F(1)=2, f(n)= f(n-1)+7

Plsss help

Answer:

Step-by-step explanation:

Our even numbers from 1-9 are:

The largest possible number using the even numbers once is 8642.

Hoped this helped

Answer:

The answer 19\24 is not reasonable because ​5\6​ is closer to 1 than to ​1\2​, and ​3\8​ is close to ​1\2​, making the difference close to 1\2. ​

Step-by-step explanation:

Answer:

1/2 1 0

Step-by-step explanation:

Answer:

x = 16

Step-by-step explanation:

If 3.75x would equal to 60, then x would equal to 60/3.75, which would be 16

Hope this helps!

Answer:

(1.7x10^13)+(0.8x10^13)

=(1.7x1e+13)+(0.8x1e+13)

=1.7e+13+0.8e+13

=2.5e+13

in actual number it would be 25,000,000,000,000

Step-by-step explanation:

Answer:

Step-by-step explanation:

We have 8 dozen bagels, or 8*12=96 bagels.  Each plate can hold 14 bagels, so we have enough bagels to fill 96/14=about 6.86 plates.  However, we cannot have a fraction of a plate, so we round up to have a total of seven plates.  To fill all seven plates fully, 7*14=98 bagels would be needed, which is two more than we have.

To summarize, Mr. Corsetti has seven plates of bagels, and would need two more bagels to fill the last one up.

Answer:

i belive its A

Step-by-step explanation: hope this helps :)

Answer:

B. 4

Step-by-step explanation:

4 times 5 is 20 and 20 divided by 5 is 4

Total pages read after 4h

[tex]\\ \sf\longmapsto 4(55)=220pages[/tex]

Pages left=350

Now

[tex]\\ \sf\longmapsto y=350+220[/tex]

[tex]\\ \sf\longmapsto y=570[/tex]

Answer: 9 triangles

Step-by-step explanation:

[tex](4a - 5b)^{2} \\ by \: \: \: using \: \: \: (x - y)^{2} = {x}^{2} - 2xy + {y}^{2} \\ = {(4a)}^{2} - 2(4a)(5b) + {(5b)}^{2} \\ = {16a}^{2} - 40ab + 25 {b}^{2} [/tex]

[tex] {16a}^{2} - 40ab + {25b}^{2} [/tex]

Hope it helps.

Do comment if you have any query.

Answer:

MQP = 138

Step-by-step explanation:

Since the angle bisector always divides the angle in half exactly, we know both of the angle measures we have are the same, and set it up like this: 5x + 19 = 2x + 49 (subtract 2x from both sides) 3x + 19 = 49 (subtract 19 on both sides) 3x = 30 (divide by 3) x = 10

Substitute that x value (10) into the original equations in place of x for each angle to get MQN = 69 and NQP = 69

Add the two values to get 138. MQP = 138

Answer:

138

Step-by-step explanation:

QN bisects ∠MQP. ⇒ ∠MQN = ∠NQP

5x + 19 = 2x + 49

Subtract 19 from both sides

5x = 2x + 49 - 19

5x = 2x + 30

Subtract 2x from both sides

5x -2x = 30

3x = 30

x = 30/3

x = 10

∠MQN = 5x + 19

            = 5*10 + 19

            = 50 +19

             = 69

∠MQP = ∠MQN +∠NQP = 69 + 69 = 138

Answer:

B. Every 16 minutes, 22 more pizzas were set out on the buffet.

Step-by-step explanation:

At 11:00 am, x would be equal to 0, and there would be 15 pizzas on the buffet line. Then, every 8 minutes, 11 more pizzas are added to the buffet line. So in 16 minutes, 22 pizzas will be added to the buffet.

[tex]\frac{16}{8} =2\\(11)(2) = 22[/tex]

Answer:

a) 2 * 3 = 6.

b) 123 is odd but contains an even digit (2).

Step-by-step explanation:

325

I hope that helps

Answer:

z

Step-by-step explanation:

hqvqhqvw karrar ras wallah a part

Answer:

15 hours

Step-by-step explanation:

20%=3 hours

times 5 both sides

100%=15 hours

Answer:

Option D)  [tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex]

Step-by-step explanation:

Given the equation, T = 15 + c × n, where:

T = represents the total monthly gym cost

c =  represents the additional charge for a group class, and

n = represents the number of group classes attended

In order to determine which equation can be used to find the number of group classes a customer attended, if there are given values for c and T, we must isolate the variable, n  algebraically.

The first step is to subtract 15 from both sides:

T = 15 + c × n

T - 15 = 15 - 15 + c × n

T - 15 = c × n

Next, divide both sides by c to isolate n :

[tex]\huge\mathsf{\frac{({T\:-\:15})}{c}\:=\:\frac{{c\:\times\:n}}{c}}[/tex]

[tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex]

Therefore, the correct answer is Option D) [tex]\huge\sf{n\:=\:\frac{(T\:-\:15)}{c}}[/tex].

Answer:

|-4| < |-5|

Step-by-step explanation:

because if modules is given sub sign will be deducate

Answer: i think it is y=−56x−4

Step-by-step explanation:

The equation in the slope intercept form which passes through the points ( 0, -4 ) and ( 6 , 9 ) is y = (-5 / 6)x - 4.

The correct answer is Option C.

Given data:

To find the equation of a line in slope-intercept form (y = mx + b) that passes through the points (0, -4) and (6, -9), we need to determine the slope (m) and the y-intercept (b).

First, calculate the slope (m):

m = (change in y) / (change in x)

m = (-9 - (-4)) / (6 - 0)

m = (-9 + 4) / 6

m = -5 / 6

Now that we have the slope, we can use one of the given points (let's use (0, -4)) to solve for the y-intercept (b):

-4 = (-5 / 6) * 0 + b

-4 = b

So, the y-intercept (b) is -4.

Now, we can write the equation of the line in slope-intercept form:

y = (-5 / 6)x - 4

Hence, the equation of the line is y = (-5 / 6)x - 4.

To learn more about equation of line, refer:

https://brainly.com/question/14200719

#SPJ3

The complete question is attached below:

Which equation, in slope-intercept form, matches the equation shown?

a line that goes through the points (0, -4) and (6, -9)

A) y = ( 4/7 )x - 4

B) y = ( 5/6 )x + 1

C) y = ( -5/6 )x - 4

D) y = ( -4/7 )x + 1

Answer:

y=3/4x-8

Step-by-step explanation:

Answer:

What is the common ratio of the sequence 3, 21, 147, …?  

7

Which formula can be used to find the nth term of the sequence 3, 21, 147, …?

c)

Use the given formula to find the indicated terms of the sequence.

a4 = 1029

a5 = 7203

Step-by-step explanation:

Answers for all three questions<3

Answer:

[tex]\frac{\sqrt{70} }{5}[/tex]

Step-by-step explanation:

[tex] = {( \frac{27}{8} )}^{ \frac{1}{3} } \times ( \frac{243}{32} )^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = { ({ (\frac{3}{2} )}^{3}) }^{ \frac{1}{3} } \times {( {( \frac{3}{2}) }^{5} )}^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{3 \times \frac{1}{3} } \times {( \frac{3}{2} )}^{5 \times \frac{1}{5} } \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = \frac{3}{2} \times \frac{3}{2} \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{1 + 1 + 2} [/tex]

[tex] = {( \frac{3}{2} )}^{4} \: or \: \frac{81}{16} [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{27}{8} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{243}{32} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

We can write as :

27 = 3 × 3 × 3 = 3³

8 = 2 × 2 × 2 = 2³

243 = 3 × 3 × 3 × 3 × 3 = 3⁵

32 = 2 × 2 × 2 ×2 × 2 = 2⁵

[tex]\sf{\longmapsto{\bigg( \dfrac{3 \times 3 \times 3}{2 \times 2 \times 2} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{{(3)}^{3}}{{(2)}^{3}} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{({3}^{5})}{{(2)}^{5}} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now, we can write as :

(3³/2³) = (3/2)³

(3⁵/2⁵) = (3/2)⁵

[tex]\sf{\longmapsto{\left\{\bigg(\frac{3}{2} \bigg)^{3} \right\}^{\frac{1}{3}} \times \Bigg[\left\{\bigg(\frac{3}{2} \bigg)^{5} \right\}^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now using law of exponent :

[tex]{\sf{({a}^{m})^{n} = {a}^{mn}}}[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{3 \times \frac{1}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{5 \times \frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{\frac{3}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{\frac{5}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times\Bigg[\bigg(\frac{3}{2} \bigg)^{1} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \times \dfrac{3}{2} \bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3 \times 3}{2 \times 2}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)\times \Bigg[\bigg(\frac{3}{2} \bigg)\times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3}{2} \times \dfrac{9}{4} \: \: \Bigg]}}\\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3 \times 9}{2 \times 4} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg(\dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{27}{8} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3}{2} \times \dfrac{27}{8}}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3 \times 27}{2 \times 8}}} \\[/tex]

[tex] \sf{\longmapsto{\dfrac{81}{16}}\: ≈ \:5.0625\:\red{Ans.}} \\[/tex]

Answer:

Answer d

Step-by-step explanation:

Answer:

12 Loaves of bread :D

Step-by-step explanation:

hope that helped , have a great day !!

Answer:

9 hours worked after school and 4 hours worked on saturday

Step-by-step explanation:

x = amount of hours worked after school

y = amount of hours worked on saturday

x + y = 13 or x = 13 - y

and

3x + 4y = 43

plug:

3 · (13 - y) + 4y = 43

39 - 3y + 4y = 43

y = 4

plug:

x = 13 - y

x = 13 - 4 = 9

9 hours worked after school and 4 hours worked on saturday

Answer:

18, 19, 20

Step-by-step explanation:

(n) + (n + 1) + (n + 2) = 57

3n + 3 = 57

3n = 57 - 3 = 54

n = 54/3 = 18

three numbers are 18, 18+1, 18+2

18, 19, 20

Answer:

18,19,20

Step-by-step explanation:

x + (x+1) + (x+2) = 57

x + x + 1 + x + 2 = 57

Combine like terms

3x + 3 = 57

subtract 3 from both sides

3x = 54

divide both sides by 3

x = 18

The answer is 18, 19, 20

Since we know the value of x, we can look at it like this:

x + (x+1) + (x+2) --- > 18 + (18+ 1) + (18+2) --> 18 + 19 + 20

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

[tex]\fbox \colorbox{black}{ \colorbox{white}{C} \:  \:  \:   \:  \:  \:  \: \: \colorbox{white}{  = }  \:  \:  \:  \:  \:   + \colorbox{white}{6/5}}[/tex]

[tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]

Let's find the value of c ~

Answer:

c= -6

Step-by-step explanation:

2c -6 = 3c

we move all to the left

2c -6 (3c)= 0

add everything and the variables

-1c - 6 = 0

move all terms containing c to the left everything else to the right

-1c = 6

c=6/-1

c=-6

the answer is -2w (i have the write at least 20 words to dont worry about this)

Answer:

-2w

Step-by-step explanation:

Like terms have same variable.

34w - 4g + 4g - 36w = 34w - 36w - 4g + 4g

                                  = - 2w + 0 = -2w

Answer:

y=1/2x

Step-by-step explanation:

Answer:

10

9

8

7

6

5

4

3

2

1

0 1 2 3 4 5 6 7 8 9 10

Step-by-step explanation:

So the arrow is pointing at 10 and  5

Answer 10 5

Not 5

its 10 5

So the answer is 10 5

Answer: The second bubble thing is correct you have picked the wrong one.