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3 years ago

Will mark brainliest and give 20 points!

Mathematics

1 answer:

jenyasd209 [6]3 years ago

3 0

Verticalset the denominator of the function equal to zero and solve for 'x', x is the vertical asymptote.Horizontalsubtract the amount of x's in the numerator from the number of x's in the denominator, then do one of the four things to find y: and if the answer is negative, then y=0 does that help any?if the answer is 0, then y= a/b (ax/bx)If the answer is greater than positive 1, then there is no horizontal asymptote
ObliqueIf the answer is 1, then it is oblique. use long division with the denominator as the divisor of the overall function given (not the fraction) and y= ax+/- b
Does this help at all
????

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At midnight, the temperature was 31.2°F. In the morning, the temperature was −6.7°F. Which statement describes the temperatures?

deff fn [24]

At midnight the temperature was 31.2 degrees above 0 and in the morning it was 6.7 degrees below 0.

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3 years ago

Read 2 more answers

Evaluate. [(12−5)⋅12]÷6

snow_lady [41]

So first you would use the PEMDAS method. You would begin by subtacting 12 and 5 which is 7 then you would 7 multiply by 12 which is 84. Lastly you would divide it by 6 which would leave you with your final answer 14

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3 years ago

What must be added to 1,316 to make 5

Ratling [72]

Answer:

-1311

Step-by-step explanation:

1316 + x = 5

Subtract 1316 from both sides:

1316 - 1316 + x = 5 - 1316

On the left side they cancel out and we are left with:

x = 5 - 1316

Which simplifies to:

x = -1311

7 0

2 years ago

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

PLEASE HELP AS SOON AS POSSIBLE

frutty [35]

Answer:

FALSE

TRUE

Step-by-step explanation:

For median,

Arrange your numbers in numerical order.

Count how many numbers you have.

If you have an odd number, divide by 2 and round up to get the position of the median number.

If you have an even number, divide by 2. Go to the number in that position and average it with the number in the next higher position to get the median.

Source: https://www.verywellmind.com/how-to-identify-and-calculate-the-mean-median-or-mode-2795785

So for April median: 2.5, 3, 3.5, 3.5 (since we have 4 numbers we divide by 2 and we count by 2 places from left. Since we have even number data, we average 3 with the next higher position to get the median so (3+3.5)/2 = 3.25

For May apply same concept and you get median to be 2.25.

Median difference is 3.25-2.25 = 1

Therefore, statement 1 is false and statement 2 is true.

For average in May, add all numbers and divide by the number of data points so May= (2.5+3+3.5+3.5)/ 4 = 3.13

Apply same concept you for April and you get 2.38 as mean.

Mean difference is 3.13-2.38 = 0.75

Therefore, statement 3 is correct (true)

4 0

2 years ago