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2 years ago

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One of the legs of a right triangle measures 1 cm and its hypotenuse measures 8 cm. Find the measure of the other leg. If necess

ary, round to the nearest tenth.

1 answer:

Artemon [7]2 years ago

4 0

Answer:

about 7.8

Step-by-step explanation:

a^2+b^2=c^2

1^2+b^2=8^2

2+b^2=64

B^2=62

B= about 7.8

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If y= 3x + 6, what is the minimum value of (x^3)(y)?

Rainbow [258]

Answer:

Given the statement: if y =3x+6.

Find the minimum value of $(x^3)(y)$

Let f(x) = $(x^3)(y)$

Substitute the value of y ;

$f(x)=(x^3)(3x+6)$

Distribute the terms;

$f(x)= 3x^4 + 6x^3$

The derivative value of f(x) with respect to x.

$\frac{df}{dx} =\frac{d}{dx}(3x^4+6x^3)$

Using $\frac{d}{dx}(x^n) = nx^{n-1}$

we have;

$\frac{df}{dx} =(12x^3+18x^2)$

Set $\frac{df}{dx} = 0$

then;

$(12x^3+18x^2) =0$

$6x^2(2x + 3) = 0$

By zero product property;

$6x^2=0$ and 2x + 3 = 0

⇒ x=0 and x = $-\frac{3}{2} = -1.5$

then;

at x = 0

f(0) = 0

and

x = -1.5

$f(-1.5) = 3(-1.5)^4 + 6(-1.5)^3 = 15.1875-20.25 = -5.0625$

Hence the minimum value of $(x^3)(y)$ is, -5.0625

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3 years ago

Prove that lines 3x-4y=12 and 3y=12-4x are perpendicular.

Svetllana [295]

Answer:

see explanation

Step-by-step explanation:

If 2 lines are perpendicular then the product of their slopes equals - 1

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Consider the given equations

3x - 4y = 12 ( subtract 3x from both sides )

- 4y = - 3x + 12 ( divide terms by - 4 )

y = $\frac{3}{4}$ x - 3 ← in slope- intercept form

with slope m = $\frac{3}{4}$

3y = 12 - 4x = - 4x + 12 ( divide terms by 3 )

y = - $\frac{4}{3}$ x + 4 ← in slope- intercept form

with slope m = - $\frac{4}{3}$

Then

$\frac{3}{4}$ × - $\frac{4}{3}$ = - 1

Since the product of their slopes = - 1 then the lines are perpendicular

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2 years ago

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pickupchik [31]

100x3=300
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