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3 years ago

Y = 4 + 6x Siope-interce What linear equation form is this equation written in?

Mathematics

1 answer:

uysha [10]3 years ago

7 0

Answer:

We can write this equation in slope-intercept form

where

is the slope and

is the

-intercept.

We know

−

and

, so we can insert these into the equation to get

−

Hope this helps!

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Helpppppppppppppppppp!

julia-pushkina [17]

Remember that the radicand (the area under the root sign) must be positive or zero for a radical with an even index (like the square root or fourth root, for example). This is because two numbers squared or to the fourth power, etc. cannot be negative, so there are no real solutions when the radicand is negative. We must restrict the domain of the square-root function.

If the domain has already been restricted to $x \geq -11$ , we can work backwards to add 11 to both sides. We see that $x+11$ must be under the radicand, so the answer is A.

7 0

3 years ago

Solve the inequality −22 + 4q > 2

Mars2501 [29]

First you had to add by twenty-two from both sides of equation form.

$-22+4q+22>2+22$

Then simplify by equation.

$4q>24$

Next, you divide by four from both sides of equation form.

$\frac{4q}{4}> \frac{24}{4}$

Finally, simplify by equation.

$24/4=6$

$24/6=4$

$6*4=24$

$4*6=24$

$q>6$

Final answer: $\boxed{q>6}$

Hope this helps!

And thank you for posting your question at here on brainly, and have a great day.

-Charlie

8 0

3 years ago

deirdre is factoring the polynomial x squared plus 2x minus 24. which combination can she use to replace the middle term, 2x ?

Shkiper50 [21]

we are given with

The polynomial x squared plus 2x minus 24

$\\\x^2+2x-24\\\\\=x^2+6x-4x-24\\\\\=x(x+6)-4(x+6)\\\\\=(x-4)(x+6)\\$

Hence the required combination is 6x and -4x

3 0

3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

Read 2 more answers

What is the measure of ∠ABC?

____ [38]

Answer:

not sure

Step-by-step explanation:

I'm not sure how to do this guys I am very sorry for the inconvenience

5 0

3 years ago

Read 2 more answers