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AysviL [449]
3 years ago
10

NEED HELP ASAP!!!

Mathematics
1 answer:
irakobra [83]3 years ago
6 0

Answer:

7.5

Step-by-step explanation:

A squared + B squared = C squared

7.2 squared + 2.1 squared = C squared

=7.5

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Is 3(–2) + 7(2) equivalent to (–3)(–3) + (–1)(1)? Let negative numbers be represented with a "–" symbol and positive numbers be
jek_recluse [69]
3(-2) + 7(2) = 8
(-3)(-3) + (-1)(1) = 8
yes
3 0
3 years ago
Which of the following is the equation of a line that is parallel to y=5x-3 and that passes through the point (2,5)
Travka [436]

parallel to the line y = 5x - 3, so the other lines must have same slope m = 5

y - y1 = m(x - x1)

y - 5 = 5(x-2)

y = 5x - 10 + 5

y = 5x - 5

3 0
3 years ago
Please help me with this question!!!!!!!!!1
sesenic [268]
Girl i have no clue, good luck broskiii
8 0
3 years ago
Read 2 more answers
Which description compares the vertical asymptotes of function A and function B correctly?
neonofarm [45]

Answer:

Option B:

Function A has a vertical asymptote at x = 1

Function B has a vertical asymptote at x = -3

Step-by-step explanation:

A function f(x) has a vertical asymptote if:

\lim_{x \to\\k^+}f(x) = \±\infty\\\\ \lim_{x \to\\k^-}f(x) = \±\infty

This means that if there is a value k for which f(x) has infinity or a -infinity then x = k is a vertical asymptote of f(x). Therefore, the closer x to k approaches, the closer the function becomes to infinity.

We can calculate the asymptote for function A.

\lim_{x \to \\1^+}(\frac{1}{x-1})\\\\ \lim_{x \to \\1^+}(\frac{1}{1^-1})\\\\ \lim_{x \to \\1^+}(\frac{1}{0}) = \infty\\\\and\\ \lim_{x \to \\1^-}(\frac{1}{x-1})\\\\\lim_{x \to \\1^-}(\frac{1}{0}) = -\infty

Then, function A has a vertical asymptote at x = 1.

The asymptote of function B can be easily observed in the graph. Note that the function b is not defined for x = -3 and when x is closest to -3, f(x) approaches infinity.

Therefore x = -3 is asintota of function B.

Therefore the correct answer is option B.

8 0
3 years ago
Graph the line given by x+y=-2 and the quadratic curve given by y=x²-4. Find all solutions to the system of equations. Verify yo
ycow [4]

Answer:

(1,-3) and (-2,0)

Step-by-step explanation:

x+y=-2

y=x^2-4

Applying the second equation to the first

x+x^2-4=-2\\\Rightarrow x+x^2-4+2=0\\\Rightarrow x^2+x-2=0

Solving the equation we get

x=\frac{-1+\sqrt{1^2-4\cdot \:1\left(-2\right)}}{2\cdot \:1}, \frac{-1-\sqrt{1^2-4\cdot \:1\left(-2\right)}}{2\cdot \:1}\\\Rightarrow x=1, -2

When x = 1

y=-2-1\\\Rightarrow y=-3

When x = -2

y=-2-(-2)\\\Rightarrow y=0

So, the line and curve will intersect at points (1,-3) and (-2,0)

5 0
3 years ago
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