What is 27% of 61?
Pls help TvT

Answer:

Let A have speed 'a.' Remember distance is rate times time, and the Pythagorean theorem. We then have (2a)^2+(2(a+5))^2=50^2. a=15, so the other rate is 20.

Step-by-step explanation:

Answer:

-2x-0.7

Step-by-step explanation:

for me that is the best answer

Answer:

$24.75

Step-by-step explanation:

Discount = Original Price x Discount %/100

Discount = 165 × 15/100

Discount = 165 x 0.15

You save = $24.75

Final Price = Original Price - Discount

Final Price = 165 - 24.75

Final Price = $140.25

Answer:

x=69/7   y=-95/7

Step-by-step explanation:

Answer:

[tex]2<p<6[/tex]

Step-by-step explanation:

If it is less than [tex]6[/tex], it must have [tex]p<6[/tex].

If it is greater than [tex]2[/tex], it must have [tex]2<p[/tex].

We can combine these inequalities to get [tex]\boxed{2<p<6}[/tex].

The graph is left as an exercise to the reader.

Step-by-step explanation:

if you really want to get answer you should follow me

Answer:

You forgot to put the picture please redo and ill edit my answer

Step-by-step explanation:

Answer:

See below:

Step-by-step explanation:

If you want to start studying but cant find the will too, set up a schedule and try to limit your mind to just studying during that time, you can start small like ~15 minutes on your first day, and gradually improve once your mind starts to adjust to it.

Its mostly figuring out how your mind works.

Funny how school teaches you, but doesnt teach you how to learn.

If this answer helped you, please consider marking as brainliest, helps a ton.

Have a nice day!

Answer:

591755.66030836 was my answer. Hope that helps.

Answer:

The number is 0.06, or 6 hundredths.

Step-by-step explanation:

If we convert 2 2/5 to a decimal, we get 2.4

Then, we can convert our percentage to a decimal by moving our digits two units down, giving us .025.

Now, we can multiply 2.4 and .025, which gives us 0.06

9514 1404 393

Answer:

  18.8 mi

Step-by-step explanation:

The Pythagorean theorem can be used to find the straight-line distance between the towns. The distances 8 mi and 17 mi form the legs of a right triangle, whose hypotenuse is the straight-line distance.

  d² = 8² +17² = 64 +289 = 353

  d = √353 ≈ 18.8  . . . miles

The two towns are about 18.8 miles apart.

Answer:

[tex]a(12) = 12 \div 3 + 2 \\ + b(1) = 12 \times 1 \\ = 4 + 2 + 12 = 18[/tex]

Given :

To find :-

Solution :

In larger triangle :-

As we know that ,

Corresponding Angles of similar triangles are same ,

y = -x + 3

y = -(x + 12)

y = -x - 12

The kilograms of peanut clusters that the clerk has will be 12 kilograms of peanut flusters.

Let the kilograms of peanut clusters = x

Let kilograms of chocolate creams = y = 3x

Therefore, from the information given, the equation to solve the question will be:

x + y = 48 kilograms

x + 3x = 48 kilograms

4x = 48 kilograms

x = 48 kilograms / 4

x = 12 kilograms

Therefore, the kilograms of peanut clusters that the clerk has will be 12 kilograms of peanut clusters.

In conclusion, the correct option is 12 kilograms.

Read related link on:

https://brainly.com/question/24959144

Applying the rules, it is found that ordered pair (24,6), option A, works for both rules.

---------------------

The rule 1 is that y is x divided by 4, that is:

[tex]y_1 = \frac{x}{4}[/tex]

The rule 2 is that y is x subtracted from 18, that is:

[tex]y_2 = x - 18[/tex]

In item A, we have that the input x = 24, thus, the results of the expressions are:

[tex]y_1 = \frac{24}{4} = 6[/tex]

[tex]y_2 = 24 - 18 = 6[/tex]

In both cases, ordered pair (24,6), thus option A is the answer.

A similar problem is given at https://brainly.com/question/22103476

Answer: 4

pls give brainliest

Answer:

A

Step-by-step explanation:

9514 1404 393

Answer:

  0.9114

Step-by-step explanation:

The multiplier for a series of changes is the product of the multipliers for each of the changes:

  (1 -2%)(1 -7%) = (0.98)(0.93) = 0.9114

(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]

Answer:

1

Step-by-step explanation:

This one once learned is clapz. Anything to the zero power is 1. ;)

Answer:

Domain = (infinity, infinity) Range = (3,0)

Step-by-step explanation:

Hard to explain without giving a 15 minute lecture lol