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3 years ago

HELPPP ITS A TESTAKAJNSDUJSWMNS

Mathematics

1 answer:

vredina [299]3 years ago

3 0

Answer:

5^6

Step-by-step explanation:

by the law of exponents,

5^2 * 5^(-8)

5^ [2+ (-8)]

5^(2-8) = 5^-6 = 1/5^6 = 5^6

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|. Identify the following Pōints of each values.Write your ans

Dmitry_Shevchenko [17]

<h2>✒️VALUE</h2>

$\\ \quad \begin{array}{c} \qquad \bold{Distance \: \green{ Formula:}}\qquad\\ \\ \boldsymbol{ \tt d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}} \end{array}\\ \begin{array}{l} \\ 1.)\: \bold{Given:}\: \begin{cases}\tt D(- 5,6), E(2.-1),\textsf{ and }F(x,0) \\ \tt DF = EF \end{cases} \\ \\ \qquad\bold{Required:}\:\textsf{ value of }x \\ \\ \qquad \textsf{Solving for }x, \\ \\ \tt \qquad DF = EF \\ \\ \implies\small \tt{\sqrt{(x -(- 5))^2 + (0 - 6)^2} = \sqrt{(x - 2)^2 + (0 - (-1))^2}} \\ \\ \implies\tt\sqrt{(x + 5)^2 + 36 } = \sqrt{(x - 2)^2 + 1 } \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt (x + 5)^2 + 36 = (x - 2)^2 + 1 \\ \\ \implies\tt x^2 + 10x + 25 + 36 = x^2 - 4x + 4 + 1 \\ \\ \implies \tt x^2 + 10x + 61 = x^2 - 4x + 5 \\ \\ \implies\tt10x + 4x = 5 - 61 \\ \\ \implies\tt14x = -56 \\ \\ \implies \red{\boxed{\tt x = -4}}\end{array} \\ \\ \\ \\\begin{array}{l} \\ 2.)\: \bold{Given:}\: \begin{cases}\tt P(6,-1), Q(-4,-3),\textsf{ and }R(0,y) \\ \tt PR = QR \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }y \\ \\ \qquad\textsf{Solving for }y, \\ \\ \qquad\tt PR = QR \\ \\ \implies \tt\small{\sqrt{(0 - 6)^2 + (y - (-1))^2} = \sqrt{(0 - (-4))^2 + (y - (-3))^2}} \\ \\ \implies\tt\sqrt{36 + (y + 1)^2} = \sqrt{16 + (y + 3)^2 } \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies \tt \: 36 + (y + 1)^2 = 16 + (y + 3)^2 \\ \\ \implies\tt 36 + y^2 + 2y + 1 = 16 + y^2 + 6y + 9 \\ \\ \implies \tt \: y^2 + 2y + 37 = y^2 + 6y + 25 \\ \\ \implies \tt \: 2y - 6y = 25 - 37 \\ \\ \implies \tt -4y = -12 \\ \\ \implies\red{\boxed{ \tt y = 3}} \end{array} \\ \\ \\ \begin{array}{l} \\ 3.)\: \bold{Given:}\: \begin{cases}\: A(4,5), B(-3,2),\textsf{ and }C(x,0) \\ \: AC = BC \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }x \\ \\ \qquad\textsf{Solving for }x, \\ \\ \qquad\tt AC = BC \\ \\ \implies\tt\small{\sqrt{(x - 4)^2 + (0 - 5)^2} = \sqrt{(x - (-3))^2 + (0 - 2)^2}} \\ \\ \implies\tt\sqrt{(x - 4)^2 + 25} = \sqrt{(x + 3)^2 + 4} \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt\:(x - 4)^2 + 25 = (x + 3)^2 + 4 \\ \\ \implies\tt\:x^2 - 8x + 16 + 25 = x^2 + 6x + 9 + 4 \\ \\ \implies\tt\:x^2 - 8x + 41 = x^2 + 6x + 13 \\ \\ \implies\tt-8x - 6x = 13 - 41 \\ \\\implies\tt -14x = -28 \\ \\ \implies\red{\boxed{\tt\:x = 2}} \end{array}$

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#BrainlyMathKnower

#5-MinutesAnswer

7 0

2 years ago

A parallelogram has verticies of (0,0) (0,6) (4,4) (4, -2). Which transformation maps the parallelogram onto itself?

gulaghasi [49]

The transformation that would map the parallelogram onto itself is the rotation of 180 degrees about the point (2, 2)

<h3>What is a transformation?</h3>

Transformation is the movement of a point from its initial location to a new location. Types of transformation are reflection, translation, rotation and dilation.

Rigid transformation is the transformation that does not change the shape or size of a figure. Examples of rigid transformations are <em>translation, reflection and rotation</em>.

The transformation that would map the parallelogram onto itself is the rotation of 180 degrees about the point (2, 2)

Find out more on transformation at: brainly.com/question/4289712

#SPJ1

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2 years ago

1 pt) find the average rate of change of the function over the given intervals. h(t)=\cot (t);

blondinia [14]

$\text{Consider the function}\\
\\
h(t)=\cot(t) \text{ on the interval }\left [ \frac{\pi}{4}, \ \frac{3\pi}{4} \right ]\\
\\
\text{we know that the average rate of change of a funtion f(x) over the }\\
\text{interval [a, b] is given by}\\
\\
f_{avg}=\frac{f(b)-f(a)}{b-a}\\
\\
\text{so using this, the average rate of change of the given function is}$

$h_{avg}=\frac{h\left ( \frac{3\pi}{4} \right )-h\left ( \frac{\pi}{4} \right )}{\frac{3\pi}{4}-\frac{\pi}{4}}\\
\\
=\frac{\cot \left ( \frac{3\pi}{4} \right )-\cot \left ( \frac{\pi}{4} \right )}{\frac{3\pi-\pi}{4}}\\
\\
=\frac{(-1)-(1)}{\frac{2\pi}{4}}\\
\\
=\frac{-2}{\frac{\pi}{2}}\\
\\
\text{Average rate of change of function}=\frac{-4}{\pi}$

6 0

4 years ago