The given expression is
[tex]106=118-x[/tex]First, we add x on each side
[tex]\begin{gathered} 106+x=118-x+x \\ 106+x=118 \end{gathered}[/tex]Then, we subtract 106 from each side
[tex]\begin{gathered} 106-106+x=118-106 \\ x=12 \end{gathered}[/tex]If 8% of the sheet aluminum is lost to scrap when forming a fuel tank, what is the weight of the tank if the raw sheets of aluminum weigh 200 pounds?
Solution:
Step 1: Calculate 8% of the raw sheets of aluminum :
[tex]200\text{ x 0.08 = 16}[/tex]This is the weight that is lost in the production of the fuel tank.
Step 2: Calculate the weight of the tank :
200 pounds - 16 pounds = 184 pounds.
So that, we can conclude that the correct answer is:
184 pounds.
I need help with this please thank you number 14
Answer:
The question is given below as
Concept:
The question will be solved using the linear pair theorem below
The Linear Pair Theorem states that two angles that form a linear pair are supplementary; that is, their measures add up to 180 degrees.
By applying the principle, we will have that
[tex]\begin{gathered} \angle x+88^0=180^0 \\ collect\text{ similar terms,} \\ subtract\text{ 88 from both sides} \\ \operatorname{\angle}x+88^0-88^0=180^0-88^0 \\ \angle x=92^0 \end{gathered}[/tex]Hence,
The value of x= 92°
Step 2:
By applying the linear pair theorem, we will also have that
[tex]\begin{gathered} \angle z+88^0=180^0 \\ collect\text{ similar terms, } \\ subtract\text{ 88 from both sides} \\ \operatorname{\angle}z+88^0-88=180^0-88 \\ \angle z=92^0 \end{gathered}[/tex]Hence,
The value of z= 92°
Step 3:
By applying the linear pair theorem also, we will have that
[tex]\begin{gathered} \angle x+\angle y=180^0 \\ 92^0+\angle y=180^0 \\ collect\text{ similar terms,} \\ substract\text{ 92 from both sides} \\ 92^0-92^0+\operatorname{\angle}y=180^0-92^0 \\ \angle y=88^0 \end{gathered}[/tex]Hence,
The value of y= 88°
When multiplying and/or dividing fractions, answer if each of the statements below are TRUE or FALSE.1: The denominators MUST be the same.2: You must convert mixed numbers into improper fractions before multiplying or dividing.3: You can keep mixed numbers when performing multiplication or division.4: Numerators must be multiplied by numerators and denominators must be multiplied by denominators.
EXPLANATION:
When multiplying and/or dividing fractions, answer if each of the statements below are TRUE or FALSE.
1.The denominators MUST be the same. (FALSE);They can have different denominators for both multiplying and dividing fractions.
2.You must convert mixed numbers into improper fractions before multiplying or dividing.(TRUE) ; This procedure is important so that the operation between fractions is as easy and correct.
3.You can keep mixed numbers when performing multiplication or division.
(FALSE) ; These mixed fractions must be converted to improper fractions to later do the correct multiplication or division.
4.Numerators must be multiplied by numerators and denominators must be multiplied by denominators.(FALSE); The numerator of the first fraction must be multiplied in a cross by the denominator of the second fraction; the denominators are multiplied by each other.
Which expression would be easier to simplify if you used the associative property to change the grouping? OA. 6+ 1; +3) OB. I(-0.2) +(-0.6)] +1.7 O c.(2+)+-) O D. (60+ 40) +-27)
A.
[tex]6+\lbrack\frac{4}{9}+(-\frac{2}{9})\rbrack[/tex]Since both fractions have the same numerator, you can factorize 1/9 aout of the parentheses, because:
[tex]\begin{gathered} \frac{1}{9}\cdot4=\frac{4}{9} \\ \text{and} \\ \frac{1}{9}\cdot2=\frac{2}{9} \end{gathered}[/tex]Then you can simplify the expression as:
[tex]6+\frac{1}{9}\lbrack4+(-2)\rbrack=6+\frac{1}{9}\lbrack4-2\rbrack[/tex]A physics student want to calculate her final grade. The grade distribution is tests = 30%, quizzes = 15%, homework=10%, labs=25%, and the final exam= 20%. Calculate her final grade if she got the following averages for each category: Tests=80%; Quizzes=70%; Homework=80%; Labs=90%; Final exam=90%.
Given:
The grade distribution is tests = 30%, quizzes = 15%, homework=10%, labs=25%, and the final exam= 20%.
Averages for each category: Tests=80%; Quizzes=70%; Homework=80%; Labs=90%; Final exam=90%.
Required:
We need to find the final grade.
Explanation:
Let the total mark for each is 100,
The mark for the test is 80.
We need to find 30% of 80.
[tex]grade\text{ for test =}\frac{30}{100}\times80=24[/tex]The mark for the quizzes is 70.
We need to find 15% of 70.
[tex]grade\text{ for quizzes =}\frac{15}{100}\times70=10.5[/tex]The mark for the homework is 80.
We need to find 10% of 80.
[tex]grade\text{ for homework =}\frac{10}{100}\times80=8[/tex]The mark for the labs is 90.
We need to find 25% of 90.
[tex]grade\text{ for labs =}\frac{25}{100}\times90=22.5[/tex]The mark for the final exam is 90.
We need to find 20% of 90.
[tex]grade\text{ for final exam=}\frac{20}{100}\times90=18[/tex]Add grade values for all the categories.
[tex]Final\text{ grade =24+10.5+8+22.5+18}[/tex]The final grade was 83 out of 100.
Divide 83 by 10.
The final grade is 8.3 out of 10.
Final answer:
[tex]Final\text{ grade =83 out of 100}[/tex][tex]Final\text{ grade =8.3 out of 10}[/tex]Suppose that an item regularly costs $100.00 and is discounted 24%. If it is then marked up 24%, is the resulting price $100.00? If not, what is it?
Given the cost of item =$ 100
If it is discounted 24%
So, the discount will be = 24% of 100 = 0.24 x 100 = $24
The cost after discount = 100 - 24 = $76
If it is marked up 24%
so, the rise will be = 24% of 76 = 0.24 x 76 = $18.24
The cost after marked up = 76 + 18.24 = $94.24
So, the answer is : No, and the cost will be = $94.24
Solve x2 + 12x + 25Hias17 by completing the square. Select all the possible solutions.-6 + 70-6 + 2.7–6 – 276-606-270-6-77
we are given the following expression:
[tex]x^2+12x+25=17[/tex]First, we will subtract 17 to both sides:
[tex]\begin{gathered} x^2+12x+25-17=17-17 \\ x^2+12x+8=0 \end{gathered}[/tex]We get an expression of the form:
[tex]ax^2+bx+c=0[/tex]To complete the square we will add and subtract the following term:
[tex]\frac{b^2}{4a}[/tex]Replacing the values:
[tex]\frac{12^2}{4(1)}=36[/tex]Therefore, we will add and subtract 36:
[tex]x^2+12x+36-36+8=0[/tex]Now we associate the first three terms:
[tex](x^2+12x+36)-36+8=0[/tex]Now we factor in the associated terms:
[tex](x+6)^2-36+8=0[/tex]Solving the operations:
[tex](x+6)^2-28=0[/tex]Now we solve for "x", first by adding 28 to both sides:
[tex](x+6)^2=28[/tex]Now we take square root to both sides:
[tex](x+6)=\sqrt[]{28}[/tex]Now we subtract 6 to both sides:
[tex]x=-6\pm\sqrt[]{28}[/tex]Now we factor 28 as 7*4:
[tex]undefined[/tex]graph the line that passes through the given point and has the given slope m(1,7); m=-5/2
Given that the line passes through the points;
[tex](1,7)[/tex]And slope;
[tex]m=-\frac{5}{2}[/tex]Solve this system of equations by graphing. First graph the equations, and then type the solution.x+3y=6y=1/2x+7
Given the system of equations;
[tex]\begin{gathered} x+3y=6---(1) \\ y=\frac{1}{2}x+7---(2) \end{gathered}[/tex]We shall first of all re-arrange the equations in the slope-intercept form;
[tex]y=mx+b[/tex]Note that the second one has already been expressed in the slope-intercept form. For the first one we would now have;
[tex]\begin{gathered} x+3y=6 \\ 3y=-x+6 \\ \text{Divide both sides by 3;} \\ \frac{3y}{3}=-\frac{x}{3}+\frac{6}{3} \\ y=-\frac{1}{3}x+2 \end{gathered}[/tex]To plot this equations on a graph we take two extreme points. We can do this by finding the value of y when x = 0, and y when x = 0.
For the first equation, we would have;
[tex]\begin{gathered} y=-\frac{1}{3}x+2 \\ \text{When x}=0 \\ y=-\frac{1}{3}(0)+2 \\ y=0+2 \\ y=2 \\ \text{That means we have the point }(0,2) \\ \text{Also, when y}=0 \\ 0=-\frac{1}{3}x+2 \\ \frac{1}{3}x=2 \\ \text{Cross multiply this and you'll have;} \\ x=2\times3 \\ x=6 \\ We\text{ now have our second point, }(6,0) \end{gathered}[/tex]Hence, for the first equation we have the two points;
[tex]\begin{gathered} A(0,2) \\ B(6,0) \end{gathered}[/tex]For the second equation;
[tex]\begin{gathered} y=\frac{1}{2}x+7 \\ \text{When x}=0 \\ y=\frac{1}{2}(0)+7 \\ y=0+7 \\ y=7 \\ \text{This means we have the point }(0,7) \\ \text{Also, when y}=0 \\ 0=\frac{1}{2}x+7 \\ -\frac{1}{2}x=7 \\ \text{Cross multiply and you'll have;} \\ x=7\times(-2) \\ x=-14 \\ \text{That means we now have the second point which is }(-14,0) \end{gathered}[/tex]For the second equation we now have the points;
[tex]\begin{gathered} A(0,7) \\ B(-14,0) \end{gathered}[/tex]We can now input both sets of coordinates and our graph would come out as shown below.
The point of intersection as we can see is at where x = -6 and y = 4. Therefore;
ANSWER:
The solution to the system of equations as shown on the graph is;
[tex](-6,4)[/tex]What happens to the graph when you change the value of a?
we have that
the value of a represent a vertical dilation
so
We stretch the graph in the vertical direction by a scale factor of
If the value of a> 1
then
we have stretching of the graph
If the value of 0 < a < 1
then
we have a compress of the graph
The dash in-front of the whole number is a negative sign, Just a little heads up :)
Okay, here we have this:
We need to solve the following expression:
[tex]\begin{gathered} -5\cdot2\text{ }\frac{1}{4} \\ =-5\cdot\frac{8+1}{4} \\ =-5\cdot\frac{9}{4} \\ =-\frac{45}{4} \\ =-11.25 \end{gathered}[/tex]Finally we obtain that the result is -11.25.
identify the terms, like terms, coefficients and constants 2c - 2b + c + 3 + b - 4
1) In this expression, we have
1.1 ) Terms: We have 6 terms in total.
2c -2b +c +3 + b -4 Combining like terms we can rewrite them as
c -b -1
1.2) Like terms are the ones that share the same variable
2c, c
-2b, b
1.3) Coefficients the numbers that multiply the variables
2c - 2b + c + 3 + b - 4
So we have:
2, -2, 1
1.4) The constants. Simply put in this case, the numbers that do not vary
-4 and 3
Holli works for the ymca and is making snack packs for the kids and afterschool program.If she has 76 cheese cracker snacks and 80 juice boxes, what is the greatest number of snack packs that she can make?How many juice boxes will go into each snack pack?
Assuming that each snack pack has 1 juice box and 1 cheese cracker snack, then Holli can make as much as the lowest quantity of juice boxes or cheese cracker snacks.
Let the triangles represent cheese cracker snacks and squares represent juice boxes:
The smallest pack possible includes one cheese cracker snack and one juice box:
The red ovals represent snack packs. Since there are not enough cheese cracker packs to make 77 snack packs, the maximum amount of packs that can be made is 76.
From the drawing, we can see that each snack pack will have one juice box.
Question 7 of 10Use the properties of logarithms to expand the following expression.(√log(x+4)5x³Your answer should not have radicals or exponents.You may assume that all variables are positive.
We have:
[tex]\begin{gathered} log(\sqrt{\frac{(x+4)^5}{x^3}}) \\ =log(\frac{(x+4)^2\sqrt{x+4}}{x\sqrt{x}} \end{gathered}[/tex]Applying properties of logarithms:
[tex]\begin{gathered} =log((x+4)^2\sqrt{x+4})-log(x\sqrt{x}) \\ =2log(x+4)+log(\sqrt{x+4})-logx-log\sqrt{x} \end{gathered}[/tex]3(y-5) = 15
Solve the following.
Answer:
y= 10
Step-by-step explanation:
look at picture for explanation
using the change-of-base formula, which of the following is equivalent to the logarithmic expression below? log6 21
Given:
Log6 21.
We are asked to use change of base formula to determine form the options which is equivalent.
let's apply the base change formula:
For C:
Log10 10
Log10 21
Log10 10
Log10 6
Apply power law of logarithm to simplify the expression:
1
LOg10 21
1
Log10 6
ivide fraction by multiplying its reciprocal:
1 x Log10 6
Log10 21
Write as a single fraction:
Log10 6
Log10 21
Apply the base chande formula: log21 6
Check its equivalency: False.
Option A is False
On your fifteenth birthday you discover you have a rich aunt sally aunt sally is a very generous women and wants to provide for your future she has decided that she will initially give you $1 then $2 the next year and so on doubling the amount each year until your 30. Use a chart to keep track how much money she will give you each year for the first 5 years.
Let's fill a table with the first two years, we already know those
So, we need to complete the chart for years 3,4, and 5. We know that in year 3, will receive double the money than we get in year 2, this is 2*$2=$4. Now we write that result on our table.
In year 4 we'll get double the money than in year 3, this is 2*$4=$8
Similarly, in year 5 we get double the money than what we got in year 4: 2*$8=$16
And we have filled in the table!
Now, a bonus, the pattern here seems to be an equation, notice this:
year 1 -> $1 = $2^0
year 2 ->$2 = $2^1
year 3 -> $4 = $2^2
year 4 -> $8 = $2^3
year 5 -> $16 = $2^4
This means that the amount of money we'll receive each year is given by
[tex]2^{t-1}[/tex]Where t is the year! (year 1, year 2, etc)
Multiply 3x11/12 and reduce to lowest terms Convert into a mix number
Multiply 3x11/12 and reduce to lowest terms Convert into a mix number
we have
3*(11/12)=11/4
convert to mixed number
11/4=8/4+3/4=2+3/4=2 3/4
answer is 2 3/4Amber has a job babysitting. She makes 7.50 per hour. What is the constant rate of change?
Remember that the constant rate of change refers to the change between the variable.
In this case, the constant rate of change is $7.50 per hour, in other words, it's 7.50. As an equation would be
[tex]y=\frac{15}{2}x[/tex]the cost of 3D movie tickets is $12 for 1 ticket, $24 for 2 tickets, and $36 for 3 tickets. determine whether the cost is proportional to the number of tickets by graphing on the coordinate plane. explain your reasoning
Explanation
1 ticket =$12
2 ticket =$24
3 ticket =$36
Point L is on line segment KM. Given LM = 5 and KL = 12, determine the length KM.
ANSWER
KM = 17
EXPLANATION
We have that point L is on the line segment KM.
Let us draw a diagram to represent it:
From the diagram, we see that the length of KM is the sum of the lengths of KL and LM.
This means that:
KM = KL + LM
KM = 12 + 5
KM = 17
That is the value of the length of KM.
Prepare an amortization schedule for the first three months on a loan of $87000 at 6% for 20 years
First, we need to calculate the value of the monthly payments. We can use the general ordinary anuity formula:
[tex]PV=\text{PMT}\cdot\lbrack\frac{1-(1+i)^{-n}}{i}\rbrack[/tex]Where PV=$87000
n=20
i=0.06
Replace the given values and solve for PMT:
[tex]\begin{gathered} 87000=\text{PMT}\cdot\lbrack\frac{1-(1+0.06)^{-20}}{0.06}\rbrack \\ 87000=\text{PMT}\cdot\lbrack\frac{1-0.3118}{0.06}\rbrack \\ 87000=\text{PMT}\cdot\lbrack\frac{0.6882}{0.06}\rbrack \\ 87000=\text{PMT}\cdot11.4699 \\ \text{PMT}=\frac{87000}{11.4699} \\ \text{PMT}=7585.06 \end{gathered}[/tex]Now that you have the payment, you can construct the table of the amortization schedule with the know information:
To calculate the missing information, start by calculating the interest component of the payment (interest paid) by multiplying the periodic interest by the remaining principal.
The monthly interest is the yearly interest divided by 12, then:
[tex]MI=\frac{0.06}{12}=0.005[/tex]And the interest paid is then:
[tex]\begin{gathered} IP=MI\cdot\text{ Remaining principal} \\ IP=0.005\cdot87000\text{ (for the first payment)} \\ IP=435 \end{gathered}[/tex]Now, calculate the principal paid by subtracting the interest paid from the payment amount:
[tex]\text{ Principal paid=7585.06-435}=7150.06\text{ (first payment)}[/tex]Then, by putting the values on the amortization schedule:
The remaining principal is 87000-7150.06 (principal paid).
Now for the second payment, calculate the interest paid with the new remaining principal:
[tex]\begin{gathered} IP=0.005\cdot79849,94\text{ (for the second payment)} \\ IP=399,25 \end{gathered}[/tex]And the principal paid is:
[tex]\text{ Principal paid=7585.06-399.25}=7185.81\text{ (second payment)}[/tex]The remaining principal is:
[tex]RP=79849.94-7185.81=72664.13[/tex]Thus:
And for the third month, you apply the same calculations and the amortization schedule is:
The waiter places a bowl of soup in front of Lacy. In a counterclockwise direction, she passes the soup 90°. The person receiving the soup passes it 30°. After the two passes, the soup is in front of which person ?○ Haifa○darcy○garret○igor
Notice that we have 12 people evenly distributed in a round table (360°).
This way, each person would be
[tex]\frac{360}{12}=30[/tex]30° from each other.
After the two passes, the soup would have moved 120°, meaning that
[tex]\frac{120}{30}=4[/tex]It would have moved 4 places.
Now, the person sitting 4 places away from Lacy, if the soup is passed counterclockwise, is Haifa
Therefore, the soup would be in front of Haifa.
The ratio of the sides of two smaller polygons is 4:5. Find the ratio of the areas.
Given that the ratio of the sides of two smaller polygons is 4:5
To Determine: The ratio of the areas
Solution:
Note that if the ratio of the sides of two similar shapes is a : b, then the ratio of their areas would be a² : b²
It was given in the question that the ratio of the sides of two smaller polygons is 4:5. Then, their areas would be
[tex]\begin{gathered} 4^2\colon5^2 \\ =16\colon25 \end{gathered}[/tex]Hence, the ratio of the areas is 16 : 25
3. The angle of depression of an aeroplane measured from a control tower, PQ, of height 88.9 m is 48°. When the plane moves along the runway from point Ato point Band stops, the angle of depression becomes 25.2°. The distance from point P to point Ais given as 119.6 m. Complete the diagram below to represent this informatior Р brid (ii) Leaving your answers correct to ONE decimal place calculate (a) Durmine the distance from the control tower to point A. (2) (b) Calculate the distance moved by the plane from its initial position.
The Solution:
Part (a)
Representing the problem fully in a diagram, we have:
Part (b)
We are required to find the length of QA= x.
We shall use Trigonometrical Ratio as below:
[tex]\begin{gathered} tan48^o=\frac{opposite}{adjacent}=\frac{88.9}{x} \\ \\ tan48=\frac{88.9}{x} \end{gathered}[/tex]Making x the subject of the formula, we get
[tex]x=\frac{88.9}{tan48}=80.0459\approx80.0m[/tex]Thus, the distance from the control tower to point A is 80.0 meters.
Part (c)
We are required to find the length of AB= y.
Considering triangle PQB, we have:
[tex]\begin{gathered} tan25.2=\frac{88.9}{x+y}=\frac{88.9}{80+y} \\ \\ 0.47056=\frac{88.9}{80+y} \end{gathered}[/tex]Solving for y, we get
[tex]\begin{gathered} 80+y=\frac{88.9}{0.47056} \\ \\ y=188.922-80 \\ y=108.922\approx108.9m \end{gathered}[/tex]Thus, the distance moved by the plane from its initial position is 108.9 meters.
Use a calculator and inverse functions to find the radian measures of a given angle around your answer to the nearest hundredth.angles whose sign is -0.26
Answer
Option C is correct.
x = -0.26 + 2pn
OR
x = -2.88 + 2pn
Explanation
Using the calculator and inverse functions
Let the unknown angle be x
Sin x = -0.26
x = Sin⁻¹ (-0.26)
x = -0.26 or -2.88 (From the calculator)
In order to generalize it, we add 2pi to both of them.
x = -0.26 + 2pn
OR
x = -2.88 + 2pn
Hope this Helps!!!
I worked on this and I can't seem to find the answer
1 liter = 1000 milliliters.
thus, 75791 liters =
[tex]75791\times1000=75791000[/tex]thus, 75791000 milliters.
I’m doing order of operation (14+16)/2-10
To solve this question, follow the steps below.
Step 01: Solve the operation inside the parentheses.
[tex]\begin{gathered} \frac{\mleft(14+16\mright)}{2}-10 \\ \frac{30}{2}-10 \end{gathered}[/tex]Step 02: Solve the division.
[tex]15-10[/tex]Step 03: Solve the subtraction.
[tex]5[/tex]Answer: 5.
A circular pool with a radius of 3 m sits in a rectangular yard that is 14 m by 8 m. What area of the yard is NOT covered by the pool?
SOLUTION:
Step 1:
In this question, we are given the following:
A circular pool with a radius of 3 m sits in a rectangular yard that is 14 m by 8 m. What area of the yard is NOT covered by the pool?
Step 2:
The area of the yard that is NOT covered by the pool is given as :
[tex]\text{Area of the Rectangular yard - Area of the circular pool}[/tex][tex]\text{( Length }X\text{ Breadth ) - ( }\pi r^2\text{ )}[/tex][tex]\begin{gathered} (\text{ 14 X 8 ) - (}\frac{22}{7}X\text{ 3 X 3 )} \\ =\text{ 112 - (}\frac{198}{7}) \\ =\text{ 112 - 28.28571429} \\ =\text{ 83.71428571} \\ \approx83.71m^2\text{ ( 2 decimal places)} \end{gathered}[/tex]CONCLUSION:
The area of the yard NOT covered by the pool is:
[tex]83.71m^2\text{ ( 2 decimaal places)}[/tex]Given the parallelogram TVWY shown above, determine how triangles TUZ and WXV can be shown to be similar. A. Since ZTU VWX and VX = XW, the triangles are similar by angle-side. B. Since TUZ WXV and TU = UZ, the triangles are similar by angle-side. C. Since TUZ VWX and ZTU WXV, the triangles are similar by angle-angle. D. Since TUZ WXV and ZTU VWX, the triangles are similar by angle-angle.
Notice that triangles TUZ and WXV have an angle with the same measure; therefore, we need two additional corresponding similar angles, or two corresponding similar sides, or one side and an angle to prove similarity.
Options A is not possible since it would imply that triangle TUZ has two 90°-inner angles.
Option B states a relation between two sides of triangle TUZ, not between two sides (one of each triangle). Option B cannot be the answer.
According to the figure, angles TUZ and WXV have to be congruent; therefore, option C cannot be the answer.
The only valid alternative is option D, option D is the answer.