the sum of three whole numbers in a row is 57. what are the three numbers?

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:

it is correct but -65 is not correct

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:

$14.40 is the discount price.

Step-by-step explanation:

0.2 x 18 = 3.6

18 - 3.6 = 14.4

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:

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:

24 I hope help you yieeeeeee

Answer:

ummmmmmmmmmm itsssssssss

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

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]

This should be right , if any doubt please comment

Answer:

There would be 12

Step-by-step explanation:

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]

9514 1404 393

Answer:

  7

Step-by-step explanation:

Of the 9 2-digit numbers ending in 6, only 2 are perfect squares: 16 and 36. The other 7 are not perfect squares.

[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:

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:

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

Answer:

3000

Step-by-step explanation:

JK=JH

[tex]\\ \sf\longmapsto 10t=7t+15[/tex]

[tex]\\ \sf\longmapsto 10t-7t=15[/tex]

[tex]\\ \sf\longmapsto 3t=15[/tex]

[tex]\\ \sf\longmapsto t=15/3=5[/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.)

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:

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:

i thig it is hj

Step-by-step explanation:

jk

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:

See answer below

Step-by-step explanation:

ABCD is a parallelogram                     Given

BC ≅ AD; AB ≅ DC                             opposite sides of a parallelogram ≅

Δ AXB = 1/2 (AB)(AD)                           Formula for the area of a Δ

Δ BYC = 1/2 (BC) (DC)                           Formula for the area of a Δ

1/2 (AB)(AD)= 1/2(BC)(DC)                      Substitution

∴ Area of Δ AXB = Area of ΔBYC

Answer:

I guess...

Step-by-step explanation:

Since QR and PN are parallels, I think what happens in MN and MP is proportional, so:

RN = x

8 is to 10 as

5 is to x

8 * x = 5 * 10

8x = 50

x = 50 / 8

x = 6.25

RN = 6.25

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: