All categories

2 years ago

What is the length of sides of the square shown below? 45 2 90 A. 2sqrt(2) B. 4sqrt(2) C. 1 OD. sqrt(2) E. 4 F. 2

Mathematics

1 answer:

sergij07 [2.7K]2 years ago

6 0

Answer:

<h2>Uh oh! It looks like your question is missing some crucial information.</h2><h2 />

Step-by-step explanation:

There is no " square shown below"

You might be interested in

Solve: 3x - 7 - 2x + 5 = 6 A) -8 B) - 18 5 C) 4 D) 8

seropon [69]

Answer:

D) 8

Step-by-step explanation:

3x - 7 - 2x + 5 = 6

3x - 2x = 6 + 7 - 5

x = 8

Note: 6 + 7 -5 = 13 -5 = 8

7 0

3 years ago

Could you guys help me ? I'll give brainliest

ExtremeBDS [4]

When you raise a number to a negative exponent, change the exponent to a positive and make the complete number a fraction with 1 as the numerator.

5 ^-2 = 1/5^2

The answer is C.

4 0

2 years ago

Solve for x: tan(180-ø).tan(90+ø)+xcos(90+ø).(cos90-ø) =sinø.(sin180-ø)

Burka [1]

Is there a picture that I’m able to look at??

5 0

2 years ago

Read 2 more answers

Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x > 0)

vfiekz [6]

Answer:

Step-by-step explanation:

given a point $(x_0,y_0)$ the equation of a line with slope m that passes through the given point is

$y-y_0 = m(x-x_0)$ or equivalently

$y = mx+(y_0-mx_0)$ .

Recall that a line of the form $y=mx+b$ , the y intercept is b and the x intercept is $\frac{-b}{m}$ .

So, in our case, the y intercept is $(y_0-mx_0)$ and the x intercept is $\frac{mx_0-y_0}{m}$ .

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph $(x_0,\frac{1}{x_0})$ . Which means that $y_0=\frac{1}{x_0}$

The slope of the tangent line is given by the derivative of the function evaluated at $x_0$ . Using the properties of derivatives, we get

$y' = \frac{-1}{x^2}$ . So evaluated at $x_0$ we get $m = \frac{-1}{x_0^2}$

Replacing the values in our previous findings we get that the y intercept is

$(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}$

The x intercept is

$\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0$

The triangle in consideration has height $\frac{2}{x_0}$ and base $2x_0$ . So the area is

$\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2$

So regardless of the point we take on the graph, the area of the triangle is always 2.

6 0

3 years ago

Im being timed so please help-

Delvig [45]

Answer:

2 and 4 I am pretty sure. hopeful that's correct.

5 0

3 years ago