Which expression is equivalent to i 233?
1
–1
i
–i

Answer:

I don't see any optional answers, but the denominator must be greater than 12.

Step-by-step explanation:

27,890*(12/X) < 27,890

Therefore (12/X) < 1

and X > 12

Answer:

1.25 kg

Explanation:

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

======

So, 1250 gram = (1250/1000) kg

We know

Now

Mass in kg:-

Option A

Answer: 21/572

[tex]\frac{5.5}{10 . 22} - \frac{4}{52}[/tex]

The dots stand for multiplication

Answer:

the first one is 0.5 inches 1.27 centimeters

the second is 0.25 inches or 0.635 centimeters

Step-by-step explanation:

Answer: £198.27

Step-by-step explanation:

Step-by-step explanation:

to figure this out you would have to take all the numbers and do the rise over run technique

Answer:

5/7.

Step-by-step explanation:

The rise = 5 and the corresponding run si 7.

Answer:

5/7

Step-by-step explanation:

so we start at the point (0,0). the slope is essentially

Rise/Run

so in this graph we see a point of (7,5)

the rise was 5 along the y axis and the run was 7 along the x axis.

so 5/7 is correct

Answer:

This is hard

Step-by-step explanation:

I'm so sorry I can't do it

Answer:

[tex]\frac{69}{20}[/tex]

Step-by-step explanation:

[tex]\frac{3}{7} -\frac{11}{4} +\frac{16}{5} +\frac{18}{7}[/tex]

= [tex]\frac{3}{7} +\frac{18}{7}-\frac{11}{4} +\frac{16}{5}[/tex]

= [tex]\frac{21}{7} -\frac{11}{4} +\frac{16}{5}[/tex] As both denominators have 7 you can add their numerators

=[tex]\frac{21}{7} -(-\frac{9}{20})[/tex]     L.C.M

=[tex]\frac{21}{7} +\frac{9}{20}[/tex]   L.C.M

=[tex]\frac{69}{20}[/tex]

Check the picture below.

let's notice something, the equation of line B is in slope-intercept form, thus

[tex]y = 2x + \underset{\stackrel{\uparrow }{b}}{2}\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]

so using that grid scale, we can just get two points from the line A to get its equation, let's use those in the picture below

[tex](\stackrel{x_1}{-6}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{(-6)}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{(-6)}}}\implies \cfrac{0+6}{6+6}\implies \cfrac{1}{2}[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{(-6)}) \\\\\\ y+6=\cfrac{1}{2}(x+6)\implies y+6=\cfrac{1}{2}x+3\implies y=\cfrac{1}{2}x-3[/tex]

Answer:

c and d

Step-by-step explanation:

because

- In f(x),evey time you find  X you write(-6x+5) instead

4x²-----------4(-6x+5)²

3x+1------------3(-6x+5) +1

-In g(x),evey time you find  X you write(-4x² +3x+1)  instead          

Answer:

22 is th area

Step-by-step explanation:

Answer:

1st option, and 4th option

Step-by-step explanation:

Is it select one, or select all that are correct?

Step-by-step explanation:

It is given that, the sum of two positive integers is 31 and the sum of the squares of these numbers is 625 and we are to find the smaller of the numbers.

So, let the two positive integers be x and y.

Therefore,

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 \: \: ........ \: (i) }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + y {}^{2} = 625 \: ...... \: (ii)}}} \\ \\[/tex]

Now, From the first equation we have,

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x + y = 31 }}} \\ \\ [/tex]

[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - x \: ...... \: (iii)}}} \\ \\ [/tex]

Now, substituting the value of y in equation (ii) we get :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 - x) {}^{2} = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (31 {}^{2} - 2.x.31 + x{}^{2} ) = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + (961 - 62x + x{}^{2} ) = 625}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} + 961 - 62x + x{}^{2} - 625 = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} + 336 - 62x = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2 x {}^{2} - 62x + 336 = 0}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad 2( x {}^{2} - 31 + 168) = 0}}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad \frac{2}{2} ( x {}^{2} - 31 + 168) = \frac{0}{2} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x {}^{2} - 31 + 168 = 0}}} \\ \\[/tex]

Now using the quadratic formula :

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }}} \\ \\[/tex]

Where,

[tex] \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ - (- 31) \pm \sqrt{ {31}^{2} - 4(1)(168)} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 961 -672} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm \sqrt{ 289} }{2(1)} }}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 \pm 17 }{2(1)} }}} \\ \\[/tex]

Now, we have two equations,

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} \: ... .....\: (iv)}}} \\ \\[/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2} ... .....\: (v)}}} \\ \\[/tex]

So, Equation (iv) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 + 17 }{2} }}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 48 }{2} }}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad x = 24 }}} \\ \\ [/tex]

Now, Equation (v) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad x = \frac{ 31 - 17 }{2}}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \frac{ 14 }{2}}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad x = \: 7}}}\\ \\ [/tex]

Now, we are to find the value of y.

Substituting the value of x (24) in equation (iii) :

[tex] \\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad y = 31 - 24 \:}}} \\ \\ [/tex]

[tex]{\longrightarrow \pmb{\sf {\qquad y = 7 \:}}} \\ \\ [/tex]

Again, Substituting the value of x (7) in equation (iii) :

[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 31 - x \:}}} \\ \\ [/tex]

[tex] {\longrightarrow \pmb{\sf {\qquad y = 31 - 7 \:}}} \\ \\ [/tex]

[tex]\\ {\longrightarrow \pmb{\sf {\qquad y = 24 \:}}} \\ \\ [/tex]

Therefore,

[tex] \\ [/tex]

So, The smaller of the numbers is 7 .

The sum of two positive integers is 31. If the sum of the squares of these numbers is 625, find the smaller of the numbers.

This question belongs to quadratic equations so we have to find the answer by making equation and solving it .

According to question , sum of two positive integers is 31 . So ,

According to question , sum of square

of these numbers is 625 . So ,

From equation 1 ( x + y = 31 ) , Value

of x :

Now , putting value of x in eq. 2 :

Solving it by using middle term

splitting :

First number ,

Second Number ,

According to question ,

Sum of numbers is equal to 31 :

Sum of squares of these numbers is equal to 625 :

Answer:

3*30 = 90

Step-by-step explanation:

3*30 = 90

Answer:

Step-by-step explanation:

x is the opposite side of angle 30°

34 is the adjacent side of 30°

So, find the value of x, we have use tan

[tex]\bf \boxed{ tan \ \theta = \dfrac{opposite \ side}{adjacent \ side}}}[/tex]

[tex]tan \ 30^\circ = \dfrac{x}{24} \\\\\\0.58 = \dfrac{x}{24}\\\\\\0.58*24=x[/tex]

x = 13.92

Answer:

p = 4

Step-by-step explanation:

Given equation:

[tex]x^2+(p-3)y^2-4x+6y-16=0[/tex]

Standard equation of a circle:

[tex](x-a)^2+(y-b)^2=r^2[/tex]

(where [tex](a,b)[/tex] is the centre of the circle, and [tex]r[/tex] is the radius)

If you expand this equation, you will see that the coefficient of [tex]y^2[/tex] is always one.

Therefore, [tex]p-3=1[/tex]

[tex]\implies p=1+3=4[/tex]

Additional information

To rewrite the given equation in the standard form:

[tex]\implies x^2+y^2-4x+6y-16=0[/tex]

[tex]\implies x^2-4x+y^2+6y=16[/tex]

[tex]\implies (x-2)^2-4+(y+3)^2-9=16[/tex]

[tex]\implies (x-2)^2+(y+3)^2=16+4+9[/tex]

[tex]\implies (x-2)^2+(y+3)^2=29[/tex]

So this is a circle with centre (2, -3) and radius √29

Answer:

A , B And D

2/4 divide both numbers by to and you get 1/2

3/6 divide both numbers by 3 and you get 1/2

4/8 divide both numbers by 4 and you get 1/2

Answer:

the answer is B five

Step-by-step explanation:

three hundred and seven minus three hundred and twelve

Answer:

5. radius= TD

inscribed angle =RQS

diameter =TA

6. radius=SY

secant=TU

Tangent=S

Answer:

60 inches

Step-by-step explanation:

Add 59+61 =120 which you divide by 2 for the mean of 60.

The total number of people attending the game is 1700.

Percentage in simple terms can be defined as a part per 100.

Percentage is a ratio or number which has to be defined as a fraction of 100.It is denoted by %.

If we are asked to calculate the percent of a number simply divide it by the whole part or number and then multiply with 100.

Percent of people supporting the home team = 92%

Number of people supporting the home team = 1564

Let x be the total number of people attending the game.

⇒ [tex]\frac{1564}{x}[/tex] × 100 = 92

⇒ [tex]\frac{1564}{x}[/tex] = 0.92

⇒ x = [tex]\frac{1564}{0.92}[/tex]

⇒ x = 1700

Hence, the total number of people attending the game will be 1700.

To learn more on Percentage, click:

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Answer:

The second option

Step-by-step explanation:

Hello!

The domain of a graph is the interval in which the graph is on the x values.

This graph has all the x values that are greater than or equal to -3, as the graph is to the right of it.

The answer would be the second option: [tex]D : x/x \geq -3[/tex]

The range is the y-values of the graph. The y value of this graph is above 0, so it would be y ≥ 0

Answer:

The correct polynomial is F. Calculation errors combining h terms and distributing -1.

Step-by-step explanation:

[tex]( - 4h + 3)( {h}^{2} + 9h - 1)[/tex]

[tex] - 4 {h}^{3} - 36 {h}^{2} + 4h + 3 {h}^{2} + 27h - 3[/tex]

[tex] - 4 {h}^{3} - 33 {h}^{2} + 31h - 3[/tex]

The ways that the policies of President Kennedy and President Johnson are similar and different is given below:

After the assassination of Kennedy, Johnson was known to have followed the footsteps of his predecessor's and went ahead on implementing some of his programs.  Johnson was known to have went ahead with the the anti-poverty program and the Civil Rights Bill.

The differences between the both presidents were:

Conclusively, these two president were said to be very were similar, and yet they were different.

Learn more about  President Kennedy from

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Answer:

well over 3 years and almost over 4 months

Answer:

Area of cube = 6l²

= 6×2²

= 24 ft²

2 feet.

how many square feet of wood he used to build the box.

each side measures 2ft.

A=L²=(2 ft)²=4 ft²

A cube has 6 sides so the total quantity of wood is:

Wood=6×A

=6×4 ft²

=24 ft²

he used 24 ft² of wood to build the box.

The system of equations that describes the situation is given by these two following equations:

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are c and x.

Carlos has 10 more friends than Xavier when they start the competition, hence:

c = 10 + x.

Carlos doubles his amount of friends and Xavier triples his. In the end they have a total of 160 friends, hence:

2c + 3x = 160.

More can be learned about a system of equations at https://brainly.com/question/24342899

Answer:

Hi! I'm not too bad at math so I hope I can help you. Except, I can't seem to find your question so please comment your question(s) to me so that I can help you. Have a great day:)