Linear Equation 24 + 6a = 6 - 4a Please help me with this ASAP

Find two geometric means between 10 and 1250.

Roman55 [17]

Finding two geometric means between 10 and 125:
a 1 = 10, a 4 = 1250
a 4 = a 1 * q^3
1250 = 10 * q^3
q^3 = 1250 : 10
q^3 = 125
q = 5
a 2 = 10 * 5 = 50
a 3 = 10 * 5² = 10 * 25 = 250
Answer: The two geometric means are: 50 and 250, since 10, 50, 250 , 1250 forms a geometric sequence.

5 0

2 years ago

Simplify √63x^15y^9/√7xy^11 Please don’t guess I need to get this right!!

podryga [215]

Answer:

3x^7 / y

Step-by-step explanation:

√63x^15y^9/√7xy^11

= √ [(63/7) x^(15-1) y^(9-11)

= √9x^14y^-2

= √9x^14 / y^2

= 3x^7 / y

4 0

2 years ago

Read 2 more answers

What is the least common multiple of 9, 17, and 51? A. 51 B. 102 C. 153 D. 867

Luden [163]

Answer:

Hope this helps

shshsgshsh

3 0

3 years ago

The median is not affected by the existence of an outlier true or false

REY [17]

True.

The median is the middle number when all numbers are placed in order, and so an outlier will not affect the median.

It well however affect the range and mean

Hope this helps

3 0

3 years ago

I need help please, I am so confused.

vazorg [7]

Answer:

$\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}$

Step-by-step explanation:

x-intercepts are when the curve intercepts the x-axis, so when y =0.

Therefore, to find the x-intercepts, substitute y = 0 and solve for x.

The vertex is the turning point: the minimum point of a parabola that opens upward, and the maximum point of the parabola that opens downward. As a parabola is symmetrical, the x-coordinate of the vertex is the midpoint of the x-intercepts.

Equation: $y=x(x-2)$

$\implies x(x-2)=0$

$\implies x=0$

$\implies (x-2)=0 \implies x=2$

Therefore, the x-intercepts are x = 0 and x = 2

The midpoint of the x-intercepts is x = 1, so the x-coordinate of the vertex is x = 1

Equation: $y=(x-4)(x+5)$

$\implies (x-4)(x+5)=0$

$\implies (x-4)=0 \implies x=4$

$\implies (x+5)=0 \implies x=-5$

Therefore, the x-intercepts are x = -5 and x = 4

The midpoint of the x-intercepts is x = -0.5, so the x-coordinate of the vertex is x = -0.5

Equation: $y=-5x(3-x)$

$\implies -5x(3-x)=0$

$\implies -5x=0 \implies x=0$

$\implies (3-x)=0 \implies x=3$

Therefore, the x-intercepts are x = 0 and x = 3

The midpoint of the x-intercepts is x = 1.5, so the x-coordinate of the vertex is x = 1.5

$\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}$

4 0

2 years ago