What is the slope of the equation Y = 5/4x-7/4

A. -7

B. -7/4

C. 5/4

D. 5

Answer:

i thig it is hj

Step-by-step explanation:

jk

This should be right , if any doubt please comment

Answer:

it is correct but -65 is not correct

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: 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:

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

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.

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:

a

Step-by-step explanation:

The transformation from Figure 1 to Figure 2 is:

The transformation of 5 units to the right and 3 units down.

Option C is the correct answer.

It is the movement of the shape in the left, right, up, and down directions.

The translated shape will have the same shape and shape.

There is a positive value when translated to the right and up.

There is a negative value when translated to the left and down.

We have,

A B C D E has points (-3, 5), (-2, 5), (-1, 4), (-2, 3), and (-5, 3).

A' B' C' D' E' has points (2, 2), (3, 2), (4, 1), (3, 0), and (0, 0).

Now,

A = (-3 + 5, 5 - 3) to A' = (2, 2)

B = (-2 + 5, 5 - 3) to B' = (3, 2)

C = (-1 + 5, 4 - 3) to C' = (4, 1)

D = (-2 + 5, 3 - 3) to A' = (3, 0)

E = (-5 + 5, 3 - 3) to E' = (0, 0)

We see that,

There is a translation of 5 units to the right and  3 units to the down.

Thus,

The transformation of 5 units to the right and 3 units down.

Learn more about translation here:

https://brainly.com/question/12463306

#SPJ1

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:

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:

Answer:

3000

Step-by-step explanation:

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°

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:

3 HOURS

Step-by-step explanation:

HE PRACTICED 3 HOURS A DAY IN BOTH SPORTS MEANING IN 1 DAY HE PRACTICED A TOTAL OF 6 HOURS 6 HOURS A DAY FOR 5 DAYS IS EQUAL TO 30 HOURS OF PRACTICE A WEEK FOR BOTH SPORTS. SO 6 HOURS A DAY TOTAL AND 3 HOURS A DAY IF YOU ONLY COUNT 1 OF THE 2 SPORTS MENTIONED.

[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:

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:

$14.40 is the discount price.

Step-by-step explanation:

0.2 x 18 = 3.6

18 - 3.6 = 14.4

Answer:

8%

Step-by-step explanation:

4/50 = 0.08 = 8%

Steps to solve "what percent is 4 of 50?" If you are using a calculator, simply enter 4÷50×100 which will give you 8 as the answer.

Answer:

use graphs

Step-by-step explanation:

Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.

Answer:

These are the answers

(-1,5)

(0,3)

(2,-1)

Step-by-step explanation:

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.

Answer:

that you can define a tangent relationship between a line and an arc. If the adjoining elements change, the tangent relationship is maintained between the elements. Geometric relationships control how a sketch changes when edits are made.

Step-by-step explanation:

i took this i guess and it was gud

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