Which simplified equation is equivalent to the equation shown below?

HELP ME I WILL GIVE YOU BRAINLIEST AND A 5.0 REVIEW

Jlenok [28]

Answer:

Base

Step-by-step explanation:

A power is the product of multiplying a number by itself. Usually, a power is represented with a base number and an exponent. The base number tells what number is being multiplied. The exponent, a small number written above and to the right of the base number, tells how many times the base number is being multiplied.

4 0

3 years ago

$5 PAYPAL + BRAINLIEST TO ANSWER THS

expeople1 [14]

Answer:

Quadratic Equation:

$3x^2=2x +5$

$\text{Standard Form: } 3x^2-2x-5=0$

From the standard form of a Quadratic Function, we get:

$a=3,\:b=-2,\:c=-5$

Discriminant:

$\Delta=\left(-2\right)^2-4\cdot \:3\left(-5\right)$

$\Delta=\left(-2\right)^2+4\cdot \:3\cdot \:5$

$\Delta=64$

From the discriminant, we conclude that the equation will have two real solutions.

State that:

$b^2-4ac$

$b^2-4ac =0:\text{The equation has 1 real solution}$

$b^2-4ac >0:\text{The equation has 2 real solutions}$

By the way, solving the equation given:

$x=\frac{2\pm\sqrt{64}}{2\cdot \:3}$

$x=\frac{2\pm\sqrt{64}}{6}$

$x=\frac{2\pm8}{6}$

$x_{1} =\frac{10}{6}=\frac{5}{3}$

$x_{2}=\frac{-6}{6} =-1$

5 0

3 years ago

Write the equation of an ellipse with a major axis of length 8 and co-vertices (0,3) and (0,-3)

alexandr1967 [171]

If the co-vertices are (0, 3) and (0, -3) where x is 0 and y has a value, then y is the minor axis. That means that the x axis is the major axis. Because of what the co-vertices are, the center of the ellipse is at the origin. The formula for an ellipse that has a horizontal major axis is $\frac{(x-h)^2}{a^2}+ \frac{(y-k)^2}{b^2}=1$ . The a value will always be larger than the b value, therefore, the a value goes under the coordinate that is the major axis. Here, its the x-axis. a is the distance that the outer edge of the ellipse is from the center. It's 8 units away from the center along the x axis and 3 units along the y axis from the center. So a = 8 and a^2 = 64; b = 3 and b^2 = 9. Our formula then is $\frac{x^2}{64}+ \frac{y^2}{9}=1$