All categories

3 years ago

BRAINLIEST ANSWER UP FOR GRABS!!

Mathematics

2 answers:

Svetach [21]3 years ago

8 0

X = sqrt(5^2 - 3^2) = 4
y = sqrt(7^2 - 3^2) = sqrt 40

Horizontal distance = x + y + 3 = 7 + sqrt40

KIM [24]3 years ago

8 0

The answer would be 13.3.

Use the Pythagorean theorem to find the other distances in the 5, 3, x triangle and the 7, 3, y triangle and add those two answers together plus 3 to get your answer.

You might be interested in

A clothing store paid an import tax of $32 on a shipment of $800 worth of silk scarves. The import tax rate is always the same.

bekas [8.4K]

This will be your formula:
$\frac{32}{800} = \frac{120}{x}$
cross multiply (32 times x) (120 times 800)
this is what you'll get:

4 0

3 years ago

What is the value of x in this simplified expression

arsen [322]

Answer:

Step-by-step explanation:

(-j)^-7 = 1/(-j)^x

Changing LHS as

1/(-j)^7 = 1/(-j)^x

Comparing LHS and RHS

x = 7

8 0

3 years ago

Look at the figure. Which step should be taken next to construct a line through point P perpendicular to BA?

alexira [117]

ANSWER:

C. Place the compass on point A. Open the compass to a point between point P and point B.

EXPLANATION:

A perpendicular is a line that would be at a right angle to line BA.

The next step is to chose a radius that is greater than PB or PA so as to construct the bisector. And this can be done by placing the compass on point A, and open the compass to a point between point P and point B.

Use this radius to draw an arc above and below the line, and repeat the same using B as the center with the same radius. This would form two intersecting arcs above and below line BA. Join the point of intersection of the arcs by a straight line through P. This is the bisector of line BA through point P.

8 0

3 years ago

Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20

Mice21 [21]

Power and chain rule (where the power rule kicks in because $\sqrt x=x^{1/2}$ ):

$\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'$

Simplify the leading term as

$\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}$

Quotient rule:

$\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}$

Chain rule:

$(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)$

$(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)$

Put everything together and simplify:

$\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}$

$=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}$

$=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}$

5 0

3 years ago

What would the radius of a circle be with a diameter of 24?

Montano1993 [528]

Answer:

r = 12

Step-by-step explanation:

using the formula

d = 2 r

solving for r

r = d/2= 24/2=12

8 0

3 years ago

Read 2 more answers