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3 years ago

CAN SOMEONE PLEASE HELP ME!!!!!!!!!

Mathematics

2 answers:

wlad13 [49]3 years ago

7 0

Answer:

12(m +1) > 18m - 3

12m + 12 > 18m - 3

12+3 > 18m - 12m

15 > 6 m.

m < 15/6

m < 2.5

4th option is correct.

Alex_Xolod [135]3 years ago

5 0

Answer:

{m| m < 2.5}

Step-by-step explanation:

m < 2 1/2,m < 2.5 , m ∑ [-∞, 5/2]

hope that helped!

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Two times the smaller of two consecutive integers is equal to 2 more than the larger. What is the LARGER integer?

LenaWriter [7]

Answer:

The larger integer is $4$ .

Step-by-step explanation:

Given: Two times the smaller of two consecutive integers is equal to $2$ more than the larger.

To find: The larger integer.

Solution:

Let the two consecutive integers be $x$ and $x+1$ , where the smaller integer is $x$ and larger integer is $x+1$ .

Now, according to question.

Two times the smaller integer $=$ $2$ more than the larger integer.

$\implies 2x=x+1+2$

$\implies 2x-x=3$

$\implies x=3$

So, the smaller integer is $3$ , and the larger integer is $3+1=4.$

Hence, the larger integer is $4.$

8 0

3 years ago

-4004 + 2023 – 24x2 – 2 / – 2x2 + 2x + 1?

Kamila [148]

Answer:

388.5/x

Step-by-step explanation:

multiply:

(-4004+2023-48-2)/ -4 + 2x +1

Simplify:

-2031/ -3 +2x

divide by -3

677/2x

388.5 / x

as simple as you can go

hope this helps!

5 0

3 years ago

Please help fast......

Tom [10]

$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\ 1.AY=BX&\text{1. Given}\\ 2.AB \cong AB&\text{2. Reflexive Property}\\ 3. AD || BC&\text{3. Property of a square}\\ 4. \angle ABE \cong \angle AXB&\text{4. Alternate Interior Angles}\\ 5. \angle BAY \cong \angle BYA&\text{5. Alternate Interior Angles}\\6. \triangle BAX \cong \triangle ABY&\text{6. Angle-Side-Angle Theorem}\\ 7. AX \cong BY&\text{7. CPCTC}\\\end{array}$

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$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\1. AB=CF&\text{1. Given}\\2.AB+BF=A'F&\text{2. Segment Addition Postulate}\\3.CF+BF=A'F&\text{3. Substitution Property}\\4.CF+BF+BC&\text{4. Segment Addition Postulate}\\5.A'F=BC&\text{5. Transitive Property}\\6. \angle AFE = \angle DBC&\text{6. Given}\\7. EF = BD&\text{7. Given}\\8. \triangle AFE \cong \triangle CBD&\text{8. Side-Angle-Side Theorem}\\\end{array}$

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$\begin{array}{c|c}\underline{Statement}&\underline{Reason}\\\text{1.AC bisects }\angle BAD&\text{1. Given}\\2. \angle BAC \cong \angle DAC&\text{2. Property of angle bisector}\\3.AC = AC&\text{3. Reflexive Property}&4. \angle ACB \cong \angle ACD&\text{4. Property of angle bisector}\\5. \triangle ABC \cong \triangle ADC&\text{5. Angle-Side-Angle Theorem}\\6.BC=CD&\text{6. CPCTC}\\\end{array}$

5 0

3 years ago

Find the distance between (-2,6) and (1,2)

olga nikolaevna [1]

Step-by-step explanation:

label the numbers (x,y) and (x2,y2)

write down the distance formula

replace the xs and ys by the numbers and then solve them

3 0

3 years ago

The school that DeShawn goes to is selling tickets to the annual dance competition. On the first day of ticket sales, the school

coldgirl [10]

Answer:

Both child tickets and senior tickets cost $14.

Step-by-step explanation:

Since the school that DeShawn goes to is selling tickets to the annual dance competition, and on the first day of ticket sales, the school sold 10 senior citizen tickets and 8 child tickets for a total of $ 252, while the school took in $ 280 on the second day by selling 10 senior citizen tickets and 10 child tickets, to determine what is the price of each of one senior citizen ticket and one child ticket, the following calculation must be performed:

10 senior tickets + 8 child tickets = 252

10 senior tickets + 10 child tickets = 280

280 - 252 = 2 child tickets

28 = 2 child tickets

28/2 = 1 child ticket

14 = 1 child ticket

14 x 10 = 140

(280 - 140) / 10 = senior tickets

140/10 = 14 = senior tickets

Therefore, both child tickets and senior tickets cost $14.

3 0

2 years ago