Which equation is an identity?
A)
3x + 4 = 5x - 1
B)
2x + 3 = 2x + 4 - 1
C)
5x = 10
D)
- 3x + 1 = -3x - 5

Answer:

RS = XY

Step-by-step explanation:

Corresponding sides will be equal. So

RS = XY

Answer:

145% increase

Step-by-step explanation:

343 - 140 = 203

203/140 = 1.45

1.45 x 100 = 145

Answer:

hey you can use scan and search

Answer:

I got : 0.42264973

Step-by-step explanation:

1/ -√3 + 1

-√3 = -1.73 ( 2 decimal place )

1/ -1.73 + 1

and found the answer

Answer: its a

Step-by-step explanation: its the same answer

Answer: the answer is 3(p - 5) + 2s

Step-by-step explanation:

its the only answer there that makes since and hope this helps im kinda bad at algebra but this was easy

Answer:

Correct choice: B

Step-by-step explanation:

Equation Solving

The area of a trapezoid with height h and bases b1 and b2 is given by:

We must solve this formula for h.

First, multiply by 2 to eliminate denominators:

Now, divide by b1+b2:

Swapping sides:

Correct choice: B

[tex]4^4(4^7)4\implies 4^4\cdot 4^7\cdot 4^1\implies 4^{4+7+1}\implies 4^{12}[/tex]

Answer:

real answer 4^-2

Step-by-step explanation:

just did on edge

Answer:

B.O.D.M.A.S

Step-by-step explanation:

B: Bracket

O: Operation

D: Division

M: Multiplication

A: Addition

S: Subtraction

Follow these steps.

Hope you understand ^^

Answer:

Step-by-step explanation:

Steps are in the attachment.

Note :-

=》2 is taken as negative & also taken out of the square root as it's the common factor.

______

⚜ Hope it helps :))

Answer:

C

Step-by-step explanation:

if you don't know it choose C

Combine the fractions on the left with a common denominator:

[tex]\dfrac a{x+1} + \dfrac b{x-1} + \dfrac c{(x+1)^2} = \dfrac{a(x+1)(x-1) + b(x+1)^2 + c(x-1)}{(x-1)(x+1)^2}[/tex]

It follows that

[tex]3x^2+3x-2 = a(x+1)(x-1) + b(x+1)^2 + c(x-1)[/tex]

Expand the right side and collect like powers of x :

[tex]3x^2+3x-2 = a(x^2-1) + b(x^2+2x+1) + c(x-1)[/tex]

[tex]3x^2+3x-2 = (a+b)x^2 + (2b + c)x -a +b - c[/tex]

Then we have the system of equations

[tex]\begin{cases}a+b=3\\2b+c=3\\-a+b-c=-2\end{cases}[/tex]

or in matrix form,

[tex]\begin{bmatrix}1&1&0\\0&2&1\\-1&1&-1\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}3\\3\\-2\end{bmatrix}[/tex]

Compute the determinant of the coefficient matrix:

[tex]\det\begin{bmatrix}1&1&0\\0&2&1\\-1&1&-1\end{bmatrix} = -4[/tex]

Then the inverse of the coefficient matrix is equal 1/det times the adjugate of the coefficient matrix (a.k.a the transpose of the cofactor matrix):

[tex]\begin{bmatrix}1&1&0\\0&2&1\\-1&1&-1\end{bmatrix}^{-1} = \dfrac1{-4} \begin{bmatrix}-3 & -1 & 2 \\ 1 & -1 & -2 \\ 1 & -1 & 2\end{bmatrix}^\top = -\dfrac14 \begin{bmatrix}3&-1&-1\\1&1&1\\-2&2&-2\end{bmatrix}[/tex]

Multiply both sides of the equation by the inverse :

[tex]\begin{bmatrix}a\\b\\c\end{bmatrix} = -\dfrac14 \begin{bmatrix}3&-1&-1\\1&1&1\\-2&2&-2\end{bmatrix} \begin{bmatrix}3\\3\\-2\end{bmatrix} = \begin{bmatrix}2\\1\\1\end{bmatrix}[/tex]

So, we have a = 2 and b = c = 1, and the partial fraction decomposition is

[tex]\dfrac{3x^2+3x-2}{(x+1)^2(x-1)} = \dfrac 2{x+1} + \dfrac 1{x-1} + \dfrac 1{(x+1)^2}[/tex]

Answer:

Step-by-step explanation:

One can solve for a, b, c a little more directly than using a system of 3 equations.

If we multiply the rational expression by (x+1)², we get ...

  (3x² +3x -2)/(x -1) = (x+1)²(a/(x+1) +b/(x-1)) +c

Evaluating this for x = -1 gives ...

  (3(-1)² +3(-1) -2)/(-1 -1) = c

  -2/-2 = 1 = c

Similarly, multiplying by (x -1) gives ...

  (3x² +3x -2)/(x +1)² = (x -1)(a/(x +1) +c/(x +1)²) + b

Evaluating this for x = 1 gives ...

  (3·1² +3·1 -2)/(1 +1)² = b

  4/4 = 1 = b

Now, we need to find the value of 'a'. The identity will hold true for any value of x, so we can see what happens when we substitute x=0. We can use the values of 'b' and 'c' that we found above.

  (3·0² +3·0 -2)/((0 +1)²(0 -1)) = a/(0 +1) +1/(0 -1) +1/(0 +1)²

  -2/-1 = a -1 +1   ⇒   a = 2

_____

System of equations solution

When the terms of the right-side expansion are combined, the numerator of the result is ...

  a(x +1)(x -1) +b(x +1)^2 +c(x -1) = (a+b)x² +(2b+c)x +(-a+b-c) ≡ 3x² +3x -2

Equating the coefficients gives the system of equations whose augmented matrix is:

  [tex]\left[\begin{array}{ccc|c}1&1&0&3\\0&2&1&3\\-1&1&-1&-2\end{array}\right][/tex]

Transforming this to reduced row-echelon form using any of a variety of available tools gives ...

  [tex]\left[\begin{array}{ccc|c}1&0&0&2\\0&1&0&1\\0&0&1&1\end{array}\right][/tex]

which tells you the solution is (A, B, C) = (2, 1, 1).

Answer:

It has to be an isosceles because it has 2 congruent sides (and angles) in the same relative position. The two congruent angles both measure 30 degrees each; the total of degrees in any triangle is always 180. So, the other side has an angle measure of 120 degrees, which is more than 90. Therefore, it is

an obtuse isosceles

Answer:

i dont have a premium

Step-by-step explanation:

Answer:

Step-by-step explanation:

The scale Trent used for his elementary school which has a school yard of 85

meters wide in real life and 102 millimetre wide in drawing  is 51 : 42500

Firstly, let's do the conversion of the units

1 meters = 1000 millimetres

85 meters = ?

cross multiply

length in millimetres = 85 × 1000 = 85000 millimetres

Since the schoolyard width is 85000 millimetres in real life but it's 102 millimetres wide in drawing , the scale will be as follows:

= 102 / 85000

= 51 / 42500

Therefore, the scale is as follows:

51 : 42500

learn  more: https://brainly.com/question/25328409?referrer=searchResults

A scale is chosen to plot the large graphs on smaller paper representing each unit by a supposed unit.

1.2 mm is used to represent 1 meter.

In the scale given the 102 mm is drawn to represent 85 meters.

So dividing the graph line with the original length the scale Trent has used can be found.

102/85= 1.2

This shows that 1.2 mm is used to represent 1 meter.

Multiplying 85 with 1.2 gives 102 .

https://brainly.com/question/25798457

Answer:

So first you can find a common denominator to make this equation solvable

in this case 4 is the common denominator

2/4+y=5 1/4

Now subtract to isolate the variable (y)

4 3/4 =y

Answer: 2, D

Step-by-step explanation: f=-2, so 4x-2 would give -8 add 10 would give 2 as it is for the negative

[tex](7 + 17 - 14 - 8)q = 14[/tex]

[tex](24 - 22)q = 14[/tex]

[tex]2q = 14[/tex]

Divide both sides by 2

[tex] \frac{2q}{2} = \frac{14}{2} \\ [/tex]

[tex]q = 7[/tex]

There u go...

Have a great day ❤

Answer:

160000

Step-by-step explanation:

[tex]Hiya![/tex]

Sokka is here to help!!

Here's a explanation!

[tex]-8-(-6)[/tex]

[tex]=-8-(-6)[/tex]

[tex]=-8+6[/tex]

[tex]=-2[/tex]

The answer is B.

Here's a graph photo!

Hopefully, this helps you!!

[tex]Sokka[/tex]

Answer:

5 seconds

Step-by-step explanation:

hope it helps

correct me if i'm wrong

Answer:

AB = 23.6

BC = 18.5

CD = 23.6

DA = 18.5

Step-by-step explanation:

4.

Both sets of opposite sides are parallel, so this is a parallelogram.

In parallelogram ABCD:

AB≅CD

AD≅BC

Knowing that, you can write 2 equations with the given information:

[tex]2a-20.4=a+1.6\\a-3.5=18.5[/tex]

You can solve whichever one you want, but I'll do both to be sure.

[tex]2a-20.4=a+1.6\\2a=a+1.6+20.4\\2a-a=1.6+20.4\\a=22[/tex]

Now for the second one:

[tex]a-3.5=18.5\\a=18.5+3.5\\a=22[/tex]

a = 22, so now you can solve for the side lengths.

[tex]2(22)-20.4=23.6\\22-3.5=18.5\\22+1.6=23.6\\18.5=18.5[/tex]

Linear equations can represent parallel lines, perpendicular lines and lines with no relationship at all.

The equation of the line in form of y =mx + b is [tex]\mathbf{y = 2x + 17}[/tex]

The equation is given as:

[tex]\mathbf{y =-\frac x2 - 6}[/tex]

For a linear equation y = mx + b, the slope of the equation is m

So, by comparison:

[tex]\mathbf{m =-\frac 12}[/tex]

The relationship between the slopes of perpendicular lines is:

[tex]\mathbf{m_2 =-\frac 1{m_1}}[/tex]

So, we have:

[tex]\mathbf{m_2 =-\frac 1{-1/2}}[/tex]

[tex]\mathbf{m_2 =2 }[/tex]

This means that, the slope of the line that passes through point (-8,1) is 2

The line equation is then calculated as:

[tex]\mathbf{y = m(x - x_1) + y_1}[/tex]

This gives

[tex]\mathbf{y = 2(x + 8) + 1}[/tex]

Open brackets

[tex]\mathbf{y = 2x + 16 + 1}[/tex]

Simplify

[tex]\mathbf{y = 2x + 17}[/tex]

Hence, the equation of the line in form of y =mx + b is [tex]\mathbf{y = 2x + 17}[/tex]

Read more about linear equations at:

https://brainly.com/question/11897796

Answer: The answer would be 3.91666666667

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\mathsf{4 \dfrac{2}{3} - \dfrac{3}{4}}[/tex]

[tex]\mathsf{= \dfrac{4\times3+2}{3} - \dfrac{3}{4}}[/tex]

[tex]\mathsf{= \dfrac{12 + 2}{3} - \dfrac{3}{4}}[/tex]

[tex]\mathsf{= \dfrac{14}{3} - \dfrac{3}{4}}[/tex]

[tex]\mathsf{= \dfrac{56}{11} - \dfrac{9}{12}}[/tex]

[tex]\mathsf{= \dfrac{14\times4}{12 - 0}}[/tex]

[tex]\mathsf{= \dfrac{56 - 9}{12}}[/tex]

[tex]\mathsf{= \dfrac{47}{12}}[/tex]

[tex]\mathsf{= 3 \dfrac{11}{12}}[/tex]

[tex]\huge\textsf{Therefore, your answer should be: }\huge\boxed{\\\mathsf{\dfrac{47}{12} \ or\ 3 \dfrac{11}{12}}}\huge\checkmark[/tex]

[tex]\huge\textsf{Good luck on your assignment \& enjoy your day!}[/tex]

Answer:

Step-by-step explanation:

The area A of a rectangle is given by the formula, A=lw , where l is the length and w is the width.

so

8 * 15 = 120sq. units

Answer:

15, 4, 11, 4

Step-by-step explanation:

x=8

so 7+x, 7+8= 15

12-8 = 4

x+3 , 8+3=11

32/8 = 4

Part A

The given angle is 4pi/3. Multiply it by the factor 180/pi to convert from radians to degree mode.

Note how the given angle has pi in the numerator, while the conversion factor has pi in the denominator. The two pi terms will cancel.

(4pi/3)*(180/pi)

(4/3)*(180/1)

(4*180)/(3*1)

720/3

240

The angle 4pi/3 radians is equivalent to 240 degrees. This 240 degree angle is in quadrant 3. Any angle in this quadrant is between 180 degrees and 270 degrees, excluding both endpoints. This is the bottom left quadrant, aka the southwest quadrant.

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

Part B

To find the reference angle, we'll subtract off pi. This only works for angles in quadrant 3. This is because the first pi radians, aka 180 degrees, is taken up by the first two upper quadrants. The remaining bit in the third quadrant is all we care about to find the reference angle.

reference angle = (given angle in quadrant 3) - pi

reference angle = (4pi/3) - pi

reference angle = (4pi/3) - (3pi/3)

reference angle = (4pi - 3pi)/3

reference angle = pi/3

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

Part C

Use a calculator or a reference table to find that

tan(4pi/3) = tan(pi/3) = sqrt(3)

Alternatively, you can compute the sine and cosine values first

Dividing the two items in the order mentioned will get us the tangent value

tan = sin/cos

tan(pi/3) = sin(pi/3) divide cos(pi/3)

tan(pi/3) = sqrt(3)/2 divide 1/2

tan(pi/3) = sqrt(3)

In the jump from the second to last step, to the last step, the denominators '2' cancel out when dividing.

Answer:

a) y = 0.10(16) + 100

y = $ 101.60

b) y = 0.10(150) + 100

y = $115

c) y = 0.10(200) + 100

y = $120

d) y = 0.10(250) + 100

y = $125

Step-by-step explanation:

15 days

Step-by-step explanation:

The half-life equation is given as

[tex]m(t) = m(0)e^{-0.073t}[/tex] (1)

where m(0) is the initial mass of radioactive iodine at t = 0. To solve for the time t, let's put the exponential part on the left-hand side:

[tex]e^{-0.073t} = \dfrac{m(t)}{m(0)}[/tex]

Taking the natural logarithm of both sides to get

[tex]\ln{e^{-0.073t}} = \ln\left[\dfrac{m(t)}{m(0)}\right][/tex] (2)

Recall the following properties of logarithms:

[tex]\ln{A}^b = b\ln{A}[/tex]

[tex]\ln{e} = 1[/tex]

We can rewrite Eqn(2) as

[tex]-0.073t = \ln\left[\dfrac{m(t)}{m(0)}\right][/tex]

Solving for t, we get

[tex]t = -\dfrac{1}{0.073}\ln\left[\dfrac{m(t)}{m(0)}\right][/tex]

Since m(0) = 15 gm and m(t) = 5 gm, then the amount of time that elapses to reduce the mass down to 5 gm is

[tex]t = -\dfrac{1}{0.073}\ln\left(\dfrac{5\:\text{g}}{15\:\text{g}}\right)[/tex]

[tex]\:\:\:\:\:= 15\:\text{days}[/tex]