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8_murik_8 [283]

4 years ago

10

Which equation represents a circle with a center at (-3,5) and a radius of 6 units

1 answer:

Fittoniya [83]4 years ago

5 0

Answer:

(x+3)^2 +(y-5)^2 = 36

Step-by-step explanation:

We can write the equation of a circle as

(x-h)^2 +(y-k)^2 = r^2 where (h,k) is the center and r is the radius

(x- -3)^2 +(y-5)^2 = 6^2

(x+3)^2 +(y-5)^2 = 36

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What is x. Please show *all* the steps.

Zolol [24]

The equation 5/2 - x + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0 is a quadratic equation

The value of x is 8 or 1

<h3>How to determine the value of x?</h3>

The equation is given as:

5/2 - x + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0

Rewrite as:

-5/x - 2 + x - 5/x + 2 + 3x + 8/x^2 - 4 = 0

Take the LCM

[-5(x + 2) + (x -5)(x- 2)]\[x^2 - 4 + [3x + 8]/[x^2 - 4] = 0

Expand

[-5x - 10 + x^2 - 7x + 10]/[x^2 - 4] + [3x + 8]/[x^2 - 4] = 0

Evaluate the like terms

[x^2 - 12x]/[x^2 - 4] + [3x + 8]/[x^2 - 4 = 0

Multiply through by x^2 - 4

x^2 - 12x+ 3x + 8 = 0

Evaluate the like terms

x^2 -9x + 8 = 0

Expand

x^2 -x - 8x + 8 = 0

Factorize

x(x -1) - 8(x - 1) = 0

Factor out x - 1

(x -8)(x - 1) = 0

Solve for x

x = 8 or x = 1

Hence, the value of x is 8 or 1

Read more about equations at:

brainly.com/question/2972832

8 0

2 years ago

Triangle ABC is similar to triangle DEF. Find the value of side length DE. Side length DE is cm.

Art [367]

Answer:

??? I can't see the picture

Step-by-step explanation:

5 0

3 years ago

John stands 150 meters from a water tower and sights the top at an angle of elevation of 36º. If John's eyes are 2 meters above

horsena [70]

Answer:

The height of tower $DB=111\ meters$ .

Step-by-step explanation:

Diagram of the given scenario is shown below.

Given that,

Distance between John and tower is $CE=150 \ meters$ .

Angle of elevation to the top of the tower is $\angle DEC=36$ °.

Height of John is $CB=2\ meters$ .

To Find: Height of the tower $DB$ .

So,

In triangle ΔDCE,

$Tan$ (∠ $DEC)= \frac{DC}{CE}$

$Tan (36)= \frac{DC}{150}$

$DC= tan(36)\times 150$

$DC=108.98\ meters$

Now,

To calculate the height of tower we have

$DB=DC+CB$

$DB=108.98+2$

$DB=110.98\ meters$ ≈ $111 \ meters$

Therefore,

The height of tower $DB=111\ meters$ .

6 0

3 years ago

The area of a rectangle is found using the formula A=lw, where l is the length of the rectangle and w is the width. Multiply eac

charle [14.2K]

Answer:

A(x) = 3x² + 43x + 14

Step-by-step explanation:

Area of a rectangle = length × width

length = x + 14

width = 3x + 1

Area of a rectangle = length × width

= (x + 14)(3x + 1)

= 3x² + x + 42x + 14

= 3x² + 43x + 14

express the area of each rectangle as a single polynomial in terms of x.

A(x) = 3x² + 43x + 14

8 0

3 years ago

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago