Write the equation of the line with a slope of 2 that passes through the point (-3,
10). Write your answer in slope-intercept form.

Answer:

Step-by-step explanation:

37.203

-29.800

=36.203+1-29.800=36.2003-1+0.200=7.403

Answer:

Subtraction,... Below,...!

Step-by-step explanation:

37.203 - 29.800 = 7.403

Subtract like you where subtracting 37203 - 29800 you add the extra 0's to align the would-be decimal points...

37203 - 29800 = 7403

Answer: 7.403

Chow,...!

Answer:

[tex]25 = \frac{25}{100} [/tex]

so answer is

[tex]a) \: \: \frac{25}{100} [/tex]

Answer:

x = 3/5

Step-by-step explanation:

5th root of Y is notated in exponential notation as Y^(1/5), which is Y to the one fifth (1/5) power.

So, the 5th root of 19^3 is (19^3)^(1/5) = 19^(3 * 1/5) = 19^(3/5), since (A^a)^b = A^(a*b).

Answer:

[tex]8k + 10 + 9k + 1 = k(8 + 9) + (10 + 1) \\ = 17k + 11[/tex]

Answer:

12m perimeter

Step-by-step explanation:

all side are 3m so 3x4 = 12m

Given :

To find :-

Solution :

In larger triangle :-

As we know that ,

Corresponding Angles of similar triangles are same ,

y=mx+b

-3=3(3)+b

-3=9+b

-12=b

Answer:

Step-by-step explanation:

12*30=360

360/6=60

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:

ok?

Step-by-step explanation:

Answer:

9762

Step-by-step explanation:

Step-by-step explanation:

if you really want to get answer you should follow me

Answer:

it is a

Step-by-step explanation:

2-5

Answer:

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

Step-by-step explanation:

(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]

9514 1404 393

Answer:

  (x, y, z) = (-9, -1, 8)

Step-by-step explanation:

My calculator tells me the reduced row-echelon form of ...

  [tex]\left[\begin{array}{ccc|c}-3&-2&-3&5\\1&2&-1&-19\\-2&1&-2&1\end{array}\right][/tex]

is ...

  [tex]\left[\begin{array}{ccc|c}1&0&0&-9\\0&1&0&-1\\0&0&1&8\end{array}\right][/tex]

This means the solution is (x, y, z) = (-9, -1, 8).

_____

Additional comment

In case you haven't learned how to use your graphing calculator to solve matrix equations, you can solve this system easily as follows:

Subtract 2 times the first equation from 3 times the third equation. This eliminates the x and z terms and gives you the value of y. (y=-1)

Substitute that value into the first and second equations. This gives you two equations in the sum and difference of x and z. (You may need to divide the coefficients by the common factor.) Add these sum and difference equations to find one of the variables, then substitute to find the other.

Answer:

5, i think

Step-by-step explanation:

1. 3x = 15 and divide all by 3 (the number with a variable or letter

Answer:

f(x)=4−(x−7)

3

Let y=f(x)

So, x=f

−1

(y) ---- ( 1 )

y=f(x)

⇒ y=4−(x−7)

3

⇒ (x−7)

3

=4−y

⇒ x−7=

3

4−y

⇒ x=7+

3

4−y

⇒ f

−1

(y)=7+

3

4−y

.................. [ From ( 1 ) ]

∴ f

−1

(x)=7+

3

4−x

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

Answer:

A) = 5

B) = 3

C) = [tex]2\sqrt{3}[/tex]

D) = 2.2894

Step-by-step explanation:

A)  [tex]x^2=25[/tex]

[tex]\sqrt{x^2} = \sqrt{25} \\x= 5[/tex]

5*5 = 25

B)  [tex]x^3 = 27[/tex]

[tex]\sqrt[3]{x^3} = \sqrt[3]{27} \\x=3[/tex]

3*3*3 = 27

C)  [tex]x^2 = 12\\[/tex]

[tex]\sqrt{x^2} = \sqrt{12} \\x=2\sqrt{3} \\[/tex]

[tex]2\sqrt{3}[/tex] * [tex]2\sqrt{3}[/tex] = 12

D)  [tex]x^3 = 12[/tex]

[tex]\sqrt[3]{x^3} = \sqrt[3]{12} \\x=2.2894.....\\[/tex]

2.2894 * 2.2894 *2.2894 = 12

No exact solution

Answer:

[tex]x=-1\\y=3\\z=-4\\[/tex]

Step-by-step explanation:

Please see attached document.

Answer:

√3(√6+√15)

= 3**1/2(6**1/2 + 15**1/2)

= 3**1/2 · 6**1/2 + 3**1/2 · 15**1/2

= 18**1/2 + 45**1/2

= (2×9)**1/2 + (5×9)**1/2

= 2**1/2 · 9**1/2 + 5**1/2 · 9**1/2

= 2**1/2 · 3 + 5**1/2 · 3

= 3(√2 + √5)

Step-by-step explanation: