470.89 per month
Explanation
Step 1
Let
[tex]\begin{gathered} \text{ Principal=P}=20650 \\ Number\text{ of payments= 4}8=\text{ 4 }\cdot12=\text{ 48 monthly payment} \\ Interest\text{ rate=4.5 \% (per year)} \\ Interest\text{ rate(month)=4.5/12=0.375} \\ \end{gathered}[/tex]Now, to calculate the payment amount per month we need to use, this expression
[tex]\begin{gathered} A=P\frac{(1+r)^n}{(1+r)^n-1} \\ \text{replace} \\ A=20650\frac{(1+0.375)^{48}}{(1+.375)^{48}-1} \\ A=20650\frac{(1.375)^{48}}{(1.375)^{48}-1} \\ A=20650\frac{1.340423584}{2.574462891} \\ A=20650\frac{1.340423584}{2.574462891} \\ A=470.89 \\ \end{gathered}[/tex]I hope this helps you
very confused doing exercises practicing for GED 61-year-old woman please help any help step by step how to do this I greatly appreciate your time
SOLUTION
First we have 5% as
[tex]5\%=\frac{5}{100}=0.05[/tex]Now we will use the Zscore calculator to get the Zscore for 0.05, written as P(x>Z) = 0.05, doing this we have Z = 1.645
Now from the question, we have
[tex]\mu=50,\sigma=10[/tex]Plugging these values into the formula given we have
[tex]\begin{gathered} X=\mu+Z\sigma \\ X=50+1.645\times10 \\ X=50+16.45 \\ X=66.45 \end{gathered}[/tex]Hence the lowest score is 66.45
The perimeter of a rectangle is 33.6 m, and it’s diagonal length is 12 m. Find its length and width
The perimeter of a rectangle is 33.6 m, and it’s diagonal length is 12 m. Find its length and width
we have that
The perimeter of a rectangle is equal to
P=2(L+W)
the diagonal of a rectangle is equal to
d^2=L^2+W^2
The length and width of the rectangle would be 9.6 m and
7.2 m.
What is an expression? What is equation Modelling? What is the perimeter of a rectangle?A mathematical expression is a combination of terms separated by mathematical operators. For example : 2x + 3y + 6. Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis, observations and results of the given problem. The perimeter of the rectangle is given by -
P = 2(L + B)
We have the perimeter of a rectangle is 33.6 m, and it’s diagonal length is 12 m.
Assume that the length and breadth of the rectangle be [x] and [y] repsectively.
We can write the perimeter as -
2(x + y) = 33.6
x + y = 16.8
and
we can also write -
x² + y² = 144
Refer to graph. For rectangle [x] > [y], we get
x = 9.6
y = 7.2
Therefore, the length and width of the rectangle would be 9.6 m and
7.2 m.
To solve more questions on Equation Modelling, visit the link below -
brainly.com/question/14441381
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find the lenghth of hypotonuse if one lengh is 6 , leave it simplified redical form
We are given a right triangle and we are asked to determine the length of the hypotenuse. To do that we will use the Pythagorean theorem:
[tex]h^2=a^2+b^2[/tex]Where "h" is the hypothenuse and a and b the other two sides. In this case, the two sides are equal since the triangle is 45 - 45 - 90. We have the following values:
[tex]\begin{gathered} a=6 \\ b=6 \end{gathered}[/tex]Substituting:
[tex]h^2=(6)^2+(6)^2[/tex]Solving the operations:
[tex]\begin{gathered} h^2=36+36 \\ h^2=2(36) \end{gathered}[/tex]Taking square root to both sides:
[tex]h=\sqrt[]{2(36)}[/tex]Now we will use the following property:
[tex]\sqrt[]{ab}=\sqrt[]{a}\times\sqrt[]{b}[/tex]Applying the property:
[tex]h=\sqrt[]{2}\times\sqrt[]{36}[/tex]Solving the operations:
[tex]h=6\sqrt[]{2}[/tex]And thus we have found the hypothenuse in its most simplified form.
4|2x+6|=-40Whats the solution? :)
4|2x+6|=-40
divide both-side of the equation by 4
|2x+6| = -10
Since the right hand-side of the equation is -10
Then the equation has no solution
Whats the percentage change of the first quantity withreference to the second quantity: 480 units, 560units.
We have to calculate the percentage change of the first quantity with reference to the second quantity.
The quantities are:
[tex]\begin{gathered} X_1_{}=480 \\ X_2=560 \end{gathered}[/tex]We then can calculate the percentage change as:
[tex]\begin{gathered} p=\frac{480-560}{560}\cdot100\% \\ p=\frac{-80}{560}\cdot100\% \\ p=-0.1429\cdot100\% \\ p=-14.29\% \end{gathered}[/tex]Answer: the percentage change of the first quantity with reference to the second quantity is -14.29%.
Suppose that the function H is defined, for all real numbers, as followers
SOLUTION
From the question, we have that
[tex]\begin{gathered} h(x)=-3 \\ \text{if }x\le-2 \\ \end{gathered}[/tex]so for
[tex]\begin{gathered} h(-4) \\ x\text{ here is -4 and -4 is less than -2, so it satisfies that equation above} \\ \end{gathered}[/tex]Hence
[tex]h(-4)=-3[/tex]Also for
[tex]\begin{gathered} h(-2) \\ x\text{ here is -2 and -2 is equal to -2, so it satisfies the equation} \end{gathered}[/tex]Hence
[tex]h(-2)=-3[/tex]Finally, we have that
[tex]\begin{gathered} h(x)=-(x+1)^2+2 \\ \text{if -2}So [tex]\begin{gathered} h(1) \\ x\text{ here is 1 and 1 is greater than -2, but less than 2} \\ so\text{ it satisfies the equation.} \end{gathered}[/tex]Hence we have that
[tex]\begin{gathered} h(x)=-(x+1)^2+2 \\ h(1)=-(1+1)^2+2 \\ h(1)=-(2)^2+2 \\ =-4+2 \\ -2 \end{gathered}[/tex]Hence
[tex]h(1)=-2[/tex]Rays DA and DC are perpendicular. Point B lies in the interior of angle ADC. If m angle ADB=(3a+10)degrees and m angle BDC=13adegrees, find a, m angle ABD, and m angle DBC.
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data
∠ ADB = 3a + 10
∠ BDC = 13a
DA and DC are perpendicular
Step 02
Graph
Step 03:
(3a + 10) + 13a = 90
16a + 10 = 90
16a = 90 - 10
a = 80 / 16
a = 5
m ∠ ADB = 3a + 10 = 3*5 + 10 = 15 + 10 = 25°
m ∠ BDC = 13a = 13 * 5 = 65°
The solution is:
a = 5
m ∠ ADB = 25°
m ∠ BDC = 65°
32. Solve for x: ln() + ln( − 1) = ln(4)a. x=0 or 5b. x=0c. x=5d. x=3e. There is no solution
To solve this question, we will proceed thus:
[tex]\ln (x)+\ln (x-1)=\ln (4x)[/tex]Simplifying further: (Applying log rules)
[tex]x(x-1)=4x_{}[/tex][tex]x^2-x=4x[/tex][tex]\begin{gathered} x^2-x-4x=0 \\ x^2-5x=0 \\ x(x-5)=0 \\ \text{This means that:} \\ x=0\text{ or } \\ x=5 \\ \\ To\text{ verify our solutions, we will put baxk the value of x into the initital } \\ expression\colon \\ \text{When x=0} \\ \ln (0)+\ln (0-1)=\ln (0)\text{ will result in math error:} \\ \text{When x=5} \\ \ln (5)+\ln (5-1)=\ln (4\times5)\text{ this will result in actual solutions} \\ \\ So\text{ with this above test, we can conclude that the correct solution is} \\ x=5 \\ \text{The correct answer therefore is C.} \end{gathered}[/tex]Answer = Option C.
The equation of a parabola is 12y = (x-1)^2 - 48 . Identify the vertex, focus, and directrix of the parabola.show each step
Given: The equation of a parabola below
[tex]12y=(x-1)^2-48[/tex]To Determine: The vertex, focus, and directrix of the parabola
Let us re-write the given equation
[tex]\begin{gathered} 12y=(x-1)^2-48 \\ (x-1)^2=12y+48 \\ (x-1)^2=12(y+4) \end{gathered}[/tex]The general equation of the parabola with vertex (h, k) is of this form
[tex](x-h)^2=4p(y-k)[/tex][tex]\begin{gathered} focus=(h,k+p) \\ directrix=y=k-p \\ vertex=(h,k) \end{gathered}[/tex]Let us compare the general equation with the given equation
[tex]\begin{gathered} (x-1)^2=12(y+4) \\ (x-h)^2=4p(y-k)_{} \\ h=1,k=-4 \end{gathered}[/tex]Therefore
[tex]\begin{gathered} vertex=(h,k)=(1,-4) \\ vertex=(1,-4) \end{gathered}[/tex][tex]\begin{gathered} 4p=12 \\ p=\frac{12}{4}=3 \\ p=3 \end{gathered}[/tex][tex]\begin{gathered} focus=(h,k+p)=(1,-4+3)=(1,-1) \\ focus=(1,-1) \end{gathered}[/tex][tex]\begin{gathered} directrix=y=k-p,y=-4-3=-7 \\ directrix=y=-7 \end{gathered}[/tex]Hence,
Vertex = (1, -4)
Focus = (1, -1)
Directrix, y = -7
The terminal of 0 is (1/2,3/2). What is the cos 0?
Answer:
Option B
Explanation:
Given that:
[tex]\text{Terminal point of }\theta=(\frac{1}{2},\frac{\sqrt[]{3}}{2})[/tex]If P(x, y) is the terminal point of an angle, then x is the length of its adjacent side and y is the length of its opposite side.
Here,
[tex]\begin{gathered} \text{Length of adjacent side =}\frac{1}{2} \\ \text{Length of opposite side =}\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]First, find the length of hypotenuse using the Pythagorean theorem.
[tex]\begin{gathered} \text{Hypotenuse}^2=Adjacentside^2+Oppositeside^2 \\ =(\frac{1}{2})^2+(\frac{\sqrt[]{3}}{2})^2 \\ =\frac{1}{4}+\frac{3}{4} \\ =1 \\ \text{Hypotenuse}=\sqrt[]{1}=1 \end{gathered}[/tex]Using the trigonometric ratio,
[tex]cos\theta=\frac{\text{Adjacent side}}{Hypotenuse}[/tex]Plug the values into the formula.
[tex]\begin{gathered} cos\theta=\frac{\frac{1}{2}}{1} \\ =\frac{1}{2} \end{gathered}[/tex]Hence, option B is correct.
What is the slope of the line represented by the equation y --VE15x?-55
Explanation
we have
[tex]\begin{gathered} y=-\frac{2}{3}-5x \\ \text{reorder} \\ y=-5x-\frac{2}{3} \end{gathered}[/tex]the equation is writen in slope-intercept form
[tex]\begin{gathered} y=mx+b \\ \text{where m is the slope} \\ \text{and b is the y -intercept} \end{gathered}[/tex]then
[tex]undefined[/tex]A bag of 11 marbles contains 7 marbles with red on them, 4 with blue on them, 6 with green on them and 6 with red and green on them. what is the probability that a randomly chosen marble has either green or red on it? note that these events are not mutually exclusive. express your answer as a fraction.
Answer:
7/11
Explanation:
Given;
Total number of marbles = 11
Number of marbles with red on them = 7
Number of marbles with blue on them = 4
Number of marbles with green on them = 6
Number of marbles with red and green on them = 6
We'll use the below formula to determine the probability that a randomly chosen marble has either green or red on it;
[tex]P(G\cup R)=P(G)+P(R)-P(G\cap R)[/tex]where;
[tex]\begin{gathered} P(G\cup R)=\text{ the probabilty of choosing a marble with green or red on it = ?} \\ P(G)=\text{ the probability of choosing marbles with gre}en\text{ on them }=\frac{6}{11} \\ P(R)=\text{the probability of choosing marbles with red on them}=\frac{7}{11} \\ P(G\cap R)=\text{the probability of choosing marbles with gre}en\text{ and red on them }=\frac{6}{11} \end{gathered}[/tex]Let's go ahead and substitute the above values into the formula and evaluate;
[tex]\begin{gathered} P(G\cup R)=\frac{6}{11}+\frac{7}{11}-\frac{6}{11} \\ P(G\cup R)=\frac{7}{11} \end{gathered}[/tex]
Allre segment with endpoints R (3,5) and (5,5) is reflected across the line y=-x and translated 2 units down Determine whether each cholce is the Image of an endpoint of the line segment Select Yes or No for A-C. No AR(-5,-3) Yes VA BR'(-5,-5) No BE Yes No CS(-5, -7)
To solve this problem we can grph the dot and the replection line so:
So the reflecton will change the sign of component x and y of the points so the tranformation will be:
[tex]\begin{gathered} (3,5)\to(-3,-5) \\ (5,5)\to(-5,-5) \end{gathered}[/tex]now if we translate the point 2 units doun we have to rest 2 units to the y component so:
[tex]\begin{gathered} (-3,-5)\to(-3,-7) \\ (-5,-5)\to(-5,-7) \end{gathered}[/tex]What is the purpose of testosterone
we have that
the number 7 is greater than zero
therefore
7 > 0
the answer is
>
8) The measure of three angles of a triangle are xº,(x +29) ° and (x - 5)º. Find the measures of all threeangles. (3 pts)
Let's name each of them:
[tex]\begin{gathered} a=xº \\ b=(x+29)º \\ c=(x-5)º \end{gathered}[/tex]The sum of the angles of a triangle is always equal to 180º, so:
[tex]\begin{gathered} a+b+c=180º \\ xº+(x+29)º+(x-5)º=180º \\ x+x+29+x-5=180 \\ 3x+24=180 \\ 3x=180-24 \\ 3x=156 \\ x=\frac{156}{3} \\ x=52 \end{gathered}[/tex]Now that we know tha value of x, we can find the values of all three angles:
[tex]\begin{gathered} a=52º \\ b=(52+29)º=81º \\ c=(52-5)º=47º \end{gathered}[/tex]Find the mean and median of the following set:-25 10 40 -15 0 35 15 -10
We are given the following data set
[tex]-25,10,40,-15,0,35,15,-10[/tex]We are asked to find the mean and median of the data set.
Recall that the mean of a set is given by
[tex]\operatorname{mean}=\frac{\text{sum of numbers}}{\text{total numbers}}[/tex]In this case, there are a total of 8 numbers in the set.
[tex]\operatorname{mean}=\frac{-25+10+40-15+0+35+15-10}{8}=\frac{50}{8}=6.25[/tex]Therefore, the mean of the set is 6.25
To find the median, first, arrange the set from least to greatest.
[tex]-25,-15,-10,0,10,15,35,40[/tex]The median is the number in the middle of the set.
But in this case, there are two middle numbers (0 and 10) so we will find the mean of these two numbers and that will be our median.
[tex]\operatorname{median}=\frac{0+10}{2}=\frac{10}{2}=5[/tex]Therefore, the median of the set is 5
Use the remainder theorem to find which of the following is not a factor of x3 – 4x2 – 4x + 16. 1) x – 42) x – 123) x – 24) x + 2
As given by the question
There are given that the equation
[tex]x^3-4x^2-4x+16[/tex]Now,
According to the concept of the remainder theorem, the remainder theorem is the approach of education division of polynomial.
If we divide a polynomial P(x) by a factor (x-a).
Then,
From the given option, not a factor is (x-12).
Hence, the correct option is B.
Simplify the expression and assume that X, y, and z denote any positive real numbers
Let's simplify the following radical expression:
[tex]\text{ }\sqrt[8]{x^8y^4z^4}[/tex]We get,
[tex]\text{ }\sqrt[8]{x^8y^4z^4}[/tex][tex]\text{ (}\sqrt[8]{x^8})(\sqrt[8]{y^4})(\sqrt[8]{z^4})[/tex][tex](\text{ x}^{\frac{8}{8}})(y^{\frac{4}{8}})(z^{\frac{4}{8}})[/tex][tex]\text{ (x)(y}^{\frac{1}{2}})(z^{\frac{1}{2}})[/tex][tex]\text{ x}\sqrt[]{yz}[/tex]Therefore, the answer is:
[tex]\text{ x }\sqrt[]{yz}[/tex]A coin purse contains 4 quarters and 6 dimes. Two coins are drawn at random without replacement. Find the probability that at least one of the two coins is a quarter.
P(at least one of the two coins is a quarter) = 2/3
Explanation:Number of quarters = 4
Number of dimes = 6
Total number of coins = 10
P(At least one of the two coins is a quarter) = P(one is a quarter) + P(the two coins are quarters)
P(one of the coins is a quarter) = (4/10)(6/9) + (6/10)(4/9)
P(one of the coins is a quarter) = 24/90 + 24/90
P(one of the coins is a quarter) = 48/90
P(one of the coins is a quarter) = 8/15
P(the two coins are quarters) = (4/10)(3/9)
P(the two coins are quarters) = 12/90
P(the two coins are quarters) = 2/15
P(At least one of the two coins is a quarter) = 8/15 + 2/15
P(At least one of the two coins is a quarter) = 10/15
P(at least one of the two coins is a quarter) = 2/3
The radius of your tire is 1.5 ft. You need a snow chain to surround your wheel tire. Will a 6ft long chain be long enough? Explain.
First, find the circumference of the tire to estimate the length of chain that is needed.
The wheel tire takes the shape of a circle, hence use the formula for the circumferece of a circle:
[tex]C=2\pi r[/tex]Substitute r=1.5 into the formula and use π=22/7:
[tex]\begin{gathered} C=2\times\frac{22}{7}\times1.5 \\ \approx9.43\text{ ft} \end{gathered}[/tex]Notice that the circumference of the wheel is about 9.43ft which is more than 6ft. Hence, a 6ft long chain will not be enough to surround the wheel tire.
Please help inserting the numbers into formula I need help with this exercise practicing for ged test in 10 days im a 61-year-old woman in need of help really need help please help me with exercise five
Given:
[tex]Z=\frac{X-\mu}{\sigma}[/tex]a)
[tex]\begin{gathered} X=\text{ \$}70 \\ \mu=\text{ \$}100\text{ } \\ \sigma=\text{ \$12} \\ \\ Therefore, \\ Z=\frac{70-100}{12}=-2.5 \end{gathered}[/tex]Hence,
[tex]P\left(xTherefore, the answer is 0.0062097.b)
[tex]\begin{gathered} X_1=\text{ \$90,}X_2=\text{ \$120} \\ \\ Z_1=\frac{90-100}{12}=-0.8333 \\ Z_2=\frac{120-100}{12}=1.6667 \end{gathered}[/tex]Therefore,
[tex]P\left(-0.8333Hence, the answer is 0.74988.c)
[tex]\begin{gathered} X=\text{ \$140} \\ \\ Z=\frac{140-100}{12}=3.3333 \end{gathered}[/tex]Therefore,
[tex]P\left(x>Z\right)=0.00042911[/tex]Hence, the answer is 0.00042911.
I need some help on this gcse percentages question please
Donald's annual wage is
[tex]23500\text{ pound}[/tex]He does not have to pay tax first 6400 pound.
He have to pay tax
[tex]23500-6400=17100\text{ pound}[/tex]At 25% the tax will be 25% of 17100 which is
[tex]\frac{25}{100}\times17100=4275\text{ pound}[/tex]At 40% the tax will be 40% of 17100 which is
[tex]\frac{40}{100}\times17100=6840pound[/tex]x<=5 graph the inequality
Given inequality is
[tex]x\le5[/tex]Mark the point (5,0) on the x-axis and draw a vertical line to graph x=5.
The value of x is less than 5, so the region is taking the left side of the line.
Shade left side of line x=5 to graph x<=5.
The graph is
During a winter cold spell the temperature change was -1.2 f per hour for a period of 4.5 hours which expressions can be used to find the overall change in the temperature during that time period.A.) 4.5 divided (-1.2) degrees FahrenheitB.) 4.5x (-1.2) degrees FahrenheitC.) 4.5 - (-1.2) degrees FahrenheitD.) (-1.2)+(-1.2)+(1.2)+(1.2) degrees FahrenheitE.) 4(-1.2)+(0.5)(-1.2) degrees Fahrenheit
Given:
The change in temperature = -1.2 f per hour
For a period of 4.5 hours
The expression will be :
[tex]4.5\times(-1.2)[/tex]So, the answer is option : B.) 4.5x (-1.2) degrees Fahrenheit
Complete the conversion using an equivalent rate.3,000 mL = LClick the icon to view the metric units of capacity.3,000 mL = 21LXL1,000 mL x 3,000 mL(Type integers or decimals.)
1000 milliliter = 1 liter
3000 milliliter = x
To find x, we cross-multiply, this leaves us with:
[tex]\begin{gathered} 3000ml\text{ }\times1l=1000ml\times x \\ \text{Dividing both sides by 1000ml gives x to be:} \\ x=\frac{3000ml\text{ }\times1l}{1000ml}=3l \end{gathered}[/tex]3000ml = 3l
The number meant to be in all the boxes is 3
How do I solve and what would the answer be?
Given the function below
[tex]\begin{gathered} f(x)=\frac{2}{x-5} \\ \therefore y=\frac{2}{x-5} \end{gathered}[/tex]The inverse of the function is denoted by
[tex]f^{-1}(x)_{}[/tex]It can be obtained through the process below
[tex]y=\frac{2}{x-5}[/tex]Swap the position of y and x in the equation above
[tex]x=\frac{2}{y-5}[/tex]Solve for y in the resulting equation from the swap
[tex]\begin{gathered} By\text{ cross multiplying} \\ x(y-5)=2 \\ \text{Divide both sides by x} \\ \frac{x(y-5)}{x}=\frac{2}{x} \end{gathered}[/tex][tex]\begin{gathered} y-5=\frac{2}{x} \\ \text{Add -5 to both sides} \\ y-5+5=\frac{2}{x}+5 \end{gathered}[/tex][tex]y=\frac{2}{x}+5[/tex]Hence, the inverse of f(x) is
[tex]f^{-1}(x)=\frac{2}{x}+5[/tex]Which ordered pair is a solution to the system of inequalities? y 2-X + 2 y> x-5
A(3,2)
Explanation
[tex]\begin{gathered} y\ge-x+2 \\ y>x-5 \end{gathered}[/tex]to find this replace each pair in the inequalities and check if it is tru
Step 1
A(3,2)
[tex]\begin{gathered} y\ge-x+2 \\ 2\ge-3+2 \\ 2\ge-1,\text{ true} \\ and \\ y>x-5 \\ 2>3-5 \\ 2>-2,\text{true} \end{gathered}[/tex]Step 2
B(-5,-2)
[tex]\begin{gathered} y\ge-x+2 \\ -2\ge-(-5)+2 \\ -2\ge5+2 \\ -2\ge7\text{ False} \end{gathered}[/tex]Step 3
C(0,0)
[tex]\begin{gathered} y\ge-x+2 \\ 0\ge-0+2 \\ 0\ge2,false \end{gathered}[/tex]Step 4
D(-1,1)
[tex]\begin{gathered} y\ge-x+2 \\ 1\ge-(-1)+2 \\ 1\ge1+2 \\ 1\ge3,false \end{gathered}[/tex]so, the answer is A(3,2)
Find BC.law of cosine 4A. 29 inB. 22.5 inC. 30 inD. 28 in
Use the law of cosine as the statement suggest you:
[tex]BC^2=AB^2+AC^2-2(AB)(AC)\cos(A)[/tex]Replace and solve:
[tex]BC^2=19^2+12^2-(2*19*12*\cos(138))[/tex][tex]BC^2=843.87[/tex][tex]BC=\sqrt{843.87}[/tex][tex]BC=29.05[/tex]Fifth grade Y.11 Multi-step problems with customary unit conversions MJ9 You have pr Sasha's science class is making slime from cornstarch and glue. Sasha is helping pass out materials. She gives each student 4 ounces of cornstarch from a 5-pound bucket. If Sasha's class has a total of 18 students, how many ounces of cornstarch will she have left? ounces Submit 0
She will have 8 ounces of cornstarch left
Explanation:Given that there is a 5-pound bucket of cornstarch, and 18 students are to get 4 ounces each.
The total of 4 * 18 = 72 ounces will be given to them
what is left is:
5 pounds - 72 ounces
But 1 pound = 16 ounces
5 pounds = 5 * 16 ounces
= 80 ounces
Therefore, what is left is:
80 - 72 = 8 ounces
Hey I need help answering this question. So can someone help me?
.Explanation:
We are given a cyclic quadrilateral.
The task involves obtaining the value of x and y
To get the values, we will use the theorem:
Opposite angles of a cyclic quadrilateral are supplementary
Therefore
[tex]105^0+(5y+10)^0=180^0[/tex]The value of y will be obtained as follow
[tex]\begin{gathered} 105^0+5y+10=180^0 \\ \text{colllecting like terms} \\ 5y=180-105-10 \\ 5y=65 \\ y=\frac{65}{5} \\ y=13^0 \end{gathered}[/tex]To get the value of x, we will have
[tex]\begin{gathered} 70^0+x+30=180 \\ x+100^0=180^0 \\ x=180-100 \\ x=80^0 \end{gathered}[/tex]Therefore
x= 80 degrees
y= 13 degrees