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3 years ago

The first one asks for the value of x. I need help on both. I am struggling to understand how to do this, please help

Mathematics

1 answer:

zzz [600]3 years ago

6 0

(1) since the horizontal lines are parallel, and you know the vertical intersecting line crosses at right angle, it must be that the two angles with variable x sum up to 90 degrees. So same rewritten as an equation:

x+8+6x-58 = 90

7x = 140

x = 20 (degrees)

(2) here you have two variable so you will need to find two equations.

The first one comes from the triangle and its angles summing up to 180:

63 + (3x+45) + (y-6) = 180

The second one comes from the fact that the two horizontal lines are parallel and so the angles (7x-7) and (3x+45) must be the same:

7x-7=3x+45

Let's solve the last one first:

4x = 52

x = 13

then plug this solution into the first equation and solve for y

63 + (3x+45) + (y-6) = 180

3x + y = 78

y = 78-39 = 39

so the value of y is 39 (degrees)

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sasho [114]

Answer is 54^2.

since there are triangles in this one, it is easier to do this. all you have to do is make boxes in and since there are angles, outside the lines. like this picture.

the orange is 1
red and dotted is 2
blue is 3
purple is 4(a regular box which should be easy to count.

all you have to do is add up all the boxes within the boundaries.
boundary 1: 12
boundary 2: 18
boundary 3: 6
boundary 4: 36

now when the boundary has a triangle in it (1, 2, & 3) divide the number you got in half or 2.
boundary 1: 6
boundary 2: 9
boundary 3: 3

the ractangle box doesn't not get divided.
boundary 1: 6
boundary 2: 9
boundary 3: 3
boundary 4: 36

add all the numbers you got now for each boundary and that would be your area squared.

6+9+3+36=54^2

so your final answer is 54^2.

i hope this helps you.

3 0

3 years ago

Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

4 0

3 years ago

The _____ justifies the conclusion that ∆ABD ≅ ∆AEC.

elena55 [62]

It is the AAS Postulate.

Explanation: You’re given two right angles (because perpendicular sides make right angles).

This also means that the angle next to them are also right angles, because they are linear pairs (180-90=90)

The right angles are:
This is your first A.

Both triangles also share a similar angle:
This is your second A.

You’re given side BD is congruent to EC.

This is your S.

It’s AAS and not ASA because the order refers to how they are connected: they are congruent in the order that the first set of congruent angles connect to the next, then to the congruent side.

3 0

3 years ago

Identify the coefficient of 17xy3z12

expeople1 [14]

Answer: Coeffcient is 17

Explaination:-

Given expression is :

17xy³z¹²

To find the coefficient of given expression

In mathematics, a coefficient is a multiplicative factor in some term of a polynomial, a series, or any expression; it is usually a number, but may be any expression.

For example : In 2x⁵

Coeffcient is 2

Now come to the question:

Coeffcient is 17.

8 0

3 years ago

The length of the side of square A is 50% of the length of the side of square b

galben [10]

Answer:

25%

Step-by-step explanation:

let length of 1st square be x and second be y

then

A/q

x = y/2

area of first square = x^2 = (y/2)^2 = (y^2)/4

and

area of second square = y^ 2

so from sbove two lines

area of first square = 1/4 * area of second square

= 1/4 *100%

so..

area of first sqare = 25% of area of second square

8 0

3 years ago