Find the least common denominator (LCD) of 5/8
and 1/4

Answer:

75 times

Step-by-step explanation:

Answer:

∆K = 98°

∆M = 82°

∆G = 118°

∆H = 62°

Step-by-step explanation:

∆K & ∆G is by corresponding angle

and

∆M & ∆H is by linear pairs

Answer:

82 - 2i

Step-by-step explanation:

(5 - 2i) + (-4 + 81)

5 - 2i + 77

= 82 - 2i

1 + 6i

we open the brackets to give us

5 - 2i - 4 + 8i

collecting like terms, we have,

-2i + 8i + 5 - 4

6i + 1

or

1 + 6i

Answer:

The points would be: (-3, 3) (-4, -1) (2, -1) (1, 3)

Step-by-step explanation:

Because when you move 4 units left, all of the x values in the points will go down by 4, and when you move 2 units down, the y values will go down by 2 units. So when applied to the points shown, you get the answer I put.

I believe this is correct? Hope it helps!

Answer:

2nd option is the correct answer

Answer:

Step-by-step explanation:

300/100 = 3 = 1%

if 3 = 1% then 3/2 or 1.5 = 0.5%

5 x 3 = 15 = 5%

15 + 1.5 = 16.5 = 5.5%

Answer:

[tex]- 5 {x}^{3} - {x}^{2} - 11x + 7 \\ [/tex]

Step-by-step explanation:

Before solving that we have to know that,

[tex]( + ) \times ( + ) = ( + ) \\ ( - ) \times (- ) = ( + ) \\ ( + ) \times ( - ) = (- )[/tex]

Let's solve now.

[tex]7 {x}^{3} - 11x - 1 - (12 {x}^{3} + {x}^{2} - 8) \\ 7 {x}^{3} - 11x - 1 - 12 {x}^{3} - {x}^{2} + 8 \\ 7 {x}^{3} - 12 {x}^{3} - {x}^{2} - 11x - 1 + 8 \\ = - 5 {x}^{3} - {x}^{2} - 11x + 7[/tex]

Hope this helps you.

Let me know if you have any other questions :-):-)

Answer:

we can do it again and I am a great time with you and your family and friends and family is not good

Answer:

The other angle's measure is 2/3 (54) = 108/3 = 36 degrees.

Answer:

Step-by-step explanation:

You should add 9 fl. oz. of oil to your 3.75 gallons of gas.

Puch Maxi Moped from the mid-Eighties now since this moped doesn't run on straight gasoline, you need to create a special fuel mixture. This involves combining oil and gas in a specific ratio. The magic ratio here is 2.4 fl. oz. (fluid ounces) of oil for every gallon of gasoline.

You mentioned you've got 3.75 gallons of gas. To figure out how much oil you need, we can use some math. First, let's find out how much oil is required for one gallon of gas:

2.4 fl. oz. of oil/gallon x 1 gallon = 2.4 fl. oz. of oil.

So, for 3.75 gallons of gas, you'll need:

2.4 fl. oz. of oil/gallon x 3.75 gallons = 9 fl. oz. of oil.

You should add 9 fluid ounces of oil to your 3.75 gallons of gas to create the proper fuel mixture for your Puch Maxi Moped.

To know more about Ratio here

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Answer:

N+5

Step-by-step explanation:

4,9,14,19

4+5=9+5=14+5=19….

Answer:

None of the option is correct.

The right answer is 18/7 or 2.57

Step-by-step explanation:

As we know :

now let's plug the value of given terms :

Answer:

babawi lang sana

Step-by-step explanation:

naul pi

Answer:

y= 80/9

Step-by-step explanation:

Given

" y varies directly as x" and "as the square of z" so y = x* z²* some number

if y =25/9 when x= 5, z= 1 and y = x*z²* some number

then 25/9  = 5 * 1² * some number

      5*(5/9) = 5 * some number

therefore y = x* z²* (5/9)

y= ? when x=1 , z= 4

y = x * z² * (5/9)

y = 1 * 4² * (5/9)

y = 16*5 / 9

y = 80/9

Answer:

2/3multiply by20 u get 13.3 pounds

Answer:

x = 36 , y = 64 , z = 80

Step-by-step explanation:

Exterior angle equals the sum of opposite interior angle.

3x +  8 = x +  z

3x - x - z = -8

2x  -  z = -8   ------------ (I)

z + 3x - 8 = 180 {linear pair}

3x + z = 180 + 8

3x + z = 188 --------------(II)

Add (I) and (II) and z will be eliminated and we can find the value of 'x'

(I)            2x - z = -8

(II)           3x + z = 188   {Now add}

               5x      = 180

x = 180/5

x = 36

Plugin x = 36 in equation (II)

     3*36 +z = 188

        108 + z = 188

                z = 188- 108

z = 80

x + y + z = 180 ----------------(III)  {angle sum property of triangle}

36 + 80 +y = 180

       116 + y = 180

               y = 180 - 116

y = 64

Answer: 0.8333

Step-by-step explanation:

Answer: 5h+9

Step-by-step explanation:

10h + (-5h)

5h+(6(+3))

Answer:

x=[tex]\frac{60}{121}[/tex],z=[tex]\frac{535}{121}[/tex],y=-10

Step-by-step explanation:

Answer:

[tex]1) \: \: 12x + 6y + 7z = - 35 \\ 6y = - 35 - 12x + 7z \\ y = \frac{ - 35 - 12x + 7z}{6} \\ \\ 2) \: \: 7x - 5y - 6z = 200 \\ - 5y = 200 - 7x + 6z \\ y = \frac{200 - 7x + 6z}{ - 5} \\ \\ 3) \: \: x + y = - 10 \\ y = - 10 - x \\ [/tex]

Answer:

keep the car at a friends house untill you find a place to get it fixed at... or keep it at a trustworthy family members house.

If these don't help then im sorry.

(1)

(a) Use the fact that [tex]\sqrt{x^2} = |x|[/tex] for all [tex]x[/tex]. Since [tex]x\to+\infty[/tex], we have [tex]x>0[/tex] and [tex]|x| = x[/tex].

[tex]\displaystyle \lim_{x\to\infty} \frac{\sqrt{9x^2 - 2}}{x + 4} = \lim_{x\to\infty} \frac{\sqrt{x^2} \sqrt{9 - \frac2{x^2}}}{x + 4} \\\\= \lim_{x\to\infty} \frac{ x \sqrt{9 - \frac2{x^2}}}{x+4} \\\\= \lim_{x\to\infty} \frac{\sqrt{9 - \frac2{x^2}}}{1 + \frac4x} \\\\= \frac{\sqrt9}1 = \boxed{3}[/tex]

(b) This time [tex]x\to-\infty[/tex], so [tex]x < 0[/tex] and [tex]|x| = -x[/tex].

[tex]\displaystyle \lim_{x\to\infty} \frac{\sqrt{9x^2 - 2}}{x + 4} = \lim_{x\to\infty} \frac{\sqrt{x^2} \sqrt{9 - \frac2{x^2}}}{x + 4} \\\\= \lim_{x\to\infty} \frac{ \boxed{-x} \sqrt{9 - \frac2{x^2}}}{x+4} \\\\= (-1) \times \lim_{x\to\infty} \frac{\sqrt{9 - \frac2{x^2}}}{1 + \frac4x} \\\\= -\frac{\sqrt9}1 = \boxed{-3}[/tex]

(c) We immediately have

[tex]\displaystyle \lim_{x\to\infty} (x - \sqrt x) = \boxed{\infty}[/tex]

since [tex]x > \sqrt x[/tex] for all [tex]x > 1[/tex].

(d) Introduce a difference of squares by factoring in the limand's conjugate. The rest mirrors what we did in (a)/(b).

[tex]\displaystyle \lim_{x\to\infty} \left(\sqrt{x^2 + 12x} - x\right) = \lim_{x\to\infty} \frac{\left(\sqrt{x^2+12x} - x\right) \left(\sqrt{x^2+12x} + x\right)}{\sqrt{x^2 + 12x} + x} \\\\ = \lim_{x\to\infty} \frac{\left(\sqrt{x^2+12x}\right)^2 - x^2}{\sqrt{x^2+12x}+x} \\\\ = \lim_{x\to\infty} \frac{12x}{\sqrt{x^2+12x}+x} \\\\ = \lim_{x\to\infty} \frac{12}{\sqrt{1 + \frac{12}x} + 1} = \frac{12}{\sqrt1 + 1} = \boxed{6}[/tex]

(e) Divide through by the highest-degree exponential term.

[tex]\displaystyle \lim_{x\to\infty} \frac{12e^{2x} - 3e^{3x}}{2e^{2x} + 4e^{3x}} = \lim_{x\to\infty} \frac{12e^{-x} - 3}{2e^{-x} + 4} = \frac{0 - 3}{0 + 4} = \boxed{-\frac34}[/tex]

(2) By definition of the derivative, we have

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h) - f(x)}h[/tex]

For [tex]f(x) = \sqrt{x^2+1}[/tex], the limit becomes

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{\sqrt{(x+h)^2+1} - \sqrt{x^2+1}}h[/tex]

Factor in the conjugate.

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{\left(\sqrt{(x+h)^2+1} - \sqrt{x^2+1}\right) \left(\sqrt{(x+h)^2+1} + \sqrt{x^2+1}\right)}{h \left(\sqrt{(x+h)^2+1} + \sqrt{x^2+1}\right)}[/tex]

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{\left(\sqrt{(x+h)^2+1}\right)^2 - \left(\sqrt{x^2+1}\right)^2}{h \left(\sqrt{(x+h)^2+1} + \sqrt{x^2+1}\right)}[/tex]

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{\bigg((x+h)^2+1\bigg) - (x^2+1)}{h \left(\sqrt{(x+h)^2+1} + \sqrt{x^2+1}\right)}[/tex]

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{2xh + h^2}{h \left(\sqrt{(x+h)^2+1} + \sqrt{x^2+1}\right)}[/tex]

[tex]f'(x) = \displaystyle \lim_{h\to0} \frac{2x + h}{\sqrt{(x+h)^2+1} + \sqrt{x^2+1}}[/tex]

[tex]\implies \boxed{f'(x) = \displaystyle \lim_{h\to0} \frac{x}{\sqrt{x^2+1}}}[/tex]

(3) The tangent line to

[tex]y = \frac1{x^2+1}[/tex]

at the point (2, 1/5) has slope equal to the derivative [tex]\frac{dy}{dx}[/tex] when [tex]x = 2[/tex]. Compute the derivative; since [tex]y = \frac1{f(x)^2}[/tex] where [tex]f(x)[/tex] is the function from the previous problem, using the chain rule gives

[tex]y = \dfrac1{f(x)^2} \implies \dfrac{dy}{dx} = -\dfrac{2f'(x)}{f(x)^3} = -\dfrac{2 \times \frac{x}{\sqrt{x^2+1}}}{\left(\sqrt{x^2+1}\right)^3} \\\\ \implies \dfrac{dy}{dx} = -\dfrac{2x}{(x^2+1)^2}[/tex]

The tangent line at (2, 1/5) then has slope

[tex]\dfrac{dy}{dx}\bigg|_{x=2} = -\dfrac{2\times2}{(2^2+1)^2} = -\dfrac4{25}[/tex]

Using the point-slope formula, the equation of the tangent line is

[tex]y - \dfrac15 = -\dfrac4{25} (x - 2) \implies \boxed{y = -\dfrac{4x - 13}{25}}[/tex]

The sequence of the number of girl attendees at each Barangay is nonzero, given by multiplying the number of attendees in the previous Barangay by 2

The required values are;

a. The completed table is presented as follows;

[tex]\underline{\begin{array}{|l|c|c|c|c|c|c|c|}\mathbf{Barangay}&1&2&3&4&5&6&7\\\mathbf{Number \ of \ attandees}&4&8&16&32&64&128&256\end{array}}[/tex]

b. The sequence of the number of girls attendees aₙ = 4×2⁽ⁿ⁻¹⁾

c. The sum of beneficiaries of the lecture from the twelve Barangay is 16,384 girls

Reason:

Known parameters are;

Number of girls that attended the lecture at Barangay 1 = 4 girls

Number of girls that attended from Barangay 2 = 8 girls

Number that attended from Barangay 3 = 16 girls

a. The required table with the assumption that the number of girls that attended at each Barangay from Barangay 1 to Barangay 7 doubles, is presented as follows;

b. The sequence of the number of girls who attended the lecture from Barangay 1 to Barangay 7 is presented as follows;

The sequence is a geometric progression having the form, aₙ = a·r⁽ⁿ⁻¹⁾

The first term of the sequence, a = 4

The common ratio, r = 2 (each term is twice the previous term)

n = The number of terms

The sequence is therefore;

c. The total number of girls in the geometric progression, (G. P.) is given by the sum of a G. P. as follows

[tex]S_n = \dfrac{a \cdot (r^n - 1 )}{r - 1}[/tex]

Where;

n = The number of terms = 7 + 5 = 12

Which gives;

The total number of girls who will benefit from the lectures, S₁₂ = 16,384 girls

Learn more about geometric progression here:

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Answer:

2700 more eggs were ordered

Answer:

Depends, rise over run could be considered a fraction, or if it's a number that has a whole number like one and 3/4 then you just plot the point to the best of your ability. In other words guess.

Answer:

[tex]D)\:(3x^{2} +2)(x+4)[/tex]

Step-by-step explanation:

[tex]3x^{3} + 12x^{2} + 2x + 8[/tex]

Group, [tex]3x^3+12x^2+2x+8[/tex]  [tex]=(3x^{3} +12x^{2} )+(2x+8)[/tex]factor out [tex]3x^{2}[/tex]in the first and 2 in the second group.

[tex]3x^{2} (x+4)+2(x+4)[/tex]

Factor out common term [tex]x+4[/tex] by using distributive property.

polynomial [tex]3x^{2} +2[/tex] is not factored since it does not have any rational roots.

[tex](3x^{2} +2)(x+4)[/tex]

_____________________________

Answer:d

Step-by-step explanation: