Who can help me finish a course on edge ive done half of the course already its algebra !!

5x + 3 = 7x – 1. Find x a. 1/3 b. ½ c. 1 d. 2

weqwewe [10]

Answer:

Step-by-step explanation:

First you need to get like items on the same side(put the variables on one side and the non-variables on the other). Remember anything you do on one side of the equal, you have to do on the other side, too.

5X+3=7X-1 I'm going to move the -1 first, but you don't have to as long as you keep the equation equal.

5X+3+1=7X-1+1 We know that +1-1 is zero and so is -1+1. So

5X+3+1=7X We don't need to keep that zero.

5X+4=7X Now, we want to move the 5X over.

5X-5X+4=7X-5X Simplify and get rid of the zero.

4=2X Last, we have to divide to get the 2 away from the X.

4/2=2X/2 Not everyone know that slash mean divide, :-).

2=X

3 0

3 years ago

T - 6.2 < 4 help me please

enyata [817]

Answer:

t < 10.2

Step-by-step explanation:

Step 1: Add 6.2 to both sides.

$t - 6.2 + 6.2 < 4 + 6.2$
$t < 10.2$

Therefore, the answer is t < 10.2.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

6 0

2 years ago

What is the volume of a sphere if its radius is 30

ASHA 777 [7]

Answer:

About 113,097.34 units

Step-by-step explanation:

The volume of a sphere can be found with the equation V = $\frac{4}{3}$ πr³ when r is the radius.

V = $\frac{4}{3}$ πr³

V = $\frac{4}{3}$ π(30)³

V ≈ 113,097.34 units

3 0

2 years ago

Find the critical points, domain endpoints, and local extreme values for the function

aliina [53]

Answer:

a) $x = -3$ , b) $x = 0$ , $x = -6$ , c) $x = 0$ , d) $x = -6$

Step-by-step explanation:

a) Let derive the function:

$f'(x) = \frac{10\cdot x \cdot (x+3)-5\cdot x^{2}}{25\cdot (x+3)^{2}}$

$f'(x)$ is undefined when denominator equates to zero. The critical point is:

$x = -3$

b) $f'(x) = 0$ when numerator equates to zero. That is:

$10\cdot x \cdot (x+3) - 5\cdot x^{2} = 0$

$10\cdot x^{2}+30\cdot x -5\cdot x^{2} = 0$

$5\cdot x^{2} + 30\cdot x = 0$

$5\cdot x \cdot (x+6) = 0$

This equation shows two critical points:

$x = 0$ , $x = -6$

c) The critical points found in point b) and the existence of a discontinuity in point a) lead to the conclusion of the existence local minima and maxima. By plotting the function, it is evident that $x = 0$ corresponds to a local maximum. (See Attachment)

d) By plotting the function, it is evident that $x = -6$ corresponds to a local minimum. (See Attachment)