We have the equation
[tex]y=1.37x+75[/tex]Where y = time in minutes and x = number of miles
Then, for every mile, this is x = 1, we have:
[tex]y=1.37(1)+75=1.37+75=76.37[/tex]Answer: the driving time increase in 76.37 minutes
What is the ordered pair for A' if A(2,-3) is reflected over the line y = x?
We are asked to determine the coordinates of a point that is reflected over the line y = x. Let's remember that when we reflect a point (x,y) over y = x, we get the following result:
[tex]R_{y=x}(x,y)=(y,x)[/tex]Therefore, when we reflect the point (2,-3) we get:
[tex]R_{y=x}(2,-3)=(-3,2)[/tex]The expression 3(-2m + 5) can be written in the form am + b. What is the value of a + b ?
We will solve as follows:
[tex]3(-2m+5)=-6m+15[/tex]So, the values of a & b are -6 & 15 respectively. So a + b = 9.
Find an equation of the line tangent to the curve y=1/2(ln(sin^2(x))) at the point (pi/4 , -1/2ln2)
Given:
Equation of Curve is:
[tex]y=\frac{1}{2}\ln (\sin ^2x)[/tex]To find: Equation of tangent to the curve at point
[tex](\frac{\pi}{4},\frac{-1\ln 2}{2})[/tex]Equation of tangent to the curve is given by:
[tex]y-y_1=m(x-x_1)[/tex]where, m is slope of line.
Now, m is derivative of y with respect to x at given point.
Hence,
[tex]\begin{gathered} m=\frac{1}{2}\text{ }\times\frac{2\sin x\cos x}{\sin^2x} \\ m=\frac{\cos \text{ x}}{\sin \text{ x}} \\ m=\cot \text{ x} \end{gathered}[/tex]At given point, the slope m is:
[tex]\begin{gathered} m=\cot (\frac{\pi}{4}) \\ m=1 \end{gathered}[/tex]Therefore,the equation of tangent to the curve is given as:
[tex]\begin{gathered} y-y_1=1(x-x_1) \\ y-(\frac{-1(\ln \text{ 2))}}{2})=x-\frac{\pi}{4} \\ \frac{2y+\ln \text{ 2}}{2}=\frac{4x-\pi}{4} \\ 2y+\ln 2=\frac{4x-\pi}{2} \end{gathered}[/tex][tex]\begin{gathered} 4y+2\ln 2=4x-\pi \\ 4y=4x-\pi-2\ln 2 \end{gathered}[/tex]Thus the required equation of tangent to the curve is
[tex]4y=4x-\pi-2\ln 2[/tex]what's the domain of [tex]f(x) = {7}^{x} [/tex]
The function is given to be:
[tex]f(x)=7^x[/tex]The domain of a function is the set of input or argument values for which the function is defined.
The function has no undefined points or domain constraints.
Therefore, the domain is given to be:
[tex]\begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:Fine the distance d(A,B)between points A and B A(4,-3); B(4,5)
The distance between points A and B can be calculated with this formula:
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex][tex]\begin{gathered} A=(4,-3) \\ x_1=4 \\ y_1=-3 \\ B=(4,5) \\ x_2=4 \\ y_2=5 \end{gathered}[/tex]Now inputting the values above into the formula, we will proceed thus:
[tex]\begin{gathered} d(A,B)=\sqrt[]{(4-4)^2+(5-(-3))^2} \\ d(A,B)=\sqrt[]{0^2+8^2} \\ d(A,B)=\sqrt[]{64} \\ d(A,B)=8 \\ \text{The distance betw}een\text{ points A and B is 8 units.} \end{gathered}[/tex]1. Write down the quadratic formula.
The quadratic equation formula is shown below;
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=coefficientofx^2 \\ b=\text{coefficient of x} \\ c=\text{constant} \end{gathered}[/tex]3. Which of the following binomials is a factor of x³+4x²+x-6?
O x-2
O x + 2
O x-4
O x + 4
Answer:
It is x+2
Step-by-step explanation:
For the first option, x-2, use synthetic division to solve. Since it has a remainder (20/x-2), this is not the correct option. For the second solution, x+2, we can also use synthetic division. When you write it out, you can see that there is no remainder. Therefore, your answer is x+2.
2. Emily spent 4 1/2 hours cleaning her room. Aubrey spent 2 3/5 hours cleaning her room. a. How much longer did Emily spend cleaning her room?
Emily spent 1 9/10 longer cleaning her room
Explanation:The difference between the number of hour Emily and Aubrey spent cleasing is how much longer Emily spent.
4 1/2 = 9/2, and
2 3/5 = 13/5
So
4 1/2 - 2 3/5 = 9/2 - 13/5
= (45 - 26)/10
= 19/10
= 1.9
OR
1 9/10
Pls explain this goof I have a final for it number 4
The value of x is 10/7
Explanation:Given:
4) Triangle ABC is similar to triangle DEF
side AB = 5, side BC = x
side EF = x + 2. side DE = 12
To find:
to draw the triangles, write the proportion and solve for x
First, we need to draw the triangles using the given information:
In similar triangles, the ratios of the corresponding sides are equal
For Triangle ABC similar to triangle DEF
AB corresponds to DE
BC corresponds to EF
AC corresponds to DF
The ratio of the corresponding sides will give the proportion:
[tex]\begin{gathered} \frac{AB}{DE}\text{ = }\frac{BC}{EF}\text{ = }\frac{AC}{DF} \\ \\ Since\text{ we don't have values for AC abd DF, we will use the first two ratios:} \\ \frac{AB}{DE}\text{ = }\frac{BC}{EF}\text{ } \end{gathered}[/tex][tex]\begin{gathered} The\text{ proportion:} \\ \frac{5}{12}\text{ = }\frac{x}{x+2} \end{gathered}[/tex]Finally, we will solve for x:
[tex]\begin{gathered} \frac{5}{12}\text{ = }\frac{x}{x+2} \\ cross\text{ multiply:} \\ 5(x\text{ + 2\rparen = x\lparen12\rparen} \\ 5x\text{ + 10 = 12x} \\ 10\text{ = 12x - 5x} \\ 10\text{ = 7x} \\ \\ divide\text{ both sides by 7:} \\ x\text{ = 10/7} \end{gathered}[/tex]For Exercises 35–48, identify p, q, and r if necessary. Then translate each argument to symbols and use a truth table to decide if the argument is valid or invalid.If it snows, I can go snowboarding.It did not snow.∴ I cannot go snowboarding.
SOLUTION
Step 1 :
In this question, we are meant to translate each argument to symbols and use a truth table to decide if the argument is valid or invalid.
Step 2 :
Let p represents: If it snows
Let q represents: I can go snowboarding
Then
[tex]\begin{gathered} \approx\text{ p represents : It did not show} \\ \approx\text{ q represents : I cannot go snowboarding} \end{gathered}[/tex]Hence, the argument is VALID
Answer:
Step-by-step explanation:
For the figure below, give the following. (a) one pair of vertical angles (b) one pair of angles that form a linear pair (c) one pair of angles that are supplementary 3/4
Answer:
(a)∠1 and ∠6.
(b)∠5 and ∠6.
(c)∠3 and ∠4.
Explanation:
(a)Vertical Angles are angles that form an X-Shape.
A pair of vertical angles is ∠1 and ∠6.
(b)Linear pairs are two angles on a straight line that adds up to 180 degrees.
A linear pair is ∠5 and ∠6.
(c)A pair of supplementary angles are two angles that add up to 180 degrees.
An example is ∠3 and ∠4.
If Jimmy's age is one year less than the sum of his ages of his siblings serena and tyler. which equation represents Jimmy's age?
Jimmy's age = (serena age + tyler age) - 1
Guys help
I don’t know what 2+2 is
Answer:
It's 4
Step-by-step explanation:
2+2 =
1+1+1+1 =
0.5 + 0.5 + 0.5 + 0.5 + 0.5+ 0.5 + 0.5 + 0.5
and so on.
It's a really complicated proof, so you're gonna have to take my word for it. Sorry about that.
Okay I'm gonna help you out. Let's make things easier.
Let A = {a,b}
and B = {x,y}
Assuming that a ≠ b ≠ x ≠ y
If I see their cardinality, (note that there only 2 elements in either of the sets which is finite) then,
|A| = 2......(1)
|B| = 2........(2)
From 1. and 2,
|A| = |B|
By the definition of addition,
|A| + |B| = 2 + 2 = 4
Hope you understood the underlying concept, i.e., if you take 2 2s and put them together they'll always result in 4.
questions 1-7 calculate the scale factor from DEF to ABC
1. The scale factor is 3
2. The scale factor is 6
3. The scale factor is 4
4. The scale factor is 5
5. The scale factor is 23
Given,
Triangle DEF and Triangle ABC.
We have to find the scale factor from DEF to ABC.
Scale factor;
The ratio of the scale of an original thing to a new object that is a representation of it but of a different size is known as a scale factor (bigger or smaller). For instance, we can increase the size of a rectangle with sides of 2 cm and 4 cm by multiplying each side by, let's say, 2.
Here,
1. DEF and ABC
DE = 4, AB = 12, 12/4 = 3
DF = 5, AC = 15, 15/5 = 3
FE = 3, BC = 9, 9/3 = 3
Scale factor = 3
2. DEF and ABC
DE = 5, AB = 30, 30/5 = 6
DF = 6, AC = 36, 36/6 = 6
FE = 8, BC = 48, 48/8 = 6
Scale factor = 6
3. DEF and ABC
DF = 9, AB = 36, 36/9 = 4
EF = 10, BC = 40, 40/10 = 4
DF = 11, AC = 44, 44/11 = 4
Scale factor = 4
4. DEF and ABC
DE = 75, AB = 15, 75/15 = 5
EF = 80, BC = 16, 80/16 = 5
DF = 60, AC = 12, 60/12 = 5
Scale factor = 5
5. DEF and ABC
DE = 46, AB = 2, 46/2 = 23
EF = 92, BC = 4, 92/4 = 23
DF = 69, AC = 3, 69/3 = 23
Scale factor = 23
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7) find a polynomial with Real coefficients of degree 4, and with zeros:3, with multiplicity two; i with multiplicity one; containing the point (0, - 9).
The given polynomial is said to have a degree of 4. It has zeros 3 (multiplicity two) and i (multiplicity one) and contains the point (0,-9).
It is required to find the equation of the polynomial.
Recall the Complex Conjugate Theorem: A polynomial function or a polynomial equation with real coefficients that has a+bi as a complex zero, with b not zero, also has a-bi as a zero.
Since i is a zero of the polynomial, its conjugate -i is also a zero.
Hence, the zeros of the polynomial are 3,i,-i.
Recall from the factor theorem that if k is a zero of a polynomial with multiplicity m, then the expression,
[tex](x-k)^m[/tex]is a factor of the polynomial.
It follows that the polynomial has factors:
[tex](x-3)^2,(x-i),(x+i)[/tex]Hence, the polynomial can be written as a product of its factors and a real constant, a:
[tex]y=a(x-3)^2(x-i)(x+i)[/tex]Simplify the product on the right as follows:
[tex]\begin{gathered} \text{ Use binomial expansion and difference of two squares:} \\ y=a(x^2-6x+9)(x^2-i^2) \\ \Rightarrow y=a(x^2-6x+9)(x^2-(-1)) \\ \Rightarrow y=a(x^2-6x+9)(x^2+1) \end{gathered}[/tex]Expand the expression further:
[tex]\begin{gathered} \Rightarrow y=a(x^4+x^2-6x^3-6x+9x^2+9) \\ \Rightarrow y=a(x^4-6x^3+x^2+9x^2-6x+9) \\ \Rightarrow y=a(x^4-6x^3+10x^2-6x+9) \end{gathered}[/tex]Since it is given that the polynomial contains the point (0,-9), substitute (x,y)=(0,-9) into the equation to find the value of a:
[tex]\begin{gathered} -9=a(0^4-6(0^3)+10(0^2)-6(0)+9) \\ \Rightarrow-9=a(9) \\ \Rightarrow9a=-9 \\ \Rightarrow\frac{9a}{9}=-\frac{9}{9} \\ \Rightarrow a=-1 \end{gathered}[/tex]Substitute a=-1 back into the equation.
Hence, the required polynomial is:
[tex]\begin{gathered} y=-1(x^4-6x^3+10x^2-6x+9) \\ \Rightarrow y=-x^4+6x^3-10x^2+6x-9 \end{gathered}[/tex]find how much money there will be in the account after the given number of years and find the interest earned round to the nearest hundred as needed for both
A principal $7000, rate 5%=0.05, and time 2 years is given.
It is stated that it is compounded semiannually, that is, twice in a year.
The question requires that you calculate the amount in the account after 2 years and the interest earned.
The formula for the amount (compound interest) is given as:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Where,
• A is the final amount.
,• P is the principal
,• r is the rate
,• n is the number of times interest is compounded annually.
,• t is the time in years.
In this case P=7000, r=0.05, n=2, t=2. Substitute these values into the formula:
[tex]\begin{gathered} A=7000(1+\frac{0.05}{2})^{2(2)}=7000(1+0.025)^4 \\ =7000(1.025)^4\approx7000(1.103813) \\ \approx\$7,726.69 \end{gathered}[/tex]The equation that relates the amount, A, principal, P, and interest earned, I is given as:
[tex]A=P+I[/tex]Substitute A=7,726.69, P=7000 into the formula:
[tex]\begin{gathered} 7,726.69=7,000+I \\ \Rightarrow I=7,726.69-7,000=\$726.69 \end{gathered}[/tex]It follows that:
The amount of money in the account after 2 years is about $7,726.69.
The amount of interest earned is about $726.69.
The number of bacteria in a certain food is a function of the food’s temperature. The number is N1(T) at a temperature T degrees Celsius, described by the equation N1(T) = 15T2 + 60T + 300. Similarly, the number of bacteria in another food is given by the function, N2(T) = 20T2 – 5T + 150. Determine an equation that describes the number of bacteria in both the foods when they are mixed.30T2 + 55T + 45035T2 + 65T + 45035T2 + 55T + 450- 5T2 - 65T + 150
If we mix the two foods together, then we need to add the two equations to get the total number of bacteria. We are given the two equations:
[tex]\begin{gathered} N_1(T)=15T^2+60T+300 \\ N_2(T)=20T^2-5T+150 \end{gathered}[/tex]Thus, we add:
[tex]N_3(T)=N_1(T)+N_2(T)=(15T^2+60T+300)+(20T^2-5T+150)=35T^2+55T+450[/tex]Thus, the correct answer is the third option:
[tex]35T^2+55T+450[/tex]Consider the following function.f(x) = 6 − |x − 8|
Soution:
Given the function;
[tex]f(x)=6-|x-8|[/tex](a) Critcal points are points where the function is defined and the derivative is zero or undefined.
Absolute value functions are piecewise function;
Hence, the critical number of f is;
[tex]x=8[/tex]Find the equation of the line that passes through points A and B.
A (2,5)
B (-1,-1)
⇒The first step is to find the slope/gradient between the points using the given formula below:
[tex]m=\frac{y_{2} -y_{1} }{x_{2}-x_{1} }\\ m=\frac{-1-5}{-1-2} \\m=\frac{-6}{-3} \\m=2[/tex]
⇒Now that I have found the gradient keep in mind that the general equation of finding the equation of a straight line is y=mx+c
⇒To find the value of c I will use the point A Keep in mind you can use any point of your choice you will still finds the correct value of the same as any way
[tex]5=2(2)+c\\5=4+c\\5-4=c\\c=1[/tex]
⇒This just simply means that the equation of the line of the points A and B is:
[tex]y=2x+1[/tex]
GOODLUCK!!
The number of members in the band was 82 in 2000 and has decreased by 8% each year. Which model shows the band's membership in terms of t, the number of years since 2000?
Given:
The number of members in the band was 82 in 2000.
The number of members decreased by 8% each year.
Required:
We need to find the model for the given situation.
Explanation:
Let t be the number of years since 2000.
Let y be the number of members in t number of years since 2000.
The initial value is 82.
The number of members is decreasing.
Consider the exponential decay equation.
[tex]y=82(1-\frac{8}{100})^t[/tex][tex]y=82(1-0.08)^t[/tex][tex]y=82(0.92)^t[/tex]Final answer:
[tex]y=82(0.92)^t[/tex]It cost $110 to plant watermelons on an acre of farmland. How much would it cost to only plant 2/3 of the acre?
No solve this you have to multiply the cost per acre by the total are you want to plant:
[tex]110\cdot\frac{2}{3}=73.3[/tex]It'll cost $73.3 to plant 2/3 of the acre.
Solve each system of equations by graphing Y=x+4Y=-2x-2
Solution:
If we graph the following lines, on the same plane
Y= x+4
Y= - 2x-2
we obtain:
According to the above image, we can see that both lines intersect at the point (x, y) = (-2, 2)
So that, we can conclude that the solution of the given system of linear equation is:
(x, y) = (-2,2)
Help! Also even though it looks blurry, just zoom in to see the full question
The matrix transformation of triangle XYZ with coordinates
[tex]\begin{gathered} X=(-2,-2) \\ Y=(4,1) \\ Z=(0,6) \end{gathered}[/tex]When dilated by a factor of 2.5 now becomes;
[tex]\begin{gathered} 2.5\begin{bmatrix}{-2} & 4{} & 0{} \\ {-2} & {1} & {6}{}\end{bmatrix}=\begin{bmatrix}{2.5(-2)} & {2.5(4)} & {2.5(0)} \\ {2.5(-2)} & {2.5(1)} & {2.5(6)} \\ & {} & {}\end{bmatrix} \\ =\begin{bmatrix}{-5} & {10} & {0} \\ {-5} & {2.5} & {15} \\ {} & {} & {}\end{bmatrix} \end{gathered}[/tex]ANSWER:
The correct vertices of the image now becomes;
[tex]\begin{gathered} X^{\prime}=(-5,-5) \\ Y^{\prime}=(10,2.5) \\ Z^{\prime}=(0,15) \end{gathered}[/tex]Martha's homemade spice mix in different size at the craft fair the graph shows a proportional relationship between teaspoons and cumin and teaspoons of chili powder in one recipe.what does the what are the coordinates of the point at x equals 1 and what do they represent? explain why the graph shows are a proportional why is the graph a solid line?
At x = 1, the coordinates are (1, 4)
They represent the proportionality of 1 teaspoon of cumin and 4 teaspoons of chili powder.
The graph is proportional because as x increases, y al
Can you please help with 6 I need to finish 8 more packets by tonight
Given
[tex]\begin{gathered} initial\text{ = \$300} \\ Final=\text{ \$575} \end{gathered}[/tex]Solution
Percentage growth is
[tex]=\frac{final-initial}{intial}\times100[/tex][tex]\begin{gathered} \frac{575-300}{300}\times100 \\ \\ 0.91666\times100 \\ =91.666 \\ \end{gathered}[/tex]The percentage growth is
[tex]91.666\text{ \%}[/tex]The final answer[tex]91.666\text{ \%}[/tex]I need help with this study guide so I can use it to study for a test.
1. Colinear points are the ones that are in the same line.
In this example, K is colinear to J and L (referring to line a).
It can be N and M, referring to line c.
2. As the line has the points H and K included, it can be named HK.
3. The intersection of a line and a plane, assuming that the line is not included in the plane, is a point. In this case, the intersection of line c and plane R is point K.
4. A point is coplanar to a plane or another point when it belongs to the plane.
In this case the point M is not coplanar to the plane R (it is outside of the plane).
5. Each face of the prism is a plane and it is can be defined by 3 points. So in this case we have 5 planes (four faces and one base, that is W).
6. We can assign a name for a plane with 3 points that belong to that plane. A possible name is ABC.
7. The intersection of plane W and ADE is the line that goes through the points A and D. We can call this line with the name AD.
8. E, C or D are no colinear to A and B. They do not belong to the line that goes through the points A and B (or the line AB).
9. DF is the sum of DE and EF, so we have:
[tex]\begin{gathered} \bar{DF}=\bar{DE}+\bar{EF} \\ \bar{DF}=(7x+1)+(4x-3)=42 \\ 11x-2=42 \\ 11x=42+2=44 \\ x=\frac{44}{11}=4 \\ \\ \bar{DE}=7x+1=7\cdot4+1=28+1=29 \end{gathered}[/tex]DE = 29
10. We know that:
[tex]\begin{gathered} JK+KL=JL \\ (5x-8)+(7x-12)=10x-2 \\ 12x-20=10x-2 \\ 12x-10x=-2+20 \\ 2x=18 \\ x=\frac{18}{2} \\ x=9 \\ \\ KL=7x-12=7\cdot9-12=63-12=51 \end{gathered}[/tex]11. If S is the midpoint of RT, then RS=ST.
We would have:
[tex]\begin{gathered} RS=ST \\ 5x+17=8x-31 \\ 5x-8x=-31-17 \\ -3x=-48 \\ x=\frac{-48}{-3}=16 \\ \\ ST=8x-31=8(16)-31=128-31=97 \end{gathered}[/tex]12. If y bisects AC, then AB=BC. Then we have:
[tex]\begin{gathered} AB=BC \\ 4-5x=2x+25 \\ -5x-2x=25-4 \\ -7x=21 \\ x=\frac{21}{-7}=-3 \\ \\ AC=AB+BC \\ AC=(4-5x)+(2x+25)=(4-5\cdot(-3))+(2\cdot(-3)+25) \\ AC=(4+15)+(-6+25)=19+19 \\ AC=38 \end{gathered}[/tex]13. AB is half the value of AC, so we have:
[tex]\begin{gathered} 2\cdot AB=AC \\ 2(3x+4)=11x-17 \\ 6x+8=11x-17 \\ 6x-11x=-17-8 \\ -5x=-25 \\ x=5 \end{gathered}[/tex]Then, we can calculate CD = AC
[tex]CD=AC=11x-17=11(5)-17=55-17=38[/tex]We can define DE as DE = CE - CD. We can calculate it as:
[tex]DE=CE-CD=49-38=11[/tex]DE = 11
PLEASE HELP ME SOLVE
Seacausus stadium has a seating capacity for 25,000 spectators. The stadium has 25 exits and can be vacated in 20 minutes. The time taken to exit the stadium varies directly with the number of spectators and inversely with the number of exits. Determine the time taken for 21,000 spectators to vacate the stadium, if only 15 exits are functional.
The time taken for 21000 spectators to vacate the stadium , if only 15 exits are functional is 28 minutes .
In the question ,
it is given that
the time taken to vacate the stadium = 20 minutes
number of exits = 25 exits
capacity of the stadium = 25000 spectators .
given that ,
time taken to exit the stadium varies directly with number of spectators and inversely with the number of exits .
time taken ∝ number of spectators ∝ 1/number of exits .
to remove the proportionality sign , we write the constant
time taken = k * (number of spectators)/(number of exits) .
20 = k * 25000/25
20 = k * 1000
k = 20/1000
k = 2/100
k = 1/50 = 0.02
So, to find the time taken for 21,000 spectators to vacate the stadium, if only 15 exits are functional , we use the formula
time taken = (0.02)*(21000/15)
= 0.02*1400
= 28
Therefore , The time taken for 21000 spectators to vacate the stadium , if only 15 exits are functional is 28 minutes .
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Find the sum of 6x+7 and 8x^2 +5x-1
Please hurry I need this asap
Answer:
11x+8x^(2)+6
Step-by-step explanation:
Answer:
8x² + 11x + 6
Step-by-step explanation:
Expression addition:Combine like terms. Like terms have same variable with same exponent.
5x and 6x are like terms. Add the coefficient of the like terms.
(-1) & 7 are constants and add (-1) & 7.
8x² + 5x - 1 + 6x + 7 = 8x² + 5x + 6x - 1 + 7
= 8x² + 11x + 6
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You have the following operation:
6x9
To solve the previous operation you consider that the expression means 6 times 9, that is, that is the sum of six 9s:
6x9 = 9 + 9 + 9 + 9 + 9 + 9 = 54
Hence, the simplied operation is 6x9 = 54
Write answer in standard form 16-(2-3i)-i
The standard form of [tex]16-(2-3i)-i[/tex] is [tex]14+2i[/tex].
The given expression is [tex]16-(2-3i)-i[/tex].
We have to write the answer is standard form.
In the question to write the answer in standard form is that we have to write the answer of this expression after solving in simplified form.
There is an variable in expression is [tex]i[/tex].
The [tex]i[/tex] shows that there is an imaginary number and the given expression is complex and complex is a type of number in the form [tex]a+ib[/tex] where [tex]a[/tex] and [tex]b[/tex] is a real number.
To solve this expression we first solve the bracket. When we multiply the variable under the bracket with minus sign then their sign will change.
The expression is [tex]16-(2-3i)-i[/tex]
Now simplifying the bracket
[tex]=16-2+3i-i[/tex]
Now we solve real number with real number and imaginary number with imaginary.
[tex]=(16-2)+i(3-1)\\=14+2i[/tex]
Hence, the standard form of [tex]16-(2-3i)-i[/tex] is [tex]14+2i[/tex].
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