Write an algebraic equation to represent the following problem:


Five friends went out to dinner. Each one of them left an $18 dollar tip and they also split the $148 bill evenly. How much did each person pay for dinner?

PLS I NEED THIS!! I'll give brainiest if correct!!!!!!

Answer:

12:5

Step-by-step explanation:

To calculate the ratio, we must first find the values of the two expenditures. The amount spent on renting a hall was 60% of £1000. To represent percentages in decimals, we can simply divide the percentage by 100, resulting in 0.6 here. We can then multiply this value by 1000 to get the amount spent on renting a hall, which comes out to £600

The ratio is then 600 : 250,  60 : 25 (divide both sides by 10) , or 12:5 (divide both sides by another 5)

Answer:

i am a duck and i can confirm water is pretty poggers

Step-by-step explanation:

Answer:

1.false

2.true

3.true

4.true

5.120

6.105

7.85

8.100

9.115

10.40

11.13

12.30

13.36

14.65

15.35

16.75

Answer:

376.8

Step-by-step explanation:

Formula Volume Cone = πr² * h/3

Volume = 3.14 * 36 * 3 1/3

Volume ≈ 376.8

If my answer is incorrect, pls correct me!

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-Chetan K

Answer: 301.6

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answer above isn't right this is, trust me.

Answer:

last one

Step-by-step explanation:

[tex]\frac{5}{7}[/tex] = [tex]\frac{10}{14}[/tex]

[tex]\frac{1}{2}[/tex] = [tex]\frac{7}{14}[/tex]

need common denominators to add fractions.

what ever is multiplied to the bottom needs to be multiplied to the top

for example : [tex]\frac{5}{7}[/tex] x [tex]\frac{2}{2}[/tex] = [tex]\frac{10}{14}[/tex]

Answer:

no bc pythagorean

Step-by-step explanation:

Answer:

volume = 12.56 cubic feet

Step-by-step explanation:

[tex]volume = \pi r^2 \times h \\\\volume = 3.14 \times 1 ^2 \times 4 = 3.14 \times 4 = 12. 56 \ cubic \ feet[/tex]

Answer:

It is not a rectangle

Step-by-step explanation:

A rectangle is a shape with 4 sides, such that we have two pairs of parallel sides, where the length of the parallel sides is equal.

so we have two parallel sides with length L and two parallel sides with length L'.

In this figure ,we can see two sides with length 20 and other two with length 21, and this is a rectangle only if the sides meet at 90°degree angles (this would mean that the opposite sides are parallel)

To see this, we can think that the triangle formed by the diagonal and two sides is a triangle rectangle, then using the Pythagorean theorem we should have that:

20^2 + 21^2 = 28^2

841 = 784

So this is not a triangle rectangle, meaning that the angle between the side of 20 and the side of 21 is not a right triangle, then the figure is not a rectangle.

Therefore, it is not a rectangle

Hope u understand

Please mark the brainliest

Thank You

Answer:

the correct answer is

C.

StartFraction 12 Over 16 EndFraction = StartFraction 18 Over 24 EndFraction = StartFraction 3 Over 4 EndFraction

Answer:

x=8

Step-by-step explanation:

assuming this is what you mean:

(5x+2)=42

try to

isolate the variable by subtracting 2 from both sides

5x=40

continue to isolate the variable

divide by 5 on both sides

5x/5=40/5

x=8

Answer:

no

Step-by-step explanation:

To determine if the point lies on the line, substitute x = - 3 into the equation and if the value obtained equals - 5 then it lies on the line.

y = 2(- 3) - 1 = - 6 - 1 = - 7 ≠ - 5

Then the point (- 3, - 5 ) is not on the line

Answer:

Option A

then you would have an angle, a side and another angle (ASA) next to each other

Opt.B would be a second sude (SAS)

Opt.C is already given

Opt.D seems to be like (A-S-S). no pun intended.

The additional information need to be added to show the given triangles congruent is angle B = angle E

Triangles are said to be congruent if their corresponding sides and angles are congruent.

Given is a pair of triangles, Δ ABC and Δ DEF, in this angle A = angle D and AB = DE = 10,

We need to find one more reason to make them congruent by ASA rule,

In ASA rule of congruence, we have two corresponding angles congruent in the triangles and a side between them need to be congruent,

Here, the pair of two angles should be ∠ A and ∠ D and ∠ B and ∠ E, because the congruent sides AB and DE are in mid of them.

Hence, the additional information need to be added to show the given triangles congruent is angle B = angle E

Learn more about triangles congruent click;

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Answer:

Logically that's not possible but if you want to prove it then, you have to use a trick.

Step-by-step explanation:

1 + 1 = 1 + [tex]\sqrt{1}[/tex]

        = 1 + [tex]\sqrt{(-1) * (-1)}[/tex]  

        = 1 + [tex]\sqrt{(-1)}[/tex] * [tex]\sqrt{(-1)}[/tex]

        = 1 + [tex]\sqrt{i^2}[/tex] * [tex]\sqrt{i^2}[/tex]

        = 1 + i * i  

        = 1 + i^2

        = 1 + (-1)

        = 1 - 1

        = 0

So, 1 + 1 = 0

Hope this will help. Please give me brainliest.

Answer:

Problem 1: 25

Problem 2: 693

Step-by-step explanation:

Problem 1:

f(x) = -5n + 1

g(x) = -6n + 2

(f + g)(-2) = f(-2) + g(-2)

f(-2) = -5(-2) + 1

f(-2) = 10 + 1

f(-2) = 11

g(-2) = -6(-2) + 2

g(-2) = 12 + 2

g(-2) = 14

(f + g)(-2) = 11 + 14

(f + g)(-2) = 25

Problem 2:

f(x) = 7 + 2x

g(x) = 5x - 2

fg(7) = f(7) × g(7)

f(7) = 7 + 2(7)

f(7) = 7 + 14

f(7) = 21

g(x) = 5(7) - 2

g(x) = 35 - 2

g(x) = 33

Therefore:

fg(7) = 21 × 33

fg(7) = 693

Answer:

[tex]-6 < k < 2[/tex]

Step-by-step explanation:

Given

[tex]y = x^2 - 2x[/tex]

[tex]y =kx -4[/tex]

Required

Possible values of k

The general quadratic equation is:

[tex]ax^2 + bx + c = 0[/tex]

Subtract [tex]y = x^2 - 2x[/tex] and [tex]y =kx -4[/tex]

[tex]y - y = x^2 - 2x - kx +4[/tex]

[tex]0 = x^2 - 2x - kx +4[/tex]

Factorize:

[tex]0 = x^2 +x(-2 - k) +4[/tex]

Rewrite as:

[tex]x^2 +x(-2 - k) +4=0[/tex]

Compare the above equation to: [tex]ax^2 + bx + c = 0[/tex]

[tex]a = 1[/tex]

[tex]b= -2-k[/tex]

[tex]c =4[/tex]

For the equation to have two distinct solution, the following must be true:

[tex]b^2 - 4ac > 0[/tex]

So, we have:

[tex](-2-k)^2 -4*1*4>0[/tex]

[tex](-2-k)^2 -16>0[/tex]

Expand

[tex]4 +4k+k^2-16>0[/tex]

Rewrite as:

[tex]k^2 + 4k - 16 + 4 >0[/tex]

[tex]k^2 + 4k - 12 >0[/tex]

Expand

[tex]k^2 + 6k-2k - 12 >0[/tex]

Factorize

[tex]k(k + 6)-2(k + 6) >0[/tex]

Factor out k + 6

[tex](k -2)(k + 6) >0[/tex]

Split:

[tex]k -2 > 0[/tex] or [tex]k + 6> 0[/tex]

So:

[tex]k > 2[/tex] or k [tex]> -6[/tex]

To make the above inequality true, we set:

[tex]k < 2[/tex] or [tex]k >-6[/tex]

So, the set of values of k is:

[tex]-6 < k < 2[/tex]

Answer:

25, 32, 40 can not form a right angle triangle.

Step-by-step explanation:

By Pythagoras theorem,

⇒ (40)² = (25)² + (32)²

⇒ 1600 = 625 + 1024

⇒ 1600 ≠ 1689

Answer:

option b

Step-by-step explanation:

according to the pythagoras theorem to be right angled triangle the sum of square of two smaller sides must be equal to the square of hypotenuse.

so ,

20 and 21 are smaller sides and hypotenuse be 29

pythagoras theorem

a^2 + b^2 = c^2

20^2 + 21^2 = 29^2

400 + 441 = 841

841 = 841  (since both sides are equal it forms right angled triangle)

25 and 32 are smaller sides and 40 be hypotenuse

a^2 + b^2 = c^2

25^2 + 32^2 = 40^2

625 + 1024 = 1600

1649 = 1600 (both sides are not equal so it does not form right angle triangle)

30 and 72 are smaller sides whereas 78 is hypotenuse

a^2 + b^2 = c^2

30^2 + 72^2 = 78^2

900 + 5184 = 6084

6084 = 6084 (since both sides are equal it forms right angled triangle )

32 and 60 are smaller sides and 68 is hypotenuse

a^2 + b^2 = c^2

32^2 + 60^2 = 68^2

1024 + 3600 = 4624

4624 = 4624  (since both sides are equal it forms right angled triangle )

The Above triangles are similar, Therefore the corresponding sides would be proportional.

that is : -

[tex]\mathfrak{best \: \: of \: \: luck \: \: for \: \: your \: \: assignment}[/tex]

Answer:

osjdjaidbsojaanod

Step-by-step explanation:

the answer is in the above image

Answer:

9

Step-by-step explanation:

Plug in 0 for y and solve for x.

0 = (-1/3)x + 3

(-1/3)x = -3

x = 9

Answer:

cumulative frequency

1

8

25

45

58

62

Answers:

Cumulative frequencies from top to bottom:

Other answers:

Refer to the diagram below

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

Explanation:

In the first row, the frequency is 1, so that's the cumulative frequency for that row. It's the total frequency so far.

In the second row, the cumulative frequency is now 8 because we add the two frequencies so far (1+7 = 8)

The third row will have a cumulative frequency of 1+7+17 = 25.

The rest of the rows will follow this pattern to fill out the table. Refer to the diagram below to see the filled out table.

----------------------------

At the very end, you should get 62 golfers total (the cumulative frequency for the bottom row).

The mode score is the most frequent value. In this case, that's the score 72 since it shows up the most times (ie has the largest frequency).

Because we have n = 62 people total, the median is between slots n/2 = 62/2 = 31 and 32

Go back to the table and note how 25 is the cumulative frequency for the third row. Since 25 is smaller than 31 and 32, this means that the median cannot be in the first three rows. Instead, it's in the fourth row because the frequency 17 here is more than enough to get us from slot 25 to slots 31 and 32.

In other words, the values 72 and 72 are in slots 31 and 32. The midpoint of which is 72.

The median is therefore 72.

Side note: the mode and median aren't always the same value.

Explanation:

Pythagorean Triples are sets of whole numbers for which the Pythagorean Theorem holds true. The most well-known triple is 3, 4, 5. This means that 3 and 4 are the lengths of the legs and 5 is the hypotenuse.

Answer:

(10) Person B

(11) Person B

(12) [tex]P(5\ or\ 6) = 60\%[/tex]

(13) Person B

Step-by-step explanation:

Given

Person A [tex]\to[/tex] 5 coins (records the outcome of Heads)

Person [tex]\to[/tex] Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

[tex]Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}[/tex]

[tex]n(Head) = 32[/tex]

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

Next, we list out the highest in each toss (sorted)

[tex]Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}[/tex]

[tex]n(Dice) = 30[/tex]

Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

[tex]Pr(5) = \frac{n(5)}{n(Head)}[/tex]

[tex]Pr(5) = \frac{1}{32}[/tex]

[tex]Pr(5) = 0.03125[/tex]

From person B list of outcomes, the proportion of 5 is:

[tex]Pr(5) = \frac{n(5)}{n(Dice)}[/tex]

[tex]Pr(5) = \frac{8}{30}[/tex]

[tex]Pr(5) = 0.267[/tex]

From the above calculations: [tex]0.267 > 0.03125[/tex] Hence, person B is more likely to get 5

Question 11: Person with Higher median

For person A

[tex]Median = \frac{n(Head) + 1}{2}th[/tex]

[tex]Median = \frac{32 + 1}{2}th[/tex]

[tex]Median = \frac{33}{2}th[/tex]

[tex]Median = 16.5th[/tex]

This means that the median is the mean of the 16th and the 17th item

So,

[tex]Median = \frac{3+2}{2}[/tex]

[tex]Median = \frac{5}{2}[/tex]

[tex]Median = 2.5[/tex]

For person B

[tex]Median = \frac{n(Dice) + 1}{2}th[/tex]

[tex]Median = \frac{30 + 1}{2}th[/tex]

[tex]Median = \frac{31}{2}th[/tex]

[tex]Median = 15.5th[/tex]

This means that the median is the mean of the 15th and the 16th item. So,

[tex]Median = \frac{5+5}{2}[/tex]

[tex]Median = \frac{10}{2}[/tex]

[tex]Median = 5[/tex]

Person B has a greater median of 5

Question 12: Probability that B gets 5 or 6

This is calculated as:

[tex]P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}[/tex]

From the sample space of person B, we have:

[tex]n(5\ or\ 6) =n(5) + n(6)[/tex]

[tex]n(5\ or\ 6) =8+10[/tex]

[tex]n(5\ or\ 6) = 18[/tex]

So, we have:

[tex]P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}[/tex]

[tex]P(5\ or\ 6) = \frac{18}{30}[/tex]

[tex]P(5\ or\ 6) = 0.60[/tex]

[tex]P(5\ or\ 6) = 60\%[/tex]

Question 13: Person with higher probability of 3 or more

Person A

[tex]n(3\ or\ more) = 16[/tex]

So:

[tex]P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}[/tex]

[tex]P(3\ or\ more) = \frac{16}{32}[/tex]

[tex]P(3\ or\ more) = 0.50[/tex]

[tex]P(3\ or\ more) = 50\%[/tex]

Person B

[tex]n(3\ or\ more) = 28[/tex]

So:

[tex]P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}[/tex]

[tex]P(3\ or\ more) = \frac{28}{30}[/tex]

[tex]P(3\ or\ more) = 0.933[/tex]

[tex]P(3\ or\ more) = 93.3\%[/tex]

By comparison:

[tex]93.3\% > 50\%[/tex]

Hence, person B has a higher probability of 3 or more

Option C with h and k

see screenshot for an example

general formula is

r² = (x - x-center)² + (y - y-center)²

Answer:

1/20

Step-by-step explanation:

only a single number can be chosen out of 20. And 1/20 is the most simplified it can get.

Answer:

Answer is 1st and Last

Step-by-step explanation:

Because if you use logic you should know why

Answer:2 and 5

Step-by-step explanation:

The ratios that are equal to [tex]cos(B)[/tex] : [tex]\bold{\frac{BC}{AB}}[/tex] and [tex]\bold{\frac{EF}{DE}}[/tex]

"Two triangles are said to be similar

- if corresponding angles are equal and the sides

- if corresponding sides of the triangles are in proportion."

In a right triangle,

cos(Ф) = (adjacent side of angle Ф) ÷ hypotenuse

For given example,

Triangles ABC and DEF are similar triangles.

This means, ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

And, [tex]\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}[/tex]               ..............(i)

For a right triangle ABC,

the cosine of angle B is,

[tex]cos(B)=\frac{BC}{AB}[/tex]

From (i),

[tex]\frac{BC}{AB}=\frac{EF}{DE}[/tex]

So, the cosine of angle B is,

[tex]cos(B)=\frac{EF}{DE}[/tex]

Therefore, the ratios that are equal to [tex]cos(B)[/tex] : [tex]\frac{BC}{AB}[/tex] and [tex]\frac{EF}{DE}[/tex]

Learn more about similar triangles here:

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Answer:

in 10 weeks because if you multiply 62.50 10 times you get 625

Answer:

The answer is 10

I hope this helps