Hey , can someone please help me with this problem.

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

Answer:

24 I hope help you yieeeeeee

Answer:

ummmmmmmmmmm itsssssssss

Step-by-step explanation:

8 x 3 = 24 for should be 24

Only because 8 times larger than three seems like a multiplication problem..

I hope this helps if it is correct!

Answer:

(8+1) x 3 = 27

Step-by-step explanation:

this is correct rather than 8 x 3 = 24 because both sides of 8x3=24 are equal rather than the sum being greater than the statement.

Answer:

15 hours

Step-by-step explanation:

20%=3 hours

times 5 both sides

100%=15 hours

Answer:

28roses 70carnations

Step-by-step explanation:

4+6+8+10=28

10+15+20+25=70

Maybe i think

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:

y = x - 5

Step-by-step explanation:

y = x - 5

10 = 5 - 5

10 is not equal 0

Answer:

y = x - 5

Step-by-step explanation:

y = x - 5

10 = 5 - 5

10 is not equal 0

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:

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

Answer:

Answers in Explanation

Step-by-step explanation:

First Question:

[tex]\sqrt{100}[/tex] + [18 ÷ 3 x 4 - 15] - (60 - 7^2 - 1)

[tex]\sqrt{100}[/tex] + [24 - 15]  - (60 - 49 - 1)

[tex]\sqrt{100}[/tex] + 9 - 10

10 + 9 - 10

Answer = 9

Second Question:

5x + 2x = 7x

5x^2 + 3x^2 = 8x^2

2x + 3x - x = 4x

2x + 3y + x + y = 3x + 4y

9x - 6x = 3x

-7y + 3x + 4x + 3y = 7x - 4y

-7x^2 + 2x^2 + 9x^2 = 8x^2

(3x^2 + 5x + 4) - (-1 + x^2) = 2x^2 + 5x + 5

(3 + 2x - x^2) + (x^2 + 8x + 5) = 10x + 8

(3x - 4) - (5x + 2) = -2x - 6

(2x^2 + 5x + 3) - (x^2 - 2x + 3) = x^2 + 7x

(3x^2 + 2x - 5) - (2x^2 - x - 4) = x^2 + 3x - 1

Third Question:

17x + 2y

(5x + 12y) + (3x + y) = 8x + 13y

17x - 8x = 9x

2y - 13y = -11y

Answer: 9x - 11y

Hello there! (:

The answer is 9.

3^4=3*3*3*3 (81)

3^2=9

81:9=9

So the answer is 9.

Hope it helps! If you have any question or query, feel free to ask! (:

~An excited gal

[tex]SparklingFlower[/tex]

Answer:

[tex]18 {m}^{2} [/tex]

Step-by-step explanation:

[tex]area \: = 6m \times 4m \\ = 24 {m}^{2} \\ \\ grass \: area = 3m \times 2m \\ = 6 {m}^{2} \\ \\ cement \: area \: = 24 {m}^{2} - 6 {m}^{2} \\ = 18 {m}^{2} [/tex]

Answer:

18 m^2

Step by step explanation:

In these types of math problems, we have two ways to solve.

1) Directly find the area of the shaded area.

2)Find the unshaded area and then minus that from the total area.

In this case, I will use the second way.

The grass area (unshaded) is 3 x 2 = 6 m^2 ( 6 square meters )

The total area (grass + cement) is 4 x 6 = 24 m^2 ( 24 square meters )

Now, we want the area of the cement part but the grass's area is also in the total.

So, we minus 6m^2 from 24m^2.

Then we get 18m^2.

And that is the answer.

I hope it helps.

(Note : because this problem is easy, you can use both ways but most use the second way. There may also be problems where we can use only the first or second way.)

Answer:

|-4| < |-5|

Step-by-step explanation:

because if modules is given sub sign will be deducate

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:

35 5/27

Step-by-step explanation:

You just have to set up an equation:

0.9/95 = 1/x

And if you cross-multiply, you get:

0.9x=95

Which means x is 95/0.9 which is 105 5/9. So, if all three t-shirts costed 105 5/9, and they all costed the same, then one t-shirt costs 35 5/27.

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

Answer:

No inverse function:  (a), (b), (c)

Inverse function exists:  (d)

Step-by-step explanation:

The graph of h(x) = (x - 2)^2 + 3 is a parabola that opens upward and has vertex at (2, 3).  If the entire graph is drawn, and the horizontal line test then applied, h(x) would not have an inverse, because the horizontal line would intersect the  parabolic graph twice.

Note that if we restricted the domain to x ≥ 2, the resulting graph would pass the horizontal line test.  This would also be true for x ≥ 3, x ≥ 4, and so on.  Not so for (a) x < 1.  False for x > -5.  True for x < 3.  True for x > 4.

No inverse function:  (a), (b), (c)

Inverse function exists:  (d)

[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

[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:

120m2

$6.25

$4.50

85

57

12

12

32

54

69

87

89

12

34

58

Answer:

-4 9/10, -4 3/4, -4.2

Step-by-step explanation:

A. -4.9

B. -4.75

C. -4.2

ANSWER:

-4 9

10

-4 3

10

-4.2

Step-by-step explanation:

HOPE IT HELPSSSS

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:

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 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:

6p=13+5

6p=18

p=18/6

p=3

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