All categories

3 years ago

Hat are the intercepts for the equation? Select all that apply. 2x + 3y = 6 *

Mathematics

1 answer:

-Dominant- [34]3 years ago

7 0

Answer:

A and D

Step-by-step explanation:

To find the x intercepts, substitute 0 into y.

2x = 6

(3,0)

To find the y intercept, substitute 0 into x;

3y = 6

(0, 2)

So a and d

You might be interested in

The school yearbook committee surveyed the student body for an article about colleges in which they are pursuing enrollment. the

Dahasolnce [82]

Can u help me with this question in need help

8 0

4 years ago

Read 2 more answers

Answer answer answer plzz

lubasha [3.4K]

Answer: g > 6

The graph is below

The graph has an open hole at 6; shading to the right

=============================================================

Explanation:

I'm assuming the variable g is <u>not</u> in the denominator.

To solve for g, we'll first multiply both sides by 3

$-\frac{2}{3}g < -4\\\\3*\left(-\frac{2}{3}g\right) < 3(-4)\\\\-2g < -12\\\\$

Then divide both sides by -2 to fully isolate g. Dividing both sides by a negative value flips the inequality sign.

$-2g < -12\\\\\frac{-2g}{-2} > \frac{-12}{-2}\\\\g > 6$

The solution for g is any number larger than 6.

The graph will have an open hole at 6 and shading to the right. See below.

7 0

3 years ago

A cake is sliced into 12 equal pieces 1/2 of the cake is: slices

klasskru [66]

Answer:

Step-by-step explanation:

1/2 of 12

1/2×12

i hope this will help

7 0

3 years ago

According to Descartes's rule of sign, how many possible positive and negative roots are there for the equation

damaskus [11]

<h2>Descartes Rule of Sign</h2><h3>Answer:</h3>

There are $3$ Real Roots for the Equation

<h3>Step-by-step explanation:</h3>

According to Descartes Rule of Sign, The maximum number of real roots a polynomial equation has is the number of times the sign of term changes from a higher power term to a lower power term.

In the given equation, $0=-4x^6-3x^5+2x^2-4x+1$ , we can already see that from $\red -4x^6$ to $\red -3x^5$ we can see that the sign didn't change so there might be no real roots. From $\red -3x^5$ to $\blue +2x^2$ , we can see that the signs change so we now know that there might be $1$ real root for equation. From $\blue +2x^2$ to $\red -4x$ , the signs changed so there might be $2$ real roots. From $\red -4x$ to $\blue +1$ , it changed again so there might be $3$ real roots. There are no other term with lesser power to compare for $\blue +1$ so we are left with $3$ real roots.