3 years ago

1

Mathematics

1 answer:

fredd [130]3 years ago

6 0

Hi there I know the answer but it was confusing so it has to be D or B

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Where do two perpendicular lines intersect

andrew-mc [135]

They intersect at a right angle

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3 years ago

If p is true and q is false, then p ∨ q is true. <br> A. True<br> B. False

Marysya12 [62]

Answer:

no its false because p v q can only be true if both are true

Step-by-step explanation:

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3 years ago

(sec(π/3) + tan(-7π/4))/(1-cos(2 π/3))

Paha777 [63]

Answer:

Step-by-step explanation:

3+3_3=3

after that 3+2+6+7+7+8+8+9+9+9+9+9+0+0+8+6+5+4+ 5=?

u will do the answer and it will be done

6 0

3 years ago

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Solve the initial value problem: y'(x)=(4y(x)+25)^(1/2) ,y(1)=6. you can't really tell, but the '1/2' is the exponent

goblinko [34]

Answer:

$y(x)=x^2+5x$

Step-by-step explanation:

Given: $y'=\sqrt{4y+25}$

Initial value: y(1)=6

Let $y'=\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\sqrt{4y+25}$

Variable separable

$\dfrac{dy}{\sqrt{4y+25}}=dx$

Integrate both sides

$\int \dfrac{dy}{\sqrt{4y+25}}=\int dx$

$\sqrt{4y+25}=2x+C$

Initial condition, y(1)=6

$\sqrt{4\cdot 6+25}=2\cdot 1+C$

$C=5$

Put C into equation

Solution:

$\sqrt{4y+25}=2x+5$

$4y+25=(2x+5)^2$

$y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$

$y(x)=x^2+5x$

Hence, The solution is $y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$ or $y(x)=x^2+5x$

4 0

3 years ago

Write y - 2x = 1/2 in slope intercept form

Kruka [31]

Y = 2x + 0.5

or
y = 2x + 1/2

6 0

3 years ago

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