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3 years ago

Triangles 14. Find the value of x.

Mathematics

1 answer:

Yanka [14]3 years ago

3 0

Answer:

X=6 .

Step-by-step explanation:

10=4X-14 .

24=4X .

Good luck ^_^

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42 inches to 54 inches

abruzzese [7]

Whats the question? :)

8 0

3 years ago

This is the graph of the function f(x) = (x-5)sin(x) +2, Using the graph, or the equation, determine the value of f(x) when x =

nirvana33 [79]

Answer:

Step-by-step explanation:

Graph: look halfway between x= 4 and x=6, the point is on the line f(x)=2

Equation: (x-5) = (5-5) =0

anything times 0 is 0

0 +2=2

8 0

3 years ago

Line K has a slope of 3. Line J is perpendicular to like K and passes through the point (3,8). Type the equation of the line in

hichkok12 [17]

I'm assuming you are finding the equation of the line of Line J

Slope-intercept form: y = mx + b

[m is the slope, b is the y-intercept or the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis]

For lines to be perpendicular, the slopes have to be negative reciprocals of each other. (basically changing the sign (+/-) and flipping the fraction/switching the numerator and the denominator)

For example:

slope = 2 or $\frac{2}{1}$

Perpendicular line's slope = $-\frac{1}{2}$ [changed sign from + to -, and flipped the fraction]

slope = $-\frac{2}{5}$

Perpendicular line's slope = $\frac{5}{2}$ [changed sign from - to +, and flipped the fraction]

Since you know the slope is 3, the perpendicular line's slope is $-\frac{1}{3}$ . Plug this into the equation

y = mx + b

$y=-\frac{1}{3} x+b$ To find b, plug in the point (3, 8)

$8=-\frac{1}{3}(3)+b$

8 = -1 + b Add 1 on both sides to get "b" by itself

8 + 1 = -1 + 1 + b

9 = b

$y=-\frac{1}{3} x+9$

8 0

3 years ago

Solve the linear differential equation 2xy' + y = 2√x

77julia77 [94]

Answer:

Step-by-step explanation:

General form of the linear differential equation can be written as:

$\frac{dy}{dx}+P(x)y=Q(x)$

For this case, we can rewrite the equation as:

$\frac{dy}{dx}+\frac{1}{2x}y=\frac{\sqrt{x}}{x}$

Here $P(x) =\frac{1}{2x}; Q(x)=\frac{\sqrt{x}}{x}$

To find the solution (y(x)), we can use the integration factor method:

$Fy(x)=\int Q(x)Fdx+C \rightarrow F=e^{\int P(x)dx$

Then $F=e^{\int \frac{1}{2x}dx}=e^{\frac{1}{2}\ln|x|\right}=\sqrt{|x|}$

So, we can find:

$y\sqrt{|x|}=\int \frac{\sqrt{x}\sqrt{|x|}}{x}dx+C$

Suppose that $x\in \double R$ , then $\sqrt{|x|}=\sqrt{x}$ , and we find:

$y\sqrt{x}=x+C \rightarrow y(x)=\sqrt{x}+\frac{C\sqrt{x}}{x}$

To check our solution is right or not, put your y(x) back to the ODE:

$y' = \frac{1}{2\sqrt{x}}-\frac{C}{2\sqrt{x^{3}}}$

$2xy'=\frac{x-C}{\sqrt{x}}$

$2xy'+y=\frac{x-C}{\sqrt{x}}+\sqrt{x}+\frac{C\sqrt{x}}{x}=2\sqrt{x}$

(it means your solution is right)

3 0

2 years ago

Leroy's total job benefits were $50,150 last year. His total job expenses for traveling and for professional development for the

Diano4ka-milaya [45]

Answer:

D. 53,650

50,150+3,500=53,650

7 0

3 years ago