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anyanavicka [17]
3 years ago
12

What does it mean to square a negative number​

Mathematics
2 answers:
igor_vitrenko [27]3 years ago
8 0
Squaring" means to multiply a number by itself. Squaring a positive number gets a positive result: (+5) × (+5) = +25. Squaring a negative number also gets a positive result: (−5) × (−5) = +25.
Andrej [43]3 years ago
5 0

Answer:

This is because to square a number just means to multiply it by itself. For example, (-2) squared is (-2)(-2) = 4. Note that this is positive because when you multiply two negative numbers you get a positive result.

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Identify all of the problems that would result in a product of 2 5 ?
mars1129 [50]
1 multiplied by 25 and 5 multiplied by 5
8 0
3 years ago
The radio of boys to girls in Miss ronilo's math class is 3 to 1 what percent of the classes is girls
9966 [12]

Imagine that the class only had 4 students (unrealistic most likely, but small numbers help much better I think)

If we had 4 students and 3 were boys, then 4-3 = 1 girl is in the class. This makes the ratio of boys to girls be 3 to 1. In other words, there are 3 times as many boys compared to girls.

Divide the number of girls (1) over the number of students total (4) to get 1/4 = 0.25 = 25%

<h3>Answer: 25%</h3>
5 0
3 years ago
Read 2 more answers
Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value
vodomira [7]

Answer:

\rm\displaystyle \displaystyle \displaystyle θ=    {60}^{ \circ} , {300}^{ \circ}

\rm \displaystyle a =    - \frac{   \sqrt{3} }{2}    - 1, \frac{\sqrt{3}}{2}  - 1

Step-by-step explanation:

we are given two <u>coincident</u><u> points</u>

\displaystyle  P( \sin(θ)+2,  \tan(θ)-2)   \: \text{and } \\  \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)

since they are coincident points

\rm \displaystyle  P( \sin(θ)+2,  \tan(θ)-2)    = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)

By order pair we obtain:

\begin{cases}  \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) =  \sin( \theta)   + 2 \\   \\  \displaystyle 3 \sin( \theta)  - 2  \cos( \theta)  + a =  \tan( \theta)  - 2\end{cases}

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using <u>substitution</u><u> method</u>

to do so, make a the subject of the equation.therefore from the second equation we acquire:

\begin{cases}  \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )=  \sin( \theta)   + 2 \\   \\  \boxed{\displaystyle  a =  \tan( \theta)  - 2 - 3 \sin( \theta)   +  2  \cos( \theta) } \end{cases}

now substitute:

\rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta)  - 2 - 3 \sin( \theta)   +  2  \cos( \theta)   \}=  \sin( \theta)   + 2

distribute:

\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)  - 6 \sin( \theta) \cos( \theta)    + 4  \cos ^{2} ( \theta)   =  \sin( \theta)   + 2

collect like terms:

\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)     + 4  \cos ^{2} ( \theta)   =  \sin( \theta)   + 2

rearrange:

\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta)  - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + =  \sin( \theta)   + 2

by <em>Pythagorean</em><em> theorem</em> we obtain:

\rm\displaystyle \displaystyle 4  - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)  =  \sin( \theta)   + 2

cancel 4 from both sides:

\rm\displaystyle \displaystyle   - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta)  =  \sin( \theta)    - 2

move right hand side expression to left hand side and change its sign:

\rm\displaystyle \displaystyle   - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2  =  0

factor out sin:

\rm\displaystyle \displaystyle  \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2  =  0

factor out 2:

\rm\displaystyle \displaystyle  \sin (θ) (- 2 \cos(θ)+1)  + 2(- 2\cos( \theta) + 1 ) =  0

group:

\rm\displaystyle \displaystyle ( \sin (θ)   + 2)(- 2 \cos(θ)+1)  =  0

factor out -1:

\rm\displaystyle \displaystyle -  ( \sin (θ)   + 2)(2 \cos(θ) - 1)  =  0

divide both sides by -1:

\rm\displaystyle \displaystyle   ( \sin (θ)   + 2)(2 \cos(θ) - 1)  =  0

by <em>Zero</em><em> product</em><em> </em><em>property</em> we acquire:

\begin{cases}\rm\displaystyle \displaystyle   \sin (θ)   + 2 = 0 \\ \displaystyle2 \cos(θ) - 1=  0 \end{cases}

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

\begin{cases}\rm\displaystyle \displaystyle   \sin (θ)     \neq  - 2 \\ \displaystyle2 \cos(θ) =  1\end{cases}

divide both sides by 2:

\rm\displaystyle \displaystyle \displaystyle \cos(θ) =   \frac{1}{2}

by unit circle we get:

\rm\displaystyle \displaystyle \displaystyle θ=    {60}^{ \circ} , {300}^{ \circ}

so when θ is 60° a is:

\rm \displaystyle a =  \tan(  {60}^{ \circ} )  - 2 - 3 \sin(  {60}^{ \circ} )   +  2  \cos(  {60}^{ \circ} )

recall unit circle:

\rm \displaystyle a =   \sqrt{3}  - 2 -  \frac{ 3\sqrt{3} }{2}   +  2   \cdot  \frac{1}{2}

simplify which yields:

\rm \displaystyle a =    - \frac{   \sqrt{3} }{2}    - 1

when θ is 300°

\rm \displaystyle a =  \tan(  {300}^{ \circ} )  - 2 - 3 \sin(  {300}^{ \circ} )   +  2  \cos(  {300}^{ \circ} )

remember unit circle:

\rm \displaystyle a =  -  \sqrt{3}   - 2  +   \frac{3\sqrt{ 3} }{2}  +  2   \cdot  \frac{1}{2}

simplify which yields:

\rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1

and we are done!

disclaimer: also refer the attachment I did it first before answering the question

5 0
3 years ago
Find the lengths of the segments with variable expressions
nlexa [21]

Answer:

EF = 7 units

AD =  3 units

BC =  11 units

Step-by-step explanation:

Since it is an isosceles trapezoid and the midsegment is x, the sum of the bases divided by 1/2 is the value of the midsegment, x.

x = 1/2(x-4 + 2x-3)

x = 1/2x - 2 + x - 1 1/2

x = 1 1/2 x - 3.5

-1/2 x = -3.5

x = 7 units

EF = 7 units

AD = 7-4 = 3 units

BC = 2(7) - 3 = 11 units

5 0
3 years ago
The daily rainfall experienced by Soaking City has been recorded on 9 randomly selected days throughout the year. This process h
belka [17]

Answer:

0.195

Step-by-step explanation:

Given:

For ; sample size, n = 9 ;

Standard Error (S. E) = 0.39

S. E = σ / sqrt(n)

σ = standard deviation

0.39 = σ / sqrt(9)

σ = 0.39 * 3

σ = 1.17

Therefore, For n = 36

S. E = 1.17 / sqrt(36)

S.E = 1.17 / 6

S.E = 0.195

3 0
3 years ago
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