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2 years ago

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The perpendicular bisectors of triangle JKL are line segment PT, line segment QT, and line segment RT. Name three isosceles tria

ngles. Write them in upper case, and alphabetical order (letter ordering as well as response ordering).

1 answer:

Olegator [25]2 years ago

7 0

9514 1404 393

Answer:

JKT, JLT, KLT

Step-by-step explanation:

Each of the given bisectors is the altitude of an isosceles triangle. Those triangles are ...

JKT

JLT

KLT

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Which rational function has zeros at x = 1 and x = 3?

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What do you mean zeroes

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if f(x)=2(x)^2+5sqrt(x+2), complete the follwoing statement ( round your answer to the nearest hundreth) : f(0)=_____

Maslowich

Answer:

$f(0)=7.07$

Step-by-step explanation:

We have the function $f(x)=2(x)^2+5\sqrt{(x+2)}$

In this case we want to find the value of f0)

To find f(0) you must replace the x in the function with the number 0 and solve as shown below

$f(0)=2(0)^2+5\sqrt{(0+2)}$

$f(0)=0+5\sqrt{(0+2)}$

$f(0)=5\sqrt{(2)}$

Therefore

$f(0)=7.07$

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3 years ago

Implicit differentiation Please help

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Answer:

$y''(-1) =8$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

<u>Algebra I</u>

Factoring

<u>Calculus</u>

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

<u>Step 1: Define</u>

-xy - 2y = -4

Rate of change of the tangent line at point (-1, 4)

<u>Step 2: Differentiate Pt. 1</u>

<em>Find 1st Derivative</em>

Implicit Differentiation [Product Rule/Basic Power Rule]: $-y - xy' - 2y' = 0$
[Algebra] Isolate <em>y'</em> terms: $-xy' - 2y' = y$
[Algebra] Factor <em>y'</em>: $y'(-x - 2) = y$
[Algebra] Isolate <em>y'</em>: $y' = \frac{y}{-x-2}$
[Algebra] Rewrite: $y' = \frac{-y}{x+2}$

<u>Step 3: Find </u><em><u>y</u></em>

Define equation: $-xy - 2y = -4$
Factor <em>y</em>: $y(-x - 2) = -4$
Isolate <em>y</em>: $y = \frac{-4}{-x-2}$
Simplify: $y = \frac{4}{x+2}$

<u>Step 4: Rewrite 1st Derivative</u>

[Algebra] Substitute in <em>y</em>: $y' = \frac{-\frac{4}{x+2} }{x+2}$
[Algebra] Simplify: $y' = \frac{-4}{(x+2)^2}$

<u>Step 5: Differentiate Pt. 2</u>

<em>Find 2nd Derivative</em>

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<u>Step 6: Find Slope at Given Point</u>

[Algebra] Substitute in <em>x</em>: $y''(-1) = \frac{8}{(-1+2)^3}$
[Algebra] Evaluate: $y''(-1) =8$

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2 years ago

Read 2 more answers

Can someone please help me on number 16-ABC

melomori [17]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

-2x < 10

-6 < -2x

<u>Part a) Is x = 0 a solution to both inequalities</u>

FOR -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

-3(0) < 10

0 < 10

TRUE!

Thus, x = 0 satisfies the inequality -2x < 10.

∴ x = 0 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 0 in -6 < -2x

-6 < -2x

-6 < -2(0)

-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

x = 0 is a solution to both inequalites.

<u>Part b) Is x = 4 a solution to both inequalities</u>

FOR -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

<u>solving -2x < 10</u>

$-2x\:$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)>10\left(-1\right)$

Simplify

$2x>-10$

Divide both sides by 2

$\frac{2x}{2}>\frac{-10}{2}$

$x>-5$

$-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}$

<u>solving -6 < -2x</u>

-6 < -2x

switch sides

$-2x>-6$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)$

Simplify

$2x$

Divide both sides by 2

$\frac{2x}{2}$

$x$

$-6$

Thus, the two intervals:

$\left(-\infty \:,\:3\right)$

$\left(-5,\:\infty \:\right)$

The intersection of these two intervals would be the solution to both inequalities.

$\left(-\infty \:,\:3\right)$ and $\left(-5,\:\infty \:\right)$

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

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Answer:

4^4= 4^-1(3r+2)

4^4= 4^(-3r-2)

-3r-2 = 4

-3r= 6

r= -2

-

Step-by-step explanation:

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3 years ago