Divide. Write your answer in simplest form. 7 -:6 3

Help? I don’t understand it mainly cause HOW ARE YOU SUPPOSED TO

Georgia [21]

Answer:

83

supplementary is two angles that add up to 180 so 180 - 97= 83

7 0

3 years ago

What is the HCF of 15and25

m_a_m_a [10]

Answer:

5

Step-by-step explanation:

HOPE IT HELPS YOU .......

4 0

3 years ago

What is the zero of the function

olga_2 [115]

Answer:

$x=6$ (or 6)

Step-by-step explanation:

Set f(x)=0:

$f(x)=\frac{x^2+x-42}{x^2+3x-28}$

$0=\frac{x^2+x-42}{x^2+3x-28}$

$0=x^2+x-42$

$0=(x+7)(x-6)$

$x=-7$ or $x=6$

Now check both solutions:

$f(-7)=\frac{(-7)^2+(-7)-42}{(-7)^2+3(-7)-28}$

$f(-7)=\frac{49-7-42}{49-21-28}$

$f(-7)=\frac{42-42}{28-28}$

$f(-7)=\frac{0}{0}$

Therefore, $x\neq-7$ because there is a hole at $(-7,0)$ . The denominator, in addition, can never be 0, so the function is undefined for $x=-7$ .

$f(6)=\frac{(6)^2+(6)-42}{(6)^2+3(6)-28}$

$f(6)=\frac{36+6-42}{36+18-28}$

$f(6)=\frac{42-42}{54-28}$

$f(6)=\frac{0}{24}$

$f(6)=0$

Since $f(6)=0$ , then $x=6$ is the only zero of the function.

4 0

3 years ago

Your friend incorrectly says the solution to the equation 11.2y - 7.4y=141.36 is y=7.6 what error did your friend make?

DerKrebs [107]

The error is clearly choice B.

11.2y - 7.4y should be subtracted not added.

Answer: Choice B

6 0

3 years ago

Read 2 more answers

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