 1 = 49 + 1 = 50

50 people are on the company retreat with 7 managers.

h(x) = x² - 3
h(1) = 1² - 3 = -2
h(2) = 2² - 3 = 1
h(3) = 3² - 3 = 6
 

Substitute x = 0 into the function:

\( f(0) = 270(0.1^0) = 270 \times 1 = 270 \)

Given:
Estimated proportion = 0.49
Margin of error = 0.04

The plausible range is:
\( 0.49 \pm 0.04 = [0.45, 0.53] \)

Maximum total weight = 4,600 pounds
Weight of trailer = 500 pounds
Each box weighs = 120 pounds

Let x be the number of boxes. Then:\[500 + 120x \leq 4600\] Subtract 500:\[120x \leq 4100\] Divide by 120:\[x \leq \dfrac{4100}{120} \approx 34.17\] So the maximum whole number of boxes is 34.

\begin{align*}
\text{Given:} \quad & -4x^2 - 7x = -36 \\
\text{Move all terms to one side:} \quad & -4x^2 - 7x + 36 = 0 \\
\text{Multiply through by } -1 \text{ to simplify:} \quad & 4x^2 + 7x - 36 = 0 \\
\text{Use the quadratic formula:} \quad & x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
\text{where } & a = 4,\ b = 7,\ c = -36 \\
x &= \dfrac{-7 \pm \sqrt{7^2 - 4(4)(-36)}}{2(4)} \\
&= \dfrac{-7 \pm \sqrt{49 + 576}}{8} \\
&= \dfrac{-7 \pm \sqrt{625}}{8} \\
&= \dfrac{-7 \pm 25}{8} \\
\text{So, } x &= \dfrac{18}{8} = \dfrac{9}{4} \quad \text{or} \quad x = \dfrac{-32}{8} = -4 \\
\text{Positive solution:} \quad & \boxed{\dfrac{9}{4}}
\end{align*}