48n + 35m - 6m = please answer asap

Can someone help me with this plss

DIA [1.3K]

Answer:

6. Quadrant four

7. Quadrant three

8. Quadrant one

Step-by-step explanation:

These questions give you points to graph and then asks which quadrants they are graphed in. Here is a visual.

6 0

2 years ago

What does a formula do in a spreadsheet program?

vekshin1

The answer is D. Think back to the quadratic formula
You would plug in values of a, b, and c to find the zeros of the equations. It would the same way in a spreadsheet program where the cells contain the a, b, and c values but you can have more values and a different equation.

4 0

3 years ago

Read 2 more answers

Can someone help me with these 4 geometry questions? Pls it’s urgent, So ASAP!!!!

blagie [28]

Question 4

1) $\overline{BD}$ bisects $\angle ABC$ , $\overline{EF} \perp \overline{AB}$ , and $\overline{EG} \perp \overline{BC}$ (given)

2) $\angle FBE \cong \angle GBE$ (an angle bisector splits an angle into two congruent parts)

3) $\angle BFE$ and $\angle BGE$ are right angles (perpendicular lines form right angles)

4) $\triangle BFE$ and $\triangle BGE$ are right triangles (a triangle with a right angle is a right triangle)

5) $\overline{BE} \cong \overline{BE}$ (reflexive property)

6) $\triangle BFE \cong \triangle BGE$ (HA)

Question 5

1) $\angle AXO$ and $\angle BYO$ are right angles, $\angle A \cong \angle B$ , $O$ is the midpoint of $\overline{AB}$ (given)

2) $\triangle AXO$ and $\triangle BYO$ are right triangles (a triangle with a right angle is a right triangle)

3) $\overline{AO} \cong \overline{OB}$ (a midpoint splits a segment into two congruent parts)

4) $\triangle AXO \cong \triangle BYO$ (HA)

5) $\overline{OX} \cong \overline{OY}$ (CPCTC)

Question 6

1) $\angle B$ and $\angle D$ are right angles, $\overline{AC}$ bisects $\angle BAD$ (given)

2) $\overline{AC} \cong \overline{AC}$ (reflexive property)

3) $\angle BAC \cong \angle CAD$ (an angle bisector splits an angle into two congruent parts)

4) $\triangle BAC$ and $\triangle CAD$ are right triangles (a triangle with a right angle is a right triangle)

5) $\triangle BAC \cong \triangle DCA$ (HA)

6) $\angle BCA \cong \angle DCA$ (CPCTC)

7) $\overline{CA}$ bisects $\angle ACD$ (if a segment splits an angle into two congruent parts, it is an angle bisector)

Question 7

1) $\angle B$ and $\angle C$ are right angles, $\angle 4 \cong \angle 1$ (given)

2) $\triangle BAD$ and $\triangle CAD$ are right triangles (definition of a right triangle)

3) $\angle 1 \cong \angle 3$ (vertical angles are congruent)

4) $\angle 4 \cong \angle 3$ (transitive property of congruence)

5) $\overline{AD} \cong \overline{AD}$ (reflexive property)

6) $\therefore \triangle BAD \cong \triangle CAD$ (HA theorem)

7) $\angle BDA \cong \angle CDA$ (CPCTC)

8) $\therefore \vec{DA}$ bisects $\angle BDC$ (definition of bisector of an angle)

8 0

2 years ago

Anyone do Connections Academy??? just wanna know.-.

Contact [7]

Answer: no but some people do

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

Factor completely 4x4 − 24x3 + 36x2.

ipn [44]

The previous answer from another person got deleted, so I'm here to put it back in my own words.

(Edit: The answer is A)

First of all, you can take 4 out of each, so simplify everything by 4.

4(x^4 - 6x^3 + 9x^2)

Next, you can see that you're able to take out x^2

4x^2(x^2 - 6x + 9)

You can see that both x^2 and 9 are perfect squares, meaning you can crunch them together like this

(x+3)^2 or (x-3)^2

Of course, only one would work, and as fate has it, (x-3)^ produces x^2 - 6x + 9

This would turn the now factored equation into :

4x^2(x-3)^2 or 4x^2(x-3)(x-3), this means the answer is A

6 0

3 years ago

48n + 35m - 6m = please answer asap​

48n + 35m - 6m = please answer asap