Can someone please give me the (Answers) to this? ... please ...



I dentify the idlowing pairs. of angles.



1.<1 and <4

2.<1 and <5

3. <2 and <8

4.<4 and <5

5. <3 and <5

6.<3 and <6

7.<2 and <6

8.<1 and <8 /
/
<————-1——2—————>
3 / 4
5 / 6
<————————————>
7 / 8
/

Answer:

Inequality: 30+12x < 100

x ≈ 5 months

Step-by-step explanation:

To solve, first create the inequality. In this problem, 30 dollars is the flat fee for joining the gym and thus a constant. Additionally, 12 is the monthly payment and thus the coefficient for a variable because it changes with the number of months. Finally, 100 is the most that can be spent (the max), so it should be on the other side of the inequality and the sign should be less than or equal to 100.

This makes the inequality: 30+12x < 100

To find the number of months solve the inequality for x.

So, the inequality equals x<5.8. However, the question asks for the number of full months. And, no more than 100 dollars can be spent. So, while you would normally round up. In this case, you must round down to 5 months.

Answer:

There would be 12

Step-by-step explanation:

Answer:

your so funny for a annoying kid

Step-by-step explanation:

A convenience store purchased a magazine and marked it up 100% from the original cost of $2.30. A week later, the store placed the magazine on sale for 50% off. What was the discount price?

Answer:

24 I hope help you yieeeeeee

Answer:

ummmmmmmmmmm itsssssssss

Step-by-step explanation:

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:

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:

[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:

120m2

$6.25

$4.50

85

57

12

12

32

54

69

87

89

12

34

58

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:

the answer is -16(t-2)(t+1)

Step-by-step explanation:

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

Step-by-step explanation:

f(x) = 2x² + 5√(x - 2)

Domain

√(x - 2) ≥ 0

x - 2 ≥ 0

x ≥ 2

x is greater than or equal to 2

Answer: Greater than

Step-by-step explanation:

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:

$14.40 is the discount price.

Step-by-step explanation:

0.2 x 18 = 3.6

18 - 3.6 = 14.4

Answer:

3000

Step-by-step explanation:

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:

Yes, This is a repeating decimal

No, This is a decimal that neither terminates or repeats

No, This is a decimal that neither terminates or repeats

Yes, This is a terminating decimal

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

This should be right , if any doubt please comment

[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:

|-4| < |-5|

Step-by-step explanation:

because if modules is given sub sign will be deducate

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:

x=6

y=2

Step-by-step explanation:

-x+3y=0 (1)

x+3y=12 (2)

(1)+(2)<=> 6y=12