Help? I need question 1-8 answers .

Step-by-step explanation:

There are two attachments to this post. The first attachment shows the numbered angles formed through the intersection of the transversal, t, into the parallel lines, m and n, which will be used to determine how each angle relates to other angles.

The corresponding angles in the first attachment are:

∠1 and  ∠5

∠2 and ∠6

∠3 and ∠7

∠4 and ∠8

Corresponding angles have the same measure.

The same-side exterior angles are the pair of angles that are outside the parallel lines, and on the same side of the transversal.  In the first attachment, the same-side exterior angles are:

∠1  and ∠8

∠2 and ∠7

These angles are supplements of each other, meaning that the sum of their measure equal 180°.

Given parallel lines, m and n that are cut by a transversal, t.

The angle that has a measure of  ∠(10x + 5)° is a supplement of ∠(5x + 25)°.  In order to solve for the value of x, we must establish the following equation:

m∠(10x + 5)° + m∠(5x + 25)° = 180°

10x° + 5° + 5x° + 25° = 180°

Combine like terms:

15x° + 30° = 180°

Subtract 30 from both sides:

15x° + 30° - 30° = 180° - 30°

15x° =  150°

Divide both sides by 15:

15x°/15 = 150°/15

x = 10°

Verify whether we have correct value for x by substituting its value into the established equation:

m∠(10x + 5)° + m∠(5x + 25)° = 180°

10(10)° + 5° + 5(10)° + 25° = 180°

100° + 5° + 50° + 25° = 180°

180° = 180° (True statement).

Therefore, we have the correct value for x = 10°.

m∠(10x + 5)°  = 105°

m∠(5x + 25)° = 75°

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

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:

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:

[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]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:

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:

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:

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:

i thig it is hj

Step-by-step explanation:

jk

Answer:

k = 2 cm

Step-by-step explanation:

3 x 15 x k = 90 cm^3

45 x k = 90 cm^3

k = 90/45 = 2 cm

[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

Answer:

3000

Step-by-step explanation:

Answer: y=(1/5)m+3

Step-by-step explanation:

Start with the standard form of an equation of a straight line:  y=mx+b, whre m is the slope and b the y-intercept (the value of y when x=0).

y=mx+b

We know b (3), so:

y=mx + 3

To find the slope, m, we can use the one given point, (5,4):

y=mx + 3 for (5,4) would be:

4 = m*5+3

1 = 5m

m = (1/5)

y=(1/5)m+3

Answer:

Option A: (It's just that the y-intercept and slope are in different spots.)

Line in (y=mx+b) form is y = 2x + 4

Slooe is 2

y-intercept is 4

Step-by-step explanation:

Slope-intercept form is written as y=mx+b, where m is slope and b is y-intercept.

By looking at the graph, we can already tell that the y-intercept is positive 4, since the line crosses the y-axis at that point.

Now use rise/run to find slope. (Remember that rise is movement up or down, and run is movement left or right.) With this in mind, the slope is 4/2, and if we simplify it, it's 2/1 (or basically just 2). Because we go upwards 4 units and to the right 2 units.

Hope this helps :)

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:

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:

There would be 12

Step-by-step explanation:

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:

This should be right , if any doubt please comment

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:

$14.40 is the discount price.

Step-by-step explanation:

0.2 x 18 = 3.6

18 - 3.6 = 14.4

Answer:

24 I hope help you yieeeeeee

Answer:

ummmmmmmmmmm itsssssssss

Step-by-step explanation:

Answer:

it is correct but -65 is not correct