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2 years ago

6. Write the slope-intercept form of the equation of the line. slope = -5/2, y y – int = 2

Mathematics

2 answers:

Zanzabum2 years ago

5 0

Answer:

A. $y=-\frac{5}{2} x+2$

P.S. Sorry, the first answer I input wasn't registered well

skelet666 [1.2K]2 years ago

5 0

The answer is:
y = -5/2x + 2
Option 1

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2 years ago

If one of the roots of the equation x²- 4x + k=0exceeds the other by 2, then find the roots and determine the value of k

BaLLatris [955]

Answer:

roots: 1 and 3

k = 3

Step-by-step explanation:

2 roots: p and p+2

(x-p) (x-p-2) = x² - 4x + k

x² -2px -2x + p² + 2p = x² - 2 (p+1)x + (p² + 2p) = x² -4x + k

-2 (p+1) = -4

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k = p² + 2p = 3

check: (x-1) (x-3) = x² - 4x + 3 = x² - 4x + k

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2 years ago

Given the parent function f(x) = 3x, which graph shows f(x) + 2?

blsea [12.9K]

The one that has positive 2 as the y intercept

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3 years ago

Ad is tangent to circle o at d. find ab. round to the nearest tenth if necessary

Naddik [55]

Check the picture below.

$\bf 0=AB^2+6AB-121\implies \stackrel{\textit{using the quadratic formula}}{AB=\cfrac{-6 \pm \sqrt{6^2-4(1)(-121)}}{2(1)}}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{520}}{2}\implies AB=\cfrac{-6 \pm \sqrt{4\cdot 130}}{2}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{2^2\cdot 130}}{2}\implies AB=\cfrac{-6 \pm 2\sqrt{130}}{2}
\\\\\\
AB=-3\pm\sqrt{130}\implies AB=
\begin{cases}
\boxed{-3+\sqrt{130}}\\\\
-3-\sqrt{130}
\end{cases}$

since the distance AB cannot be a negative value, thus is not -3-√(130).

3 0

3 years ago

Read 2 more answers