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Paha777 [63]
3 years ago
10

Solve for x? and the lengths missing side

Mathematics
2 answers:
Yuliya22 [10]3 years ago
8 0

Answer:

X = 15, the missing length (c) is 10

Step-by-step explanation:

Using Pythagorean’s Theorem we know that a^2 + b^2 = c^2

We have side a and b, represent a as 6 and b as 8.

We have 6^2 + 8^2 = c^2

Which equals 100 = c^2

10 = c

Side C is 10, and (x - 5) = 10

X = 15

Please mark BRAINLIEST!

Wittaler [7]3 years ago
6 0

Answer: x ≅ 15.95

Step-by-step explanation:

Use the pythagorean theorem for right triangles x^2 + y^2 = s^2

6^2 = 36

8^2 = 64

(x - 5)^2 requires FOIL. Unfortunately Im going to do this on paper. Its essentially First Outside Inside Last. i got x^2 - 10x + 25

Lets now add 36 and 84 and make it equal to the trinomial we just got.

120=x^2-10x+25

Now lets subtract 120 from both sides to get:

0=x^2-10x-95

You could complete the square, I think. But im going to use the quadratic formula. \frac{-b +/- \sqrt{b^2-4ac} }{2a}Im going to calculate this on paper for simplicity's sake. Im getting an answer of about \frac{10(+/-)21.909}{2}

Solve by adding and then solve by subtracting to get x ≅ 15.95 and x = -5.95. We cant have a negative side length so we can disregard that negative answer.

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PLEASE HELP 100 POINTS!!!!!!
horrorfan [7]

Answer:

A)  See attached for graph.

B)  (-3, 0)  (0, 0)  (18, 0)

C)   (-3, 0) ∪ [3, 18)

Step-by-step explanation:

Piecewise functions have <u>multiple pieces</u> of curves/lines where each piece corresponds to its definition over an <u>interval</u>.

Given piecewise function:

g(x)=\begin{cases}x^3-9x \quad \quad \quad \quad \quad \textsf{if }x < 3\\-\log_4(x-2)+2 \quad  \textsf{if }x\geq 3\end{cases}

Therefore, the function has two definitions:

  • g(x)=x^3-9x \quad \textsf{when x is less than 3}
  • g(x)=-\log_4(x-2)+2 \quad \textsf{when x is more than or equal to 3}

<h3><u>Part A</u></h3>

When <u>graphing</u> piecewise functions:

  • Use an open circle where the value of x is <u>not included</u> in the interval.
  • Use a closed circle where the value of x is <u>included</u> in the interval.
  • Use an arrow to show that the function <u>continues indefinitely</u>.

<u>First piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

\implies g(3)=(3)^3-9(3)=0 \implies (3,0)

Place an open circle at point (3, 0).

Graph the cubic curve, adding an arrow at the other endpoint to show it continues indefinitely as x → -∞.

<u>Second piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

\implies g(3)=-\log_4(3-2)+2=2 \implies (3,2)

Place an closed circle at point (3, 2).

Graph the curve, adding an arrow at the other endpoint to show it continues indefinitely as x → ∞.

See attached for graph.

<h3><u>Part B</u></h3>

The x-intercepts are where the curve crosses the x-axis, so when y = 0.

Set the <u>first piece</u> of the function to zero and solve for x:

\begin{aligned}g(x) & = 0\\\implies x^3-9x & = 0\\x(x^2-9) & = 0\\\\\implies x^2-9 & = 0 \quad \quad \quad \implies x=0\\x^2 & = 9\\\ x & = \pm 3\end{aligned}

Therefore, as x < 3, the x-intercepts are (-3, 0) and (0, 0) for the first piece.

Set the <u>second piece</u> to zero and solve for x:

\begin{aligned}\implies g(x) & =0\\-\log_4(x-2)+2 & =0\\\log_4(x-2) & =2\end{aligned}

\textsf{Apply log law}: \quad \log_ab=c \iff a^c=b

\begin{aligned}\implies 4^2&=x-2\\x & = 16+2\\x & = 18 \end{aligned}

Therefore, the x-intercept for the second piece is (18, 0).

So the x-intercepts for the piecewise function are (-3, 0), (0, 0) and (18, 0).

<h3><u>Part C</u></h3>

From the graph from part A, and the calculated x-intercepts from part B, the function g(x) is positive between the intervals -3 < x < 0 and 3 ≤ x < 18.

Interval notation:  (-3, 0) ∪ [3, 18)

Learn more about piecewise functions here:

brainly.com/question/11562909

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Can someone please help me: Slove 17+3(z-2)-11z=-7(z+2)+14
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Answer:

z = 11

Step-by-step explanation:

Simplifying

17 + 3(z + -2) + -11z = -7(z + 2) + 14

Reorder the terms:

17 + 3(-2 + z) + -11z = -7(z + 2) + 14

17 + (-2 * 3 + z * 3) + -11z = -7(z + 2) + 14

17 + (-6 + 3z) + -11z = -7(z + 2) + 14

Combine like terms: 17 + -6 = 11

11 + 3z + -11z = -7(z + 2) + 14

Combine like terms: 3z + -11z = -8z

11 + -8z = -7(z + 2) + 14

Reorder the terms:

11 + -8z = -7(2 + z) + 14

11 + -8z = (2 * -7 + z * -7) + 14

11 + -8z = (-14 + -7z) + 14

Reorder the terms:

11 + -8z = -14 + 14 + -7z

Combine like terms: -14 + 14 = 0

11 + -8z = 0 + -7z

11 + -8z = -7z

Solving

11 + -8z = -7z

Solving for variable 'z'.

Move all terms containing z to the left, all other terms to the right.

Add '7z' to each side of the equation.

11 + -8z + 7z = -7z + 7z

Combine like terms: -8z + 7z = -1z

11 + -1z = -7z + 7z

Combine like terms: -7z + 7z = 0

11 + -1z = 0

Add '-11' to each side of the equation.

11 + -11 + -1z = 0 + -11

Combine like terms: 11 + -11 = 0

0 + -1z = 0 + -11

-1z = 0 + -11

Combine like terms: 0 + -11 = -11

-1z = -11

Divide each side by '-1'.

z = 11

Simplifying

z = 11

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