Round this number 43.0129 hundredths place

Which list of numbers is arranged from least to greatest? f 0.25,17%,3 g 17%,29,0.25 h 0.25,29,17% j 17%,0.25,29?

Fed [463]

The list that is ordered least to greatest is ...
j 17%,0.25,29

4 0

3 years ago

Please assist! Limited time

Korvikt [17]

The answer would be c.

$\sqrt{28}$

7 0

3 years ago

I really need help with this! Thank you!!

photoshop1234 [79]

<h3>Answer:</h3>

x = 13; NP = 2 18/23 ≈ 2.8; NL = 5 5/23 ≈ 5.2
D. x = 17, y = 5

<h3>Step-by-step explanation:</h3>

1. In order for this problem to be workable, we need to assume PQ ║ ML. Then ΔPNQ ~ ΔLNM and ∠L = ∠P = 60°.

... ∠L = 60°

... (3x+21)° = 60°

... 3x = 39 . . . . . . . divide by °, subtract 21

... x = 13 . . . . . . . . .divide by 3

The side lengths of similar triangles are proportional, so we have ...

... (3.2 cm)/y = (6 cm)/(8-y)

Multiplying by the product of the denominators, and dividing by (cm), we have ...

... 3.2(8 -y) = 6y

... 3.2×8 = 9.2y . . . . . add 3.2y

... y = 3.2×8/9.2 = 64/23 = 2 18/23 ≈ 2.78 = NP

Then ...

... 8-y = NL ≈ 5.22

In summary: x = 13; NP ≈ 2.8; NL ≈ 5.2

_____

2. If we assume figures ABCD and PSRQ are similar, we can find the values of the variables. We assume ∠C ≅ ∠R, so ...

... 4x +27° = 95°

... 4x = 68° . . . . . subtract 27°

... x = 17° . . . . . . . divide by 4

___

AB/AD = PS/PQ . . . . . corresponding sides of similar figures are proportional

... 4y/(3y-5) = 10/5 . . . . . units of ft cancel

... 5×4y = 10(3y -5) . . . . . multiply by 5(3y-5)

... 20y = 30y -50 . . . . . . simplify

... 50 = 10y . . . . . . . . . . . add 50-20y

... 5 = y . . . . . . . . . . . . . . divide by 10

Using this value of y, we have ...

AB = 4y ft = 20 ft; AD = (3y-5) ft = 10 ft. Both these values are double the corresponding lengths on PSRQ.

In summary, x = 17°, y = 5. . . . . . (note that the ° symbol is appropriate for x)

_____

You should have your teacher show you how to work these problems <em>using only the given information</em>.

In the second problem, you cannot start with the assumption that the figures are similar, as that is what you're being asked to prove. Please note, too, that two sides and one angle of a quadrilateral are insufficient to show similarity.

7 0

3 years ago

Correct answer gets brainliest.

wolverine [178]

The answer is C based on SAS

8 0

3 years ago

Read 2 more answers

without building the graph, find the coordinates of the point of intersection of the lines given by the equation y=3x-1 and 3x+y

DaniilM [7]

<h2><u>1. Determining the value of x and y:</u></h2>

Given equation(s):

$y = 3x - 1$
$3x + y = -7$

To determine the point of intersection given by the two equations, it is required to know the x-value and the y-value of both equations. We can solve for the x and y variables through two methods.

<h3 /><h3><u>Method-1: Substitution method</u></h3>

Given value of the y-variable: 3x - 1

Substitute the given value of the y-variable into the second equation to determine the value of the x-variable.

$\implies 3x + y = -7$

$\implies3x + (3x - 1) = -7$

$\implies3x + 3x - 1 = -7$

Combine like terms as needed;

$\implies 3x + 3x - 1 = -7$

$\implies 6x - 1 = -7$

Add 1 to both sides of the equation;

$\implies 6x - 1 + 1 = -7 + 1$

$\implies 6x = -6$

Divide 6 to both sides of the equation;

$\implies \dfrac{6x}{6} = \dfrac{-6}{6}$

$\implies x = -1$

Now, substitute the value of the x-variable into the expression that is equivalent to the y-variable.

$\implies y = 3(-1) - 1$

$\implies$ $\ \ = -3 - 1$

$\implies$ $= -4$

Therefore, the value(s) of the x-variable and the y-variable are;

$\boxed{x = -1}$ $\boxed{y = -4}$

<h3 /><h3><u>Method 2: System of equations</u></h3>

Convert the equations into slope intercept form;

$\implies\left \{ {{y = 3x - 1} \atop {3x + y = -7}} \right.$

$\implies \left \{ {{y = 3x - 1} \atop {y = -3x - 7}} \right.$

Clearly, we can see that "y" is isolated in both equations. Therefore, we can subtract the second equation from the first equation.

$\implies \left \{ {{y = 3x - 1 } \atop {- (y = -3x - 7)}} \right.$

$\implies \left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.$

Now, we can cancel the "y-variable" as y - y is 0 and combine the equations into one equation by adding 3x to 3x and 7 to -1.

$\implies\left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.$

$\implies 0 = (6x) + (6)$

$\implies0 = 6x + 6$

This problem is now an algebraic problem. Isolate "x" to determine its value.

$\implies 0 - 6 = 6x + 6 - 6$

$\implies -6 = 6x$

$\implies -1 = x$

Like done in method 1, substitute the value of x into the first equation to determine the value of y.

$\implies y = 3(-1) - 1$

$\implies y = -3 - 1$

$\implies y = -4$

Therefore, the value(s) of the x-variable and the y-variable are;

$\boxed{x = -1}$ $\boxed{y = -4}$

<h2><u>2. Determining the intersection point;</u></h2>

The point on a coordinate plane is expressed as (x, y). Simply substitute the values of x and y to determine the intersection point given by the equations.

⇒ (x, y) ⇒ (-1, -4)

Therefore, the point of intersection is (-1, -4).

<h3>Graph:</h3>

5 0

2 years ago