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3 years ago

Help me please and thank you

Mathematics

2 answers:

nikklg [1K]3 years ago

6 0

The correct answer is 6.

Shkiper50 [21]3 years ago

5 0

The correct answer is 6

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A right triangle has side lengths 12 centimeters and 14 centimeters. Name two possible side lengths for the third side, and expl

madam [21]

Answer:

If 12 and 14 are both side lengths then we solve for the hypotenuse:

hypotenuse^2 = 12^2 + 14^2

hypotenuse^2 = 144 + 196

hypotenuse^2 = 340

hypotenuse = sq root (340) = 18.4390889146

If hypotenuse = 14, then third side equals:

(3rd side)^2 = 14^2 - 12^2

(3rd side)^2 = 196 -144

(3rd side)^2 = 52

3rd side = sq root (52) = 7.2111025509

Step-by-step explanation:

4 0

2 years ago

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PLEASE HELP ASAP<br> Find m

Vadim26 [7]

Answer:

x = 26

Step-by-step explanation:

3 0

2 years ago

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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

2 years ago

Please answer! <br> Solve for X

Sedbober [7]

Answer:

x = 3

Step-by-step explanation:

We can solve this by using ratios of similar triangles

3 x

------ = ------------

3+9 12

3 x

------ = ------------

12 12

x must equal 3 to keep the ratios the same

4 0

3 years ago

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Which bar diagram represents the equation 5x + 7.5 = 15? A bar diagram. The top box contains 5 x. The bottom boxes contain 7.5 a

photoshop1234 [79]

Answer:

Option C.

Step-by-step explanation:

The top box must ALWAYS contain the answer to the problem.

6 0

3 years ago

Read 2 more answers