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3 years ago

Are the two expressions equivalent when x = 5? 24x + 18 and 6 (4x + 2)

Mathematics

1 answer:

patriot [66]3 years ago

3 0

Answer:

Nope-

Step-by-step explanation:

24x + 18 = 138

6 (4x + 2) = 132

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Complete the square to rewrite y = x2 + 6x + 3 in vertex form. Then state whether the vertex is a maximum or a minimum and give

masha68 [24]

y = x² + 6x + 3

y = x² + 6x + 9 - 6

y = (x + 3)² - 6

since the coefficient of x² is positive, then min at (-3,-6). the answer is C.

3 0

3 years ago

Read 2 more answers

2 and 3 are angles. complementary vertical congruent supplementary

Lapatulllka [165]

Answer:

supplementary

Step-by-step explanation:

Angles 2 and 3 form a straight line. Straight lines are supplementary

5 0

4 years ago

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What is the base for the expression -6 to the power of 4

natima [27]

For this case we have that by definition it is fulfilled:

$a ^ b$

Where:

a: It is the base

b: It is the exponent

So, if we have the following expression:

-6 to the power of 4, is equivalent to: $(-6) ^ 4$

So we have to:

-6: It is the base

4: It is the exponent

Answer:

-6: It is the base

4: It is the exponent

8 0

3 years ago

What is the volume of the cone? Use π ≈ 3.14.

BARSIC [14]

Answer:

12.56 cubic units on edge

Step-by-step explanation:

6 0

3 years ago

In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En

mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

$\frac{dy}{dx} = 3t^{2} - 2t - 27$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$3t^{2} - 2t - 27 = 0$

$a = 3, b = -2, c = -27$

Then

$\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328$

$t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35$

$t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685$

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

$x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55$

$y = t^{2} - 4 = (3.35)^2 - 4 = 7.22$

The first point is (27.55, 7.22)

t = -2.685

$x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3$

$y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21$

The second point is (-11.3, 3.21).

8 0

3 years ago