Kaj invests money in an account paying a simple interest of 1.2% per year. If she invests $90 and no money will be added or removed from the investment, how much will she have in one year, in dollars and cents?

Answers

Answer 1

Answer:

$91.08

Step-by-step explanation:

Interest = $90 × .012 = $1.08

Total amount = $90 + $1.08 = $91.08


Related Questions

Solve the following inequality.xe^x ≥7Choose one:1. x ≤ 1.522. no solution3. x ≤ 1.954. x ≥ 1.955. x ≥ 1.52

Answers

1) Considering e =2.72

Then let's plug it in the inequality, and calculate the natural logarithm.

[tex]\begin{gathered} xe^x\ge7 \\ x2.72^x\ge7 \\ 2.72^x\ge\frac{7}{x}^{} \\ \ln 2.72^x\ge\ln (\frac{7}{x}) \\ x\text{ }\ge1.52 \end{gathered}[/tex]

2) Then option 5 is the answer

X≥ 1.52

separate 84 into two parts such that one part will be 12 less than twice the other

Answers

Answer:

32, 52

Step-by-step explanation:

84=x+y

x=2y-12

Solving the system of equations

Plug second equation into first

84=(2y-12)+y

84=3y-12

y=32

Plug y into first equation

84=x+32

x=52

Harley rides her bicycle the same distance every day for 4 days. The total distance she rides is 814 miles. How many miles does she ride each day?

Answers

Answer:

203.5

Step-by-step explanation:

just divide 814 by 4

Answer:

203.5 miles per day

Step-by-step explanation:

divide 814 by 4= 203.5

Sergio wants to hand a 40 1/2 inch by 28 2/3 inch canvas painting on a wall in his room. The width of the wall is 15 2/3 feet. The height of the wall is 3/4 the width of the wall. a. What is the area of the wall? b. How much of the wall will be uncovered after Sergio hangs the painting?

Answers

The most appropriate choice for area of rectangle will be given by-

a) Area of the wall =  [tex]26508[/tex] [tex]in^2[/tex]

b) Area of wall left uncovered = [tex]25347[/tex] [tex]in^2[/tex]

What is area of rectangle?

Area of rectangle is the total space taken by the rectangle. If l is the length of the rectangle and b is the breadth of the rectangle, then area of the rectangle is given by

Area = [tex]l \times b[/tex]

Here,

Width of the wall = [tex]15\frac {2}{3}[/tex] feet = [tex]\frac{47}{3}[/tex] feet

Height of the wall = [tex]\frac{3}{4} \times \frac{47}{3}[/tex] feet

                               = [tex]\frac{47}{4}[/tex] feet

a) Area of the wall = [tex]\frac{47}{3} \times \frac{47}{4}[/tex] [tex]ft^2[/tex]

                               =[tex]\frac{2209}{12} \times 12 \times 12[/tex] [tex]in^2[/tex]

                               = [tex]26508[/tex] [tex]in^2[/tex]

b) Length of painting = [tex]40 \frac{1}{2}[/tex]  in = [tex]\frac{81}{2}[/tex] in

Breadth of painting = [tex]28\frac{2}{3}[/tex] in = [tex]\frac{86}{3}[/tex] in

Area of painting = [tex]\frac{81}{2} \times \frac{86}{3}[/tex] [tex]in^2[/tex]

                           = [tex]1161[/tex] [tex]in^2[/tex]

Area of wall left uncovered = (26508 - 1161) [tex]in^2[/tex]

                                             = 25347 [tex]in^2[/tex]

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x square root 7x + 10
factor the equation

algerbra 2

Answers

Answer:

Step-by-step explanation:

x

=

2

x

=

5

Explanation:

x

2

+

7

x

+

10

=

0

We could right the equation in another form using brackets.

(

x

+

2

)

(

x

+

5

)

=

0

In order to get two numbers to multiply and give 0, at least one number has to be 0.

a stands for the first bracket and b for the second.

a x b = 0

a=0 or b=0 to make this equation valid.

Therefore x= -2 or x= -5 because -2+2=0 and -5+5=0 according to both brackets.

Four friends want to share a package of cookies with m number of cookies in it which expression represents the numbers of cookies each friends will get

Answers

Based on the case, we know that four friends want to share their cookies package. The number of the cookies is represent by 'm'. So, how many part for each friends will get?

We can solve the case by 'fraction'. Fractions represent the parts of a whole objects. A fraction is arranged by two parts. First, numerator is the number on the top of the line. Basically, numerator is how many equal parts of the whole objects are taken.  Second, denominator is the number below the line. In the other word, denominator is the total number.

If four friends want to share the whole package of cookies, is means that the package will be divided to 4 parts. The whole part is represented by 'm'. So, the fraction form that can describe this case is each friends will get [tex]\frac{1}{4}[/tex] m.

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a) Write 98 as the product of prime factors. Write the prime factors in ascending order.​

Answers

Step-by-step explanation:

Write 98 as the product of prime factors. Write the prime factors in ascending order.​

2 × 7 × 7

2 × 7 × 7 = 98

Answer: 2×7×7

Step-by-step explanation: 98= 2×7×7

Lets first know about what is prime factor :- A factor which is a Prime number and not a composite number.

prime factor in ascending order = 2,7

hence the required product of prime factor is 2×7×7 or 2×[tex]7^{2}[/tex]

Identify a set of parallel and a set of perpendicular lines in this image.
m
B
LL
OBFCG, BF 1 CG
OBECG, AD 1 EH
O AD
EH, EHL BE
O AD EH, BF 1 CG

Answers

A set of parallel lines and a set of perpendicular lines are: BF║CG, AD⊥BF.

What are Parallel Lines?

Parallel lines are straight lines that are of equal distance at every point away from each other, and therefore, do not intersect each other at any point. The sign "║" is used to indicate that two lines are parallel lines.

What are Perpendicular Lines?

Lines that are said to be perpendicular lines are lines that intersect each other at right angle or 90 degrees. That is, at the point of their intersection, a right angle is formed which is 90 degrees. The sign "⊥" is used to indicate that two lines are perpendicular lines.

In the image given, lines BF and CG do not intersect at any point and they are of equal distance away from each other at every point, therefore, CG and BF are perpendicular lines.

Also, we can see that a right angle is formed at the point of intersection of lines AD and BF, therefore AD and BF are perpendicular lines.

Thus, we can state that: BF║CG, AD⊥BF.

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Can someone please help answer this question on simultaneous equations?

Answers

Answer:

1 knife cost £6.6

Step-by-step explanation:

Simultaneous equation is;

4x = y ----------(equation 1)

12y + 16x = 105.64 ----------- (equation 2)

Rewrite equations:

4x=y;16x+12y=105.64

Step: Solve4x=yfor y:

4x=y

4x+−y=y+−y(Add -y to both sides)

4x−y=0

4x−y+−4x=0+−4x(Add -4x to both sides)

−y=−4x

−y/−1 = −4x/−1

(Divide both sides by -1)

y=4x

Step: Substitute 4x for y in 16x+12y=105.64:

16x+12y=105.64

16x+124x=105.64

64x=105.64(Simplify both sides of the equation)

64x/64 = 105.64/64

(Divide both sides by 64)

x=1.650625

Step: Substitute 1.650625 for x in y=4x:

y=4x

y=(4)(1.650625)

y=6.6025(Simplify both sides of the equation)

1 knife cost = 4x = 4(1.65) = 6.6

∴ 1 knife cost £6.6

Graph the image of the figure on the right under the given translation.T(3, -1) (x,y)

Answers

Answer:

The only image with edges at the given coordinates is;

Given the figure in the attached image.

we want to identify the image after a translation;

[tex]T(3,-1)(x,y)[/tex]

Using one of the edges of the figure;

[tex](x,y)=(-4,0)[/tex]

Applying the above translation we have;

[tex](-4,0)\rightarrow(-4+3,0-1)=(-1,-1)[/tex]

The edge of the image would be at;

[tex](-1,-1)[/tex]

Therefore, the only image with edges at the given coordinates is;

whats the answer? one angle of a triangle mesuares 98°. the other two angles are congruent. enter and solve an equation to find the mesuare x of the congruent angles.

Answers

Congruent angles simply means angles that are equal. A triangle has three angles . Since one of the angle is given as 98 degree, the other 2 angles which are congruents are therefore equal. The total angle of a triangle is equals to 180 degree. Therefore,

[tex]\begin{gathered} x+x+98=180 \\ 2x+98=180 \\ 2x=180-98 \\ 2x=82 \\ x=\frac{82}{2} \\ x=41 \end{gathered}[/tex]

x = 41 degree

Lisa is putting money into a savings account, she starts with $350 in the savings account, and each week she adds $60

Answers

The total amount of money in the savings account after 11 weeks is  $1010.

What is the total amount in the savings account?

The from of the equation that can be used to determine the total amount of money in the savings account is a linear equation. A linear equation is an equation that has one variable that is raised to the power of one. When a linear equation is drawn on a coordinate graph, it is usually a straight line.

The form of the linear equation is:

Total amount = amount she starts with + (number of weeks x amount of money she adds each week)

S = $350 + ($60 x W)

S = $350 + $60W

Amount she would have in 11 weeks :  $350 + ( 11 x $60)

$350 + 660 = $1010

Here is the complete question:

Lisa is putting money into a savings account. She starts with $350 in the savings account, and each week she adds $60. Let S represent the total amount of money in the savings account (in dollars), and let W represent the number of weeks Jenny has been adding money. Write an equation relating S to W. Then use this equation to find the total amount of money in the savings account after 11 weeks.

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Angle P in Triangle PQR has the same measure as Angle S in Triangle STU. Which other condition is necessary to prove that these triangles are similar?

Answers

△PQR is congruent to △STU (△PQR ≅ △STU) under ASA condition using options (B) and (D).

What is the congruency of triangles?Triangle congruence: If all three corresponding sides and all three corresponding angles are equal in size, two triangles are said to be congruent. Slide, rotate, flip, and turn these triangles to create an identical appearance. They are in alignment with one another when moved. Therefore, if all three sides of two triangles are the same, then the triangles are said to be congruent. If we have a side, an angle between the sides, and then another side that is congruent, we know they are congruent. In other words, side, angle, side.

So, to prove that △PQR ≅ △STU:

∠P = ∠S (Given)PQ = ST (Option B)∠Q = ∠T (Option D)

So, with these three conditions, △PQR ≅ △STU is under ASA condition.

Therefore, △PQR is congruent to △STU (△PQR ≅ △STU) under ASA condition using options (B) and (D).

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Suppose the function f and g are defined as follows.

Answers

In this session we will focus on calculating the composition of f and f.

given the function f, the composition f composition f is obtained by taking the definiton of f and replace the varaible x with the function f itself. So we are given that

[tex]f(x)=\frac{8}{5x}[/tex]

So

[tex]f\circ f(x)=\frac{8}{5(\frac{8}{5x})}=\frac{8}{\frac{8}{x}}=\frac{8x}{8}=x[/tex]

So we have that f composition f is the function x.

if one shirt costs £13,how many can I buy with £419?

Answers

Answer:

32 shirts

Step-by-step explanation:

419÷13

=32.230

Answer:

32

Step-by-step explanation:

419 divided by 13

= 32.230

Select all ratios equivalent to 13.14

52:56. 39:42. 3:4

Answers

52:56 and 39:42

wkskskskskkakakskskskksmwkwkwkwksskks

Which of the following is an example of a set of like terms ??

Answers

2. >>>>>>>b {18x,-x,21x} is an example of like terms because they both hold the same variables.

4.>>>>>>

[tex] = - 7(2x - 3) \\ = - 7(2x) - 7( - 3) \\ = - 14x + 21[/tex]

ATTACHED IS THE SOLUTION

a company buys equal numbers of two different card forms. it utilizes 4/5 of one kind and 6/7 of the other. what fraction of the total number is unused?

Answers

12/35 fraction of the total number is unused from two different cards.

What is a fraction?

A fraction is written in the form of p/q, where q ≠ 0.

Fractions are of two types they are proper fractions in which the numerator is smaller than the denominator and improper fractions where the numerator is greater than the denominator.

Assuming the total first kind of card form is 1 and the total second kind of card form is also one.

Given, a company buys equal numbers of two different card forms. it utilizes 4/5 of one kind and 6/7 of the other.

∴ The total unused card form is,

= (1 + 1) - (4/5 + 6/7).

= 2 - (28 + 30)/35.

= 2 - 58/35.

= (70 - 58)/35.

= 12/35.

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How is adding positive and negative integers different from adding only positive integers on a number line?

Answers

1) To understand how that works, let me draw a number line:

If we only add positive numbers like 2 +3 = 5, 6+8 =14, etc. we'll only get a positive result (right to the zero) on the number line, we are moving to the right

2) If we add positive and negative numbers, for example:

2 +(-3) =2 -3=-1

0+(-1)= -1

6+(-2)= 4

On the number line, we are starting from one point and moving to the left.

What is the solution of -8/2y-8 = 5/y+4 - 7y+8/y^2-16?

y = -4
y = -2
y = 4
y = 6

Answers

The correct option for the fraction with polynomials is y=6.

Fractions with polynomials.

Expressions of more than two algebraic terms that contain different powers of the same variable is a polynomial.

The polynomial fraction is an expression of polynomial divided by another polynomial.

We can solve the fraction to get the value of y as follows;

[-8/(2y-8)]= [5/(y+4)]-[(7y+8)/(y²-16)]

[-8/(2y-8)]= [5/(y+4)]-[(7y+8)/(y+4)(y-4)]

[-8/(2y-8)]= [5(y+4)-(7y+8)]/(y+4)(y-4)] (L.C.M) of the fraction

[-8/(2y-8)]= (5y-20-7y-8)/(y+4)(y-4)

[-8/(2y-8)]= (-2y-28)/(y+4)(y-4)

[-8/2(y-4)]= (-2y-28)/(y+4)(y-4)

[-8/2(y-4)]= [-2(y-14)]/(y+4)(y-4)

[-4/(y-4)]= [-2(y-14)]/(y+4)(y-4)

Multiply both sides of the equation by -(y-4)

4= 2(y-14)/(y+4)

Cross multiply

4(y+4)=2(y+14)

Divide through by 2

2(y+4)=y+14

2y+8=y+14

collect like terms

2y-y=14-8

y=6.

Hence, we can state that the value of y for the polynomial with fraction is 6.

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please help with this geometry question i attempted it but dont understand it

Answers

You have the following vertices of a triangle:

A(1,1)

B(4,1)

C(4,5)

For the translation four untis to the right, consider this kind of translation means that it is necessary to sum 4 units to the x-coordinate:

A(1,1) => A'(1+4,1) = A'(5,1)

B(4,1) => B'(8,1)

C(4,5) => C'(8,5)

Next, a translation three units up is done by adding 3 units to the y-coordinate of points A', B' and C':

A'(5,1) => A''(5,1+3) = A''(5,4)

B'(8,1) => B''(8,4)

C'(8,5) => C''(8,8)

Next, a reflection around y=-1 consists in subtracting to the y-coordinate units equivalent to the vertical distance to the line y =-1, just as follow:

for the point A''(5,4) you can notice that the vertical distance of the y-coordinate, which is 4, to the line y=-1 is 5 units, then, it is necessary to subtract 5 units to such line:

A''(5,4) => A'''(5,-1-5)=A'''(5,-6)

for the point B''(8,4), the distance is again 5 units, then, you have:

B''(8,4) => B'''(8,-1-5) = B'''(8,-6)

for the point C''(8,8) the distance from y-coordinate y=8 to the line y=-1 is 9 units, then, yu subtract 9 units to -1:

C''(8,8) => C'''(8,-1-9) = C'''(8,-10)

Hence, the final points are:

A'''(5,6)

B'''(8,-6)

C'''(8,-10)

Please help me with 24For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes.

Answers

Given the equation,

[tex]-9x^2+72x+16y^2+16y+4=0[/tex]

Complete squares as shown below,

[tex]\begin{gathered} -9x^2+72x-a^2=-(9x^2-72x+a^2)=-9(x^2-8x+b^2) \\ \end{gathered}[/tex]

Thus,

[tex]\begin{gathered} \Rightarrow-9x^2+72x-a^2=-9(x^{}-4)^2 \\ \Rightarrow a^2=16\cdot9=144\Rightarrow a=12 \\ \Rightarrow-9x^2+72x-144=-9(x^{}-4)^2 \end{gathered}[/tex]

Similarly,

[tex]\begin{gathered} 16y^2+16y=16(y^2+y) \\ \Rightarrow16(y+\frac{1}{2})^2=16(y^2+y+\frac{1}{4}) \end{gathered}[/tex]

Therefore,

[tex]\begin{gathered} -9x^2+72x+16y^2+16y+4=0 \\ \Rightarrow-9(x-4)^2+16(y+\frac{1}{2})^2+4=-144+4 \\ \Rightarrow-9(x-4)^2+16(y+\frac{1}{2})^2=-144 \end{gathered}[/tex]

Finally, the standard form is.

[tex]\begin{gathered} \Rightarrow-\frac{(x-4)^2}{16}+\frac{(y+\frac{1}{2})^2}{9}=-1 \\ \Rightarrow\frac{(x-4)^2}{16}-\frac{(y+\frac{1}{2})^2}{9}=1 \end{gathered}[/tex]

As for the vertices, foci, and asymptotes,

[tex]\begin{gathered} c=\pm\sqrt[]{16+9}=\pm5 \\ \text{center:}(4,-\frac{1}{2}) \\ \Rightarrow\text{foci:}(4-5,-\frac{1}{2})_{},(4+5,-\frac{1}{2})_{} \\ \Rightarrow\text{foci:}(-1,-\frac{1}{2}),(9,-\frac{1}{2}) \end{gathered}[/tex]

Foci: (-1,-1/2), (9,-1/2)

Vertices

[tex]\begin{gathered} \text{center:}(4,-\frac{1}{2}),\text{vertices:}(4\pm a,-\frac{1}{2}) \\ \text{vertices:}(4+4,-\frac{1}{2}),(4-4,-\frac{1}{2}) \\ \text{vertices:}(8,-\frac{1}{2}),(0,-\frac{1}{2}) \end{gathered}[/tex]

Vertices: (8,-1/2), (0,-1/2)

Asymptotes:

[tex]\begin{gathered} y=\pm\frac{3}{4}(x-4)-\frac{1}{2} \\ \Rightarrow y=\frac{3}{4}x-\frac{7}{2} \\ \text{and} \\ y=-\frac{3}{4}x+\frac{5}{2} \end{gathered}[/tex]

Asymptotes: y=3x/4-7/2 and y=-3x/4+5/2

3. Suppose that the scores on a statewide standardized test are normally distributed with a mean of 69 and a standard deviation of 6. Estimate the percentage of scores that were(a) between 57 and 81. %(b) above 81. %(c) below 63. %(d) between 51 and 81. %

Answers

Answer:

a) 95%

b) 2%

c) 16%

d) 98%

Explanation:

We have the following:

This is a normal distribution

Mean = 69

Standard Deviation = 6

a) Between 57 and 81%

[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ x=57 \\ z=\frac{57-69}{6}=-\frac{12}{6} \\ z=-2 \\ \\ x=81 \\ z=\frac{81-69}{6}=\frac{12}{6} \\ z=2 \\ \end{gathered}[/tex]

The probability that a score is between 57 & 81 is given by the Area between (z = -2) & (z = 2):

[tex]\begin{gathered} P=0.97725-0.02275 \\ P=0.9545 \\ P=95.45\approx95 \\ P=95\text{ \%} \\ \\ \therefore P=95\text{ \%} \end{gathered}[/tex]

b) Above 81%

[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ x>81 \\ z=\frac{81-69}{6} \\ z=\frac{12}{6}=2 \\ z=2 \end{gathered}[/tex]

The probability that a score is above 81% is given by the area of the graph greater than (z = 2):

[tex]\begin{gathered} P=0.02275 \\ P=2.275\approx2.3 \\ P=2.3\approx2 \\ P=2\text{ \%} \\ \\ \therefore P=2\text{ \%} \end{gathered}[/tex]

c) Below 63%

[tex]\begin{gathered} x<63 \\ z=\frac{63-69}{6} \\ z=-\frac{6}{6}=-1 \\ z=-1 \end{gathered}[/tex]

The probability that a score is below 63% is given by the area of the graph lesser than (z = -1):

[tex]\begin{gathered} P=0.15866 \\ P=15.866\approx16 \\ P=16\text{ \%} \end{gathered}[/tex]

d) Between 51 and 81

[tex]\begin{gathered} 51\le x\le81 \\ z=\frac{51-69}{6} \\ z=-\frac{18}{6}=-3 \\ z=-3 \\ \\ z=\frac{81-69}{6} \\ z=\frac{12}{6}=2 \\ z=2 \end{gathered}[/tex]

The probability that a score is between 51 & 81 is given by the Area between (z = -3 & (z = 2):

[tex]\begin{gathered} P=0.97725-0.00135 \\ P=0.9759 \\ P=97.59\approx98 \\ P=98\text{ \%} \end{gathered}[/tex]

The number 24 is 4 times as many as 6.
Write this comparison as a multiplication equation.
+
Compare with multiplication

Answers

24/4=6 6*4 = 24. This should help

Let x represent the unknown value, then write an algebraic expression for: double a quantity increased by nine

Answers

Answer:

2(x + 9)

Explanation:

Let x represent the unknown value.

Double a quantity increased 9 means;

*We'll multiply the quantity by 2

*Increased by signifies addition

*Quantity means we'll put the expression after it in brackets

Let's go ahead and write the required algebraic expression;

[tex]2(x+9)[/tex]

n a tournament, a professional golfer knows that she is 200 yards from the hole. A spectator is watching her play and is 140 yards away from the golfer.

Answers

We can use the sine rule to find the hole angle, and then find the golfer angle:

[tex]\frac{200}{\sin(115)}=\frac{140}{\sin (Hole)}[/tex][tex]\text{Hole Angle = }\sin ^{-1}(\frac{140\times\sin (115)}{200})[/tex][tex]\text{Hole Angle = }39.37664303\text{ degre}es[/tex]

Now we can find the golfer angle:

[tex]\text{Golfer = 180 - 115 - 39.38 }\cong25.6\text{ degre}es[/tex]

Answer: 25.6 degrees.

Determine the midpoint of segment Ed coordinates e and d are 2 and -2

Answers

Answer: Midpoint = 0

Explanation:

The midpoint of segment ED can be calculated as:

[tex]\text{Midpoint}=\frac{E+D}{2}[/tex]

Where E and D are the coordinates of E and D respectively.

So, replacing E by 2 and D by -2, we get:

[tex]\text{MIdpoint}=\frac{2+(-2)}{2}=\frac{2-2}{2}=\frac{0}{2}=0[/tex]

Therefore the midpoint of segment ED is equal to 0.

jerry purchased a single family lot that measured 90' x 120'. the city requires a 20' setback on the front, a 15' setback along the sides and a 10' setback at the back of the lot. how big can jerry build his house?

Answers

The best measure of how big Jerry can build his house as required in the task content is; 60' by 90'.

Dimensions.

It follows from the task content that the dimension of the family lot purchased by Jerry in a bid to build his house has dimensions; 90' × 120'.

Since the city requires 20' setback on the front and 10' setback at the back, it follows that the vertical length remaining and available to build his house is; 120' - 20' - 10' = 90'.

Also, the horizontal width available for him to build his house after the 15' setback along the sides is; 90' - 15' - 15' = 60'.

Therefore, the space available for Jerry to.build his house is; 60' × 90'.

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SLook at point in the coordinate grid.If a line contains both points and the origin, which point would the line also contain? *(2 points)DS(12,8)(14,7)(15,9)(20, 12)5tvAle--AR

Answers

the point which will fall in the line is (14, 7)

Explanation:

The coordinates o f S = (2, 1)

When we draw a line from the origin (0, 0) touching the point S, we will see the next point will be (4, 2). Followed by (6, 3).

So when we look at the trend, we see that the x axis is the double of the y axis:

(2y, y) = (x, y)

(2,1 ) , (4, 2), (6, 3)

From the above, we can say the point which will fall in the line will follow that trend:

When we check the options, the only point with that trend is (14, 7).

This because (2(7), 7) = (2y, y)

Hence, the point which will fall in the line is (14, 7)

Determine the point (x, y) on the unit circle associated with the following real numbers. Write the exact answer as an ordered pair. Do not round.S = 30

Answers

Remember that

In a unit circle, the radius of the circle is 1

see the figure below to better understand the problem

we have that

[tex]\sin (30^o)=\frac{y}{1}[/tex]

and we know that

[tex]\sin (30^o)=\frac{1}{2}[/tex]

so

[tex]\begin{gathered} \frac{y}{1}=\frac{1}{2} \\ y=\frac{1}{2} \end{gathered}[/tex][tex]\cos (30^o)=\frac{x}{1}[/tex]

and we know that

[tex]\cos (30^o)=\frac{\sqrt[]{3}}{2}[/tex]

so

[tex]x=\frac{\sqrt[]{3}}{2}[/tex]

therefore

the coordinates (x,y) are

[tex](\frac{\sqrt[]{3}}{2},\frac{1}{2})[/tex]

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