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2 years ago

Identify the solution set of the inequality, using the given replacement set.

2 answers:

7 0

Thank you for posting your question here at brainly. Feel free to ask more questions.

The best and most correct answer among the choices provided by the question is C.{–10, –4.3} .
 
Hope my answer would be a great help for you.

irga5000 [103]2 years ago

3 0

Hello there.

Identify the solution set of the inequality, using the given replacement set.

x < –4; {–10, –4.3, –4, –3.9, 2, 6.5}

C.{–10, –4.3}

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Find the area of this semi-circle with diameter, d = 58cm. Give your answer rounded to 2 DP.

Anastasy [175]

Answer:

The area of the semi-circle

A = 1320.37 cm²

Step-by-step explanation:

Explanation:-

Given the diameter of the circle 'd' = 58cm

The radius of the circle 'd' = 2r

$r = \frac{d}{2} = \frac{58}{2} = 29$

The area of the semi-circle

$A = \frac{1}{2} \pi r^{2}$

$A = \frac{1}{2} (3.14) ((29))^{2}$

A = 1320.37 cm²

5 0

3 years ago

10×(6×5) +9^3 ×8 <img src="https://tex.z-dn.net/?f=10%20%5Ctimes%20%286%20%5Ctimes%205%29%20%2B%20%20%7B9%7D%5E%7B3%7D%20%20%

Minchanka [31]

Answer: 6132

Step-by-step explanation:

10x(6x5)+9^3 x8

10x30+9^3 x8

10x30+729x8

300+729x8

300+5832

6132

Hope this helps!

3 0

3 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

What is the probability of spinning praline wafer and then spinning oven baked apple & lavender calzone on the next spin pls

kolezko [41]

Answer:

Can you be a bit more specific?

Step-by-step explanation:

I need to know a bit more to help

8 0

3 years ago

The point C(5, 4) is translated 9 units down.

Naily [24]

Answer:

The new point is at (5, -5)

Step-by-step explanation:

When a point is being translated down, they are being subtracted on the y-value. So all you have to do is subtract 9 from 4 and we end up with -5, our new y-value.

(5, -5)

3 0

3 years ago

Read 2 more answers