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2 years ago

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Find the least common denominator

a1" title="\frac{3}{9} \frac{8}{4}" alt="\frac{3}{9} \frac{8}{4}" align="absmiddle" class="latex-formula">

1 answer:

MA_775_DIABLO [31]2 years ago

8 0

Answer:

36 is your least common denominator. Hope this helps

Step-by-step explanation:

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Explain how to find the value of x. Be sure to include the postulates, definitions and theorems that justify your answer.

Black_prince [1.1K]

Answer: x = 12

Step-by-step explanation:

We can see two triangles rectangles, we will use the angle S to solve this, as this angle is common to both triangles.

We know that:

cos(S) = adj cathetus/hipotenuse.

Then we have, for the triangle SRT.

Cos(S) = 9/SR.

for the big triangle, SQR, we have:

Cos(S) = SR/(9 + 16)

now we can find the quotient of those two equations and get:

cos(S)/cos(S) = 9/Sr*(9+16)/SR

SR^2 = 9*(25)

SR = 15.

Now with this side we can find the value of x.

We can use the Pythagorean theorem in the triangle SRT, where the sum of the squared cathetus is equal to the hypotenuse:

9^2 + x^2 = 15^2

x = √(15^2 - 9^2) = 12

7 0

3 years ago

X/2-7=9 solve what “x” mean

Elden [556K]

Answer:

x = 32

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

x /2 −7=9

1 /2 x + (−7) = 9

1 /2 x − 7 = 9

Step 2: Add 7 to both sides.

1 /2x − 7 + 7 = 9 + 7

1 /2 x = 16

Step 3: Multiply both sides by 2.

2*( 1 /2 x) = (2) * (16)

x = 32

7 0

3 years ago

A guitar is on sale for 20% off the original price. Let p represent the original price. What expressions can be used to calculat

Gala2k [10]

20% off so, u are paying 80%
80% = 0.8
0.8p

5 0

2 years ago

Read 2 more answers

1)The sum of a number and 81 is at least 102.

Katyanochek1 [597]

Answer:

any number grater than or equal to 21

Step-by-step explanation:

3 0

2 years ago

Can someone thoroughly explain this implicit differentiation with a trig function. No matter how many times I try to solve this,

Anton [14]

Answer:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

Step-by-step explanation:

So we have the equation:

$\tan(x-y)=\frac{y}{8+x^2}$

And we want to find dy/dx.

So, let's take the derivative of both sides:

$\frac{d}{dx}[\tan(x-y)]=\frac{d}{dx}[\frac{y}{8+x^2}]$

Let's do each side individually.

Left Side:

We have:

$\frac{d}{dx}[\tan(x-y)]$

We can use the chain rule, where:

$(u(v(x))'=u'(v(x))\cdot v'(x)$

Let u(x) be tan(x). Then v(x) is (x-y). Remember that d/dx(tan(x)) is sec²(x). So:

$=\sec^2(x-y)\cdot (\frac{d}{dx}[x-y])$

Differentiate x like normally. Implicitly differentiate for y. This yields:

$=\sec^2(x-y)(1-y')$

Distribute:

$=\sec^2(x-y)-y'\sec^2(x-y)$

And that is our left side.

Right Side:

We have:

$\frac{d}{dx}[\frac{y}{8+x^2}]$

We can use the quotient rule, where:

$\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}$

f is y. g is (8+x²). So:

$=\frac{\frac{d}{dx}[y](8+x^2)-(y)\frac{d}{dx}(8+x^2)}{(8+x^2)^2}$

Differentiate:

$=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

And that is our right side.

So, our entire equation is:

$\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

To find dy/dx, we have to solve for y'. Let's multiply both sides by the denominator on the right. So:

$((8+x^2)^2)\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}((8+x^2)^2)$

The right side cancels. Let's distribute the left:

$\sec^2(x-y)(8+x^2)^2-y'\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy$

Now, let's move all the y'-terms to one side. Add our second term from our left equation to the right. So:

$\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy+y'\sec^2(x-y)(8+x^2)^2$

Move -2xy to the left. So:

$\sec^2(x-y)(8+x^2)^2+2xy=y'(8+x^2)+y'\sec^2(x-y)(8+x^2)^2$

Factor out a y' from the right:

$\sec^2(x-y)(8+x^2)^2+2xy=y'((8+x^2)+\sec^2(x-y)(8+x^2)^2)$

Divide. Therefore, dy/dx is:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)+\sec^2(x-y)(8+x^2)^2}$

We can factor out a (8+x²) from the denominator. So:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

And we're done!

8 0

3 years ago