please answer quick
10 points

Answer:

27s

Step-by-step explanation:

d = distance to rendezvous point in miles

t = time of arrival at 10mph, i.e. 3 mins later than appointed time

10 mph = 10 miles per hour = ¹/₆ miles per min

12 mph = ¹/₅ miles per min

t = d/(¹/₆)

t = 6d

5 min = ¹/₁₂ hr

t - ¹/₁₂ = d/(¹/₅)

t = 5d + ¹/₁₂

Elimination of t:

6d = 5d + ¹/₁₂

d = ¹/₁₂ mi

t = 6(¹/₁₂)

t = ¹/₂ min

Time left to appointed time = ¹/₂ - ³/₆₀

= ¹⁰/₂₀ - ¹/₂₀

= ⁹/₂₀.min → 27 seconds

Answer:

27 min

Step-by-step explanation:

rsm

Answer:

rotation

Step-by-step explanation:

Answer:

Reflection

Step-by-step explanation:

I think its reflection because if you reflect them across the y axis then the x axis I believe they will be in those positions.

This should be right , if any doubt please comment

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](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:

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

Step-by-step explanation:

je........porte

tu.......portes

nous.......portons

vous........portez

ils......portent

Answer:

120m2

$6.25

$4.50

85

57

12

12

32

54

69

87

89

12

34

58

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:

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:

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:

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

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 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:

no

yes

yes

no

no

yes

Step-by-step explanation:

a prime number cant be devided by any thing other than itself or one

Answer:

Step-by-step explanation:

Is 2 is 3 is 11 is 18 is 20 is 27

Answer:

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

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

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:

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)

Answer:

|-4| < |-5|

Step-by-step explanation:

because if modules is given sub sign will be deducate

Answer:

24 I hope help you yieeeeeee

Answer:

ummmmmmmmmmm itsssssssss

Step-by-step explanation:

Answer:

6p=13+5

6p=18

p=18/6

p=3

Answer:

9 = [tex]\frac{n}{54}[/tex] or n ÷ 54

Step-by-step explanation:

nine equals the quotient of a number and 54

       /\

9 = the quotient of a number and 54

             /\

9 =       will be division of a number and 54

                                /\

                    9 = [tex]\frac{n}{54}[/tex] or n ÷ 54

Have a nice day!

    I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

Answer:

$14.40 is the discount price.

Step-by-step explanation:

0.2 x 18 = 3.6

18 - 3.6 = 14.4