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3 years ago

10

(please zoom / zoom out if needed to see question. (: )

2 answers:

Llana [10]3 years ago

8 0

To find the measure of x, we can use Pythagoras' Theorem of c² - b² = a². In this case, c would be 26 (AB), and b would be 22 (AC), so you can plug in the numbers and write it as
26² - 22² = x²
676 - 484 = x²
192 = x²
√
x = 13.86 (2 d.p)
So your answer is x = 13.86, I hope this helps!

PIT_PIT [208]3 years ago

7 0

Answer: 13.86

Hypotenuse Formula : a^2 + b^2 = c^2

a^2 + b^2 = 26^2

484 + b^2 = 676

b^2 = 676 - 484

b^2 = 192

b = 13.86

x = 13.86

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Read 2 more answers

Use the intermediate value theorem to find the value of c such that f(c) = M. f(x) = x^2 - x + 1 text( on ) [1,8]; M = 21 c =

DanielleElmas [232]

Answer:

$c = 5$

Step-by-step explanation:

Given

$f(c) = M$

$f(x) = x^2 - x + 1$

$Interval: [1,8]$

$M = 21$

Required

Find c using Intermediate Value theorem

First, check if the value of M is within the given range:

$f(x) = x^2 - x + 1$

$f(1) = 1^2 - 1 + 1$

$f(1) = 1$

$f(x) = 8^2 - 8 + 1$

$f(x) = 57$

$1 \le M \le 57$

$1 \le 21 \le 57$

M is within range.

Solving further:

We have:

$f(c) = f(x) = M$

$f(x) = 21$

Substitute 21 for f(x) in $f(x) = x^2 - x + 1$

$21 = x^2 - x + 1$

Express as quadratic function

$x^2 - x + 1 - 21 = 0$

$x^2 - x - 20 = 0$

Expand

$x^2 + 4x - 5x - 20$

$x(x+4)-5(x+4)=0$

$(x - 5)(x+4) = 0$

$x - 5 = 0$ or $x + 4= 0$

$x = 5$ or $x = -4$

The value of $x = -4$ is outside the $Interval: [1,8]$

So:

$x = 5$

$f(c) = f(x) = M$

$f(c) = f(5) = 21$

By comparison:

$c = 5$

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3 years ago