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3 years ago

6

WHATS THE VALUE IF X HELP ASAP‼️‼️

1 answer:

chubhunter [2.5K]3 years ago

7 0

Answer:

31

Step-by-step explanation:

If it says 31 on the other side of the triangle triangle then it should be 31

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Complete the equation of variation where y varies inversely as x and y = 80 when x = 0.7.

andre [41]

Y varies inversely with x
means that y=k/x
with x=0.7 and y=80
we have 80=k/0.7
k=0.7x80=56
so the equation is
y=56/x

3 0

3 years ago

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Round to the nearest thousand, 5870

andrew11 [14]

Answer: 5870 rounded to the nearest thousand is 6000

Step-by-step explanation:

Remember, when we round we have to be careful with the place value they are asking you to round to. Here, they want you to round to the thousands place. So, you have to look at the hundreds.

5870. 800 is the hundreds, and it is greater than “5”. So, you round to the next thousands. It happens to be 6000.

<em>Hence, 5870 rounded is 6000.</em>

4 0

2 years ago

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)

ELEN [110]

Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

$d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

3 0

3 years ago

In two or more complete sentences, compare the number of x-intercepts in the graph of f(t) = 12 to the number of x-

alina1380 [7]

Answer:

Both functions have one x-intercept each.

Step-by-step explanation:

The first function is

$f(t )= {t}^{2}$

This is a parabola with vertices at the origin and has one x-intercept at t=0.

The transformed function is

$g(t) = {(t + 3)}^{2}$

The function g(t) is obtained by shifting the graph of f(t) to the left by 3 units.

This graph also has one x-intercept at x=-3.

Therefore both functions has the same number of x-intercepts

8 0

3 years ago

Please help if i get this wrong i will fail

LUCKY_DIMON [66]

The answer is either b or c

3 0

3 years ago

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