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2 years ago

What is the value of x? _________ *

Mathematics

1 answer:

Flura [38]2 years ago

7 0

Look at the picture.

The angle y° and 71° are Supplementary Angles. Therefore y° + 71° = 180°.

$y^o=180^o-71^o=109^o$

The sum of measures of these three angles of any triangle is invariably equal to 180°.

Therefore we have the equation:

$x^o+109^o+28^o=180^o\\\\x^o+137^o=180^o\ \ \ |-137^o\\\\x^o=43^o$

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The dimensions of four rectangles are given below. Which has an area different from the area of each of the other three?

storchak [24]

To determine which of the rectangles has a different area, the areas of the four rectangles must be calculated. The rectangle with a different area is rectangle b

The dimension of the 4 rectangles are:

a. length: 4x and width: 4

b. length: 11 and width: x

c. length: 2 and width: 8x

d. length: 16x and width: 1

The area of a rectangle is:

$Area = Length \times Width$

Rectangle (a)

$Area = 4x \times 4$

$Area = 16x$

Rectangle (b)

$Area = 11 \times x$

$Area = 11x$

Rectangle (c)

$Area = 2 \times 8x$

$Area = 16x$

Rectangle (d)

$Area = 16x \times 1$

$Area = 16x$

Rectangles (a), (c) and (d) have the same area (i.e. 16x) while rectangle (b) has 11x as its area.

Hence, the rectangle with a different area is rectangle (b).

Read more about areas of rectangles at:

brainly.com/question/14383947

7 0

2 years ago

In ΔUVW, the measure of ∠W=90°, the measure of ∠U=41°, and VW = 5.1 feet. Find the length of UV to the nearest tenth of a foot.

Reil [10]

Answer:

7.8

Step-by-step explanation:

becuase

7 0

2 years ago

Power Series Differential equation

KatRina [158]

The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for $y$

$\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0$

which indeed gives the recurrence you found,

$a_{n+3}=-\dfrac{n-3}{n+3}a_n$

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that $a_2=0$ , and substituting this into the recurrence, you find that $a_2=a_5=a_8=\cdots=a_{3k-1}=0$ for all $k\ge1$ .

Next, the linear term tells you that $6a_0+6a_3=0$ , or $a_3=a_0$ .

Now, if $a_0$ is the first term in the sequence, then by the recurrence you have

$a_3=a_0$
$a_6=-\dfrac{3-3}{3+3}a_3=0$
$a_9=-\dfrac{6-3}{6+3}a_6=0$

and so on, such that $a_{3k}=0$ for all $k\ge2$ .

Finally, the quadratic term gives $6a_1-12a_4=0$ , or $a_4=\dfrac12a_1$ . Then by the recurrence,

$a_4=\dfrac12a_1$
$a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1$
$a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1$
$a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1$

and so on, such that

$a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}$

for all $k\ge2$ .

Now, the solution was proposed to be

$y=\displaystyle\sum_{n\ge0}a_nx^n$

so the general solution would be

$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots$
$y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)$
$y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)$

4 0

2 years ago

X=-7y+5 5x=4y=-6 solve by substitution

lyudmila [28]

5x (-7)y=4
y=-4/35
5x=4
x=4/5
(x, y) = (4/5, 4/35)

4 0

2 years ago

which one of the following uses estimation to check 267-154=113? a. 300-200=100. b. 113+154=267. c. 300-150=150. d. 270-150=120

Vlada [557]

A. 300-200=100
You round 267 to 300, 154 to 200, and 113 to 100.

8 0

2 years ago

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