4x−6y=36
Put the following equation of a line into slope-intercept form, simplifying all fractions.

Answer:

30 [tex](1 + \frac{.18}{12})^{5*12}[/tex]

Step-by-step explanation:

Answer:

It will still be 9.  y=9 is a horizontal line.  No matter the x value, the answer will be the same.

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

The triangles do have equal angles and proportional sides, because they are bot right triangles formed on parallel lines.

:)ur welcome

Answer: Choice B

Yes. Right triangles formed on parallel lines have proportional side lengths and congruent angles.

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

Explanation:

Parallel lines always have equal slopes, but different y intercepts.

Recall that slope = rise/run.

The "rise" is the vertical portion of the triangle, while the "run" is the horizontal portion.

So let's say we had a slope of rise/run = 2/3. This means rise = 2 and run = 3.

If we had something like rise/run = 4/6, then that reduces to 2/3, showing that 4/6 = 2/3. They are the same slope.

We can use the SAS similarity theorem to prove the triangles formed in the diagram are congruent based on what is discussed above.

Because the triangles are similar, this means the corresponding angles are congruent. One such pair is the pair of right angles shown.

Answer:

17,100

Step-by-step explanation:

A light bulb consumes 3600 watts per day

Therefore the number of watts consumed in 4 days 18 hours can be calculated as follows

3600= 24 hours

x= 1 hour

= 3600/24

= 150 watts in an hour

3600×4

= 14,400

150×18

= 2,700

Total watts

= 14,400+2700

= 17,100

Hence the number if watss produced in 4 days 18 hours is 17,100 watts

Answer:

(Altitude)^2+(Base)^2=(Hypotenuse)^2

Answer:

x = 15

Step-by-step explanation:

-3x+20+7x=80

Combine like terms

4x+20 = 80

Subtract 20 from each side

4x+20 -20 = 80-20

4x = 60

Divide by 4

4x/4 = 60/4

x = 15

Answer:

- [tex]\frac{1}{3}[/tex]

Step-by-step explanation:

Given a line with slope m then the slope of a line perpendicular to it is

[tex]m_{perpendicular}[/tex] = - [tex]\frac{1}{m}[/tex]

Here m = 3 , then

[tex]m_{perpendicular}[/tex] = - [tex]\frac{1}{3}[/tex]

Answer:

So, the step1 is correct.

Step-by-step explanation:

The expression is

[tex](6 x^3 + 8 x^2 - 7 x)-(2x^2 + 3)(x- 8)\\\\(6 x^3 + 8 x^2 - 7 x) - (2x^3 - 16 x^2 + 3 x - 24)\\\\6 x^3 + 8 x^2 - 7 x - 2 x^3 - 16 x^2 + 3 x - 24\\\\4 x^3 - 8 x^2 - 4 x - 24[/tex]

So, the step 1 is correct.

Answer:

7 3/7

Step-by-step explanation:

Answer:

39,800+x0.18= his total pay

Your Welcome

Biffy Out!!!

Answer:

198/7

Step-by-step explanation:

perimeter of semi circle= (pi) x r + d

                                     = 22/7 x 3 + 6

                                     = 198 cm

Answer:

15.42 cm

Step-by-step explanation:

perimeter of semicircle = πr + 2r

=3.14*3  + 2*3

=9.42 + 6

=15.42

Answer

16.25

Step-by-step explanation: To The Nearest Hundredth

(-1, 2), (3, 1)

√17

(3, 1), (7, 2)

√17

(3, 1), (7, 2)

√8

So our answer is =

8 + √17 + √17 = 16.25

hope it works for you!

Answer:

11.8 ft

Step-by-step explanation:

Since the length of one side, l = 5.3 ft, the longer base b = 12.7 ft and the one brace, d form a triangle with angle between the longer base and side being the angle facing the brace is θ = 68.4°, we use the cosine rule to find the length of thee brace.

So d² = l² + b² -2lbcosθ

So, substituting the values of the variables into the equation, we have

d² = (5.3 ft)² + (12.7 ft)² -2(5.3)(12.7)cos68.4°

d² = 28.09 ft² + 161.29 ft² - 134.62(0.3681)

d² = 189.38 ft² - 49.56 ft²

d² = 139.82 ft²

taking square-root of both sides, we have

d = √139.82 ft²

d = 11.82 ft

d ≅ 11.8 ft to the nearest tenth of a foot.

Answer:

8a³b

Step-by-step explanation:

Answer:

I got 45 degrees

Step-by-step explanation:

knowing that where T is is 90 degrees and a triangle always= 180 degrees I just split 90 in half and got 45. I could be wrong but I hope this helps :)

Scale factor from figure a to figure b = DIVIDE BY 3

{Check:- 33/3 = 11 & 15/3 = 5}

hope this helps :)

Answer:

1/3

Step-by-step explanation:

These two quadrilaterals are similar because figure A's sides are 3 times figure B's sides. Figure A's bottom is 15, while figure B's bottom is 5. To get from 15 to 5, we multiply by 1/3. This is the scale factor. All the other sides of figure A are also 3 times bigger than figure B's.

Answer:

In mathematics,reciprocal is an expression or function so related to another that their product is unity; the quantity obtained by dividing the number one by a given quantity.

Answer:Rh

Step-by-step explanation:mama

Answer:

288 cm^2

Step-by-step explanation:

If you have 2 congruent squares, side by side, the perimeter is 6*sidelength.

We divide 72 by 6 to find sidelength, and get 12.

We square 12 to find the area of a square, and get 144.

Multiply by 2 to find the area of the rectangle, and get 288 cm^2

Answer:

12

Step-by-step explanation:

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Factors of 180: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180

Answer: 14

Step-by-step explanation:

y -1 0 = -6

Ans -2 (x-10) = 14

Hope I could help :)

Answer:

[tex] \displaystyle - \frac{1}{2} [/tex]

Step-by-step explanation:

we would like to compute the following limit:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{1}{ \ln(x + \sqrt{ {x}^{2} + 1} ) } - \frac{1}{ \ln(x + 1) } \right) [/tex]

if we substitute 0 directly we would end up with:

[tex] \displaystyle\frac{1}{0} - \frac{1}{0} [/tex]

which is an indeterminate form! therefore we need an alternate way to compute the limit to do so simplify the expression and that yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right) [/tex]

now notice that after simplifying we ended up with a rational expression in that case to compute the limit we can consider using L'hopital rule which states that

[tex] \rm \displaystyle \lim _{x \to c} \left( \frac{f(x)}{g(x)} \right) = \lim _{x \to c} \left( \frac{f'(x)}{g'(x)} \right) [/tex]

thus apply L'hopital rule which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \dfrac{d}{dx} \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right) [/tex]

use difference and Product derivation rule to differentiate the numerator and the denominator respectively which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \frac{1}{x + 1} - \frac{1}{ \sqrt{x + 1} } }{ \frac{ \ln(x + 1)}{ \sqrt{ {x}^{2} + 1 } } + \frac{ \ln(x + \sqrt{x ^{2} + 1 } }{x + 1} } \right) [/tex]

simplify which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \sqrt{ {x}^{2} + 1 } - x - 1 }{ (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right) [/tex]

unfortunately! it's still an indeterminate form if we substitute 0 for x therefore apply L'hopital rule once again which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \sqrt{ {x}^{2} + 1 } - x - 1 }{ \dfrac{d}{dx} (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right) [/tex]

use difference and sum derivation rule to differentiate the numerator and the denominator respectively and that is yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \frac{x}{ \sqrt{ {x}^{2} + 1 } } - 1}{ \ln(x + 1) + 2 + \frac{x \ln(x + \sqrt{ {x}^{2} + 1 } ) }{ \sqrt{ {x}^{2} + 1 } } } \right) [/tex]

thank god! now it's not an indeterminate form if we substitute 0 for x thus do so which yields:

[tex] \displaystyle \frac{ \frac{0}{ \sqrt{ {0}^{2} + 1 } } - 1}{ \ln(0 + 1) + 2 + \frac{0 \ln(0 + \sqrt{ {0}^{2} + 1 } ) }{ \sqrt{ {0}^{2} + 1 } } } [/tex]

simplify which yields:

[tex] \displaystyle - \frac{1}{2} [/tex]

finally, we are done!

9514 1404 393

Answer:

  -1/2

Step-by-step explanation:

Evaluating the expression directly at x=0 gives ...

  [tex]\dfrac{1}{\ln(\sqrt{1})}-\dfrac{1}{\ln(1)}=\dfrac{1}{0}-\dfrac{1}{0}\qquad\text{an indeterminate form}[/tex]

Using the linear approximations of the log and root functions, we can put this in a form that can be evaluated at x=0.

The approximations of interest are ...

  [tex]\ln(x+1)\approx x\quad\text{for x near 0}\\\\\sqrt{x+1}\approx \dfrac{x}{2}+1\quad\text{for x near 0}[/tex]

__

Then as x nears zero, the limit we seek is reasonably approximated by the limit ...

  [tex]\displaystyle\lim_{x\to0}\left(\dfrac{1}{x+\dfrac{x^2}{2}}-\dfrac{1}{x}\right)=\lim_{x\to0}\left(\dfrac{x-(x+\dfrac{x^2}{2})}{x(x+\dfrac{x^2}{2})}\right)\\\\=\lim_{x\to0}\dfrac{-\dfrac{x^2}{2}}{x^2(1+\dfrac{x}{2})}=\lim_{x\to0}\dfrac{-1}{2+x}=\boxed{-\dfrac{1}{2}}[/tex]

_____

I find a graphing calculator can often give a good clue as to the limit of a function.

Answer:

b

Step-by-step explanation:

because its alternation

The option (C) Corresponding trigonometric ratios are equal is correct.

It is defined as the law to prove that the two triangles have the same shape, but it is not compulsory to have the same size. The ratio of the corresponding sides is in the same proportions and the corresponding angles are congruent.

We have a statement:

Which of the following is true about similar right triangles?

As we know from the definition the two triangles have the same shape, but it is not compulsory to have the same size.

The ratio of the corresponding sides is in the same proportions

The corresponding trigonometric ratios are equal.

Thus, the option (C) Corresponding trigonometric ratios are equal is correct.

Learn more about the similarity of triangles here:

brainly.com/question/8045819

#SPJ2

Answer:

5/18

Step-by-step explanation:

conditional probability formula:

A|B (a given b)  = (A∩B)/(B)

so

Moderate|college (moderate given college)= (moderate∩college)/college

moderate∩college= 15

college= 11+15+28= 54

15/54= 5/18