Screenshot below solving equations algebra please help

Evaluate the limit as x approaches 1 of the quotient of the square root of the quantity x squared plus 3 minus 2 and the quantit

IgorLugansk [536]

A graph shows the limit to be 1/2.
https://www.desmos.com/calculator/qrf6ay47tw

Since the value of the function is the indeterminate form 0/0, L'Hôpital's rule applies. The ratio of derivatives of numerator and denominator is
.. x/ $\sqrt{x^{2}+3 }$
Evaluated at x=1, this is
.. 1/ $\sqrt{1^{2}+3 }$ = 1/2

5 0

3 years ago

Can u help me with my workk

eimsori [14]

You knew where to place the target based on the coordinates (-5,4). Starting from the origin (0,0), we know how many units to move to horizontally and vertically. We move the target 5 units to the left, because it is negative. We move the target 4 units up, because it is positive. (x,y)

6 0

3 years ago

What is the amplitude of the sinusoidal function?

saul85 [17]

Answer:

amplitude = 4

Step-by-step explanation:

the amplitude is the maximum subtract the minimum divided by 2

maximum = 1 and minimum = - 7

amplitude = $\frac{1-(-7)}{2}$ = $\frac{8}{2}$ = 4

8 0

3 years ago

Read 2 more answers

If the radius of mars is about 13.7% of neptune's radius, what is the radius of neptune?

Marianna [84]

The volumetric mean radius of Mars is about 3389.5 km. We know that it's 13.7% of Neptune's mean radius, so we can write it down as:

3389.5 = 13.7% of x (where x is Neptune's radius)
3389.5 = 13.7% x
3389.5 = 137/1000x / * 1000 (both sides)
3389500 = 137x / ÷ 137 (both sides)
x = 24740.87591240876
x ≈ 24740.88

Answer: The mean radius of Neptune is about 24740.88 km.

6 0

3 years ago

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

3 years ago