All categories

3 years ago

6(2k+5)-3k=66 show work please

Mathematics

2 answers:

erastova [34]3 years ago

7 0

12k+30-3k=66
12k-3k=36
9k=36
k=4

Katyanochek1 [597]3 years ago

5 0

6(2k+5)-3k=66

12k+30-3k=66

9k+30=66

9k=66-30

9k=36

K=36/9

K=4

Hope this helps!

You might be interested in

Determine if the following relations represent y as a function of x.

lana [24]

<h3>Answers:</h3>

A) No, it is not a function
B) Yes it's a function
C) Not a function
D) Yes it's a function
E) Yes it's a function

================================================

Explanation:

If y has an exponent of 2, 4, 6, etc (basically any even number) then it leads to having inputs with multiple outputs.

Consider something like y^2 = x. If x = 100, then y = 10 or y = -10 are possible. A function can only have exactly one y output for any valid x input. Similar issues happen for things like y^4 = x and so on. So this is why A and C are not functions.

The other equations do not have y values with such exponents, so we can solve for y and have each x input lead to exactly one y output. Therefore, they are functions.

5 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

Year 2 revenue is 5,000,000 at 32%By what percent would you need to

azamat

$x= \frac{3,000,000}{5,000,000} \cdot100\%-32\%= \frac{3}{5}\cdot100\%-32\%=60\%-32\%=28\%\\\\Ans.\ You\ need\ increase\ 28\%$

7 0

3 years ago

I have no clue how to start this problem,plz help

Maksim231197 [3]

9514 1404 393

Answer:

109°

Step-by-step explanation:

You always start a problem by taking a careful look at the information given and how it relates to what is asked. Here, the key information is in the symbols marking the lines PQ and RS. They are parallel.

This means segments QR and PS are transversals. Marked angle 41° will be an "alternate interior angle" with angle TPQ, so angle TPQ is also 41°.

The desired angle, PTR, is an exterior angle to ΔQTP, so its measure is the sum of remote interior angles TQP (68°) and TPQ (41°). That is, ...

∠PTR = ∠TQP +∠TPQ = 68° +41°

∠PTR = 109°

6 0

2 years ago

Please help me solve step by step it's urgent

NNADVOKAT [17]

Answer:

2i: 169.71

2ii: 0.17L

3a: 4×10⁻⁵

3b: 110011

Step-by-step explanation:

2i. The surface of the top and bottom of the tin is two times (top and bottom) π·r² = 2·π·3² = 18π cm².

The circumference of the circle is 2·π·r = 6π cm².

The area of the material connecting top and bottom is a rectangle of the tin height times the circumference: 6·6π = 36π cm².

This gives a total of 18π + 36π = 54π cm².

With π approximated by 22/7 the total surface area is 54*22/7 ≈ 169.71.

Notice how the calculation is simple by waiting until the very last moment to substitute π.

2ii. The volume is the area π·r² of the circle times the height of the tin: 9π*6 = 54π cm³ ≈ 169.71 cm³.

Since 1L = 1000 cm³ the volume is 0.16971 litres, which should be rounded to 0.17 L.

3a: If we rewrite P as 36 x 10⁻⁴ and realize that 36/2.25 = 16, then the fraction can be written as

16 x 10⁻⁴⁻⁶ = 16 x 10⁻¹⁰.

The square root of that is taking it to the power of 1/2, so (16x10⁻¹⁰)^0.5 = 4x10⁻⁵ = 0.00004

3b: 1111 1111 is 255 in decimal. 101 is 5 in decimal. 255/5 is 51 in decimal. 51 in binary is 110011.

6 0

3 years ago