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jek_recluse [69]

3 years ago

11

A right triangle has a hypotenuse that measures 5 inches and one leg that measures 2 inches. What is the length of the missing l

eg?

1 answer:

jolli1 [7]3 years ago

8 0

Pythagorean theorem

c2 = a2 + b2
5^2 = 2^2 + b2
25 = 4 +b2
21 = b2
b = 4.6

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Solve this system of equations by graphing. First graph the equations, and then type the solution. y= x+3 y=2x

Yuliya22 [10]

Answer:

(3,6)

Step-by-step explanation:

Y=X+3

Y=X2

X+3=2X

X-2X= -3

-x=-3

x=3

then Y=2X

Y=2×3

Y=6

(3,6)

3 0

2 years ago

Select the correct answer.

vitfil [10]

Is there an image i don’t understand what expressions you’re asking for?

6 0

2 years ago

On his first three tests, Hiram got scores of 89, 95, and 90. His final test score brought his median score to 91.5. What was I

Elena-2011 [213]

The median is the middle score, it is also the average of the two middle scores when total number is even.

original 3 scores: 89, 90, 95 ----> median = 90

with final score: 89,90,X,95 ----> median = 91.5

(90+x)/2 = 91.5
90+x = 183
x = 93

Final score was a 93.

6 0

3 years ago

Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie

melamori03 [73]

Answer:

The complete solution is

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

Step-by-step explanation:

Given differential equation is

3y"- 8y' - 3y =4

The trial solution is

$y = e^{mx}$

Differentiating with respect to x

$y'= me^{mx}$

Again differentiating with respect to x

$y''= m ^2 e^{mx}$

Putting the value of y, y' and y'' in left side of the differential equation

$3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0$

$\Rightarrow 3m^2-8m-3=0$

The auxiliary equation is

$3m^2-8m-3=0$

$\Rightarrow 3m^2 -9m+m-3m=0$

$\Rightarrow 3m(m-3)+1(m-3)=0$

$\Rightarrow (3m+1)(m-3)=0$

$\Rightarrow m = 3, -\frac13$

The complementary function is

$y= Ae^{3x}+Be^{-\frac13 x}$

y''= D², y' = D

The given differential equation is

(3D²-8D-3D)y =4

⇒(3D+1)(D-3)y =4

Since the linear operation is

L(D) ≡ (3D+1)(D-3)

For particular integral

$y_p=\frac 1{(3D+1)(D-3)} .4$

$=4.\frac 1{(3D+1)(D-3)} .e^{0.x}$ [since $e^{0.x}=1$ ]

$=4\frac{1}{(3.0+1)(0-3)}$ [ replace D by 0 , since L(0)≠0]

$=-\frac43$

The complete solution is

y= C.F+P.I

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

4 0

2 years ago

Analyze the diagram below and complete the instructions that follow. (7x - 3) (12x -7) Solve for x.

yarga [219]

Answer:

Step-by-step explanation:

(7x-3)(12x-7)

we are solving for x and we are multiplying.

we start with the x's 7x*12x= 84x

-3*-7=21

so now we have

84x+21=0

the 0 is a placer for answering the question

we minus 21 to both sides

84x=-21

now we divide

84/-21= -4

x=-4

6 0

2 years ago