The perimeter of the box is given as
12x + 32
To write an equivalent expression that shows the premier has four times the side length of the box , we would factorise the expression. It becomes
4(12x/4 + 32/4)
= 4(3x + 8
For the polyhedron, use eular's foemula to find the missing number
Given:
Edges of the polyhedron, E = 10
Vertices, V = 5
A polyhedron is a three-dimensional figure.
Let's find the number of faces using Euler's formula.
To find the number of faces of the polyhedron, we have the Euler's formula:
V + F - E = 2
Substitute values into the formula:
5 + F - 10 = 2
Combine like terms:
F + 5 - 10 = 2
F - 5 = 2
Add 5 to both sides:
F - 5 + 5 = 2 + 5
F = 7
Therefore, the number of faces of the polyhedron is 7
ANSWER:
7 faces
8. The square of a number decreased by 3 times the number is 28. Find allpossible values for the number.
We need to find a number x
We know its square (x²) decreased by 3 times it (3x) is 28. Then
x² - 3x = 28
x² - 3x - 28 = 0 [Simplifying the equation]
Since - 7 + 4 = - 3 and (-7) (4) = - 28, we factor the polynomial:
(x - 7)(x + 4) = 0 [Factoring the polynomial]
When a multiplication, like (x - 7)(x + 4), equals cero?
it equals cero if and only if
(x - 7) = 0 or (x + 4) = 0
Then, simplifying both equations
x = 7 or x = - 4
Answer: x = 7 or x = - 4
Last week, Shelly rode her bike a total of 30 miles over a three-day period. On the second day, she rode LaTeX: \frac{4}{5}45 the distance she rode on the first day. On the third day, she rode LaTeX: \frac{3}{2}32 the distance she rode on the second day
We make expressions for each afirmation
Where X is the first day, Y second day and Z the third
1. the sum of the 3 days gives us 30
[tex]X+Y+Z=30[/tex]2. Second day is 4/5 of the first day
[tex]Y=\frac{4}{5}X[/tex]3.Third day is 3/2 of the second day
[tex]Z=\frac{3}{2}Y[/tex]Whit the expressions I try to represent everything as a function of X
I must represent Z in function of X, for this I can replace Y of the second expression in the third expression
[tex]\begin{gathered} Z=\frac{3}{2}(\frac{4}{5}X) \\ Z=\frac{12}{10}X \\ Z=\frac{6}{5}X \end{gathered}[/tex]So I have:
[tex]\begin{gathered} Y=\frac{4}{5}X \\ Z=\frac{6}{5}X \\ \end{gathered}[/tex]And I can replace on the first expression
[tex]\begin{gathered} X+Y+Z=30 \\ X+(\frac{4}{5}X)+(\frac{6}{5}X)=30 \end{gathered}[/tex]I must find X
[tex]\begin{gathered} (1+\frac{4}{5}+\frac{6}{5})X=30 \\ 3X=30 \\ X=\frac{30}{3} \\ X=10 \end{gathered}[/tex]So, if I have X I can replace on this expressions to find de value:
[tex]\begin{gathered} Y=\frac{4}{5}X \\ Z=\frac{6}{5}X \end{gathered}[/tex]Where X is 10
[tex]\begin{gathered} Y=\frac{4}{5}\times10 \\ Y=\frac{40}{5}=8 \\ \\ Z=\frac{6}{5}\times10 \\ Z=\frac{60}{5}=12 \end{gathered}[/tex]To check:
[tex]\begin{gathered} X+Y+Z=30 \\ (10)+(8)+(12)=30 \\ 30=30 \\ \end{gathered}[/tex]The result is correct, therefore:
[tex]\begin{gathered} X=10 \\ Y=8 \\ Z=12 \end{gathered}[/tex]Write an equation and solve.The supplement of an angle is 63º more than twicethe measure of its complement. Find the measure ofthe angle.
The measure of the angle is 63 degrees
Explanation:Let the angle in question be x degrees.
The supplement is (180 - x) degrees, and the complement is (90 - x) degrees.
Given that the supplement is 63 degrees more than twice the measure of its complement, we have the equation:
180 - x = 2(90 - x) + 63
Solving for x in the above:
180 - x = 180 - 2x + 63
2x - x = 180 - 180 + 63
x = 63
Find the augmented matrix for the systemIt gives us 3 numbers already
It is required that we find an augmented matrix for the system.
Recall that a matrix that contains the coefficients and constant terms of a system of equations, each written in the standard form with the constant terms to the right of the equals is called an augmented matrix.
The given system of equations is:
[tex]\begin{cases}x+5y+8z=-9 \\ 3x+z=-4 \\ 7x+5y+7z=3\end{cases}[/tex]The first, second, and third equations can be rewritten to get:
[tex]\begin{cases}1x+5y+8z=-9 \\ 3x+0y+1z=-4 \\ 7x+5y+7z=3\end{cases}[/tex]Hence, the augmented matrix using the system is:
[tex]\begin{bmatrix}{1} & 5 & 8{|} & {-9} \\ {3} & {0} & {1|} & {-4} \\ {7} & {5} & {7|} & {3} \\ & {} & {} & {}\end{bmatrix}[/tex]a proton is a positively charged particle found in the nuclei of atoms a proton has a diameter of 1.5 x 10^-15 meters How is this written in standard form. A) 0.000000000000015 meters B) 0.0000000000000015 meters C) 0.00000000000000015 meters D)0.000000000000000015 meter
Jan Sara and maya ran a total of 64 miles last week. Jan and maya ran the same amount and Sarah 8 miles less then maya. how many miles did Sarah run
Let:
x = Distance traveled by Jan
y = Distance traveled by Sara
z = Distance traveled by maya
Jan Sara and maya ran a total of 64 miles last week, so:
[tex]x+y+z=64_{\text{ }}(1)[/tex]Jan and maya ran the same amount and Sarah 8 miles less then maya. therefore:
[tex]\begin{gathered} x=z_{\text{ }}(2) \\ y=z-8_{\text{ }}(3) \end{gathered}[/tex]Replace (2) and (3) into (1):
[tex]\begin{gathered} z+z-8+z=64 \\ 3z=64+8 \\ 3z=72 \\ z=24mi \end{gathered}[/tex]Replace z into (2):
[tex]\begin{gathered} y=24-8 \\ y=16mi \end{gathered}[/tex]Sara ran 16 mi
What is the domain of the function shown in the graph below? y 10 9 8 7 6 5. 4 3 2 -10 -9 -8 -7 -6 in -4 3 -2 1 6 2 8 9 10 -2 -3 -4 -5 -6 -8 9 10 W Type here to search Et TH-WL-57336
1) As the Domain is the set of inputs (x) for that function, as we can see in the graph.
There's one point in the graph x =8, where should be an asymptote i.e. a vertical or horizontal line that prevents both graphs do not trespass.
So we can write the Domain as
D =(-∞, 8) U (8, ∞)
Because in this function, the point x=8 is not included, and from point 8 on the function continues.
Graph the line.y-1= 1/5 (x+4)
We are given the following equation:
[tex]y-1=\frac{1}{5}(x+4)[/tex]Using the distributive property:
[tex]y-1=\frac{1}{5}x+\frac{4}{5}[/tex]Adding 1 to both sides
[tex]y=\frac{1}{5}x+\frac{4}{5}+1[/tex]Solving the operations:
[tex]y=\frac{1}{5}x+\frac{9}{5}[/tex]To graph this line we need two points through which the line passes. The first point can be obtained by making x = 0:
[tex]\begin{gathered} y=\frac{1}{5}(0)+\frac{9}{5} \\ y=\frac{9}{5} \end{gathered}[/tex]Therefore, the first point is (0,9/5).
The second point can be obtained by making x = 1, we get:
[tex]\begin{gathered} y=\frac{1}{5}(1)+\frac{9}{5} \\ y=\frac{10}{5}=2 \end{gathered}[/tex]Therefore, the point is (1,2). Now we plot both points and join them with a line. The graph is:
Find the slope of each line and then determine if the lines are parallel, perpendicular or neither. If a value is not an integer type it as a decimal rounded to the nearest hundredth.Line 1: passes through (-8,-55) and (10,89) the slope of this line is Answer.Line 2: passes through (9,-44) and (4,-14) the slope of this line is Answer.The lines are Answer
For line 1 =
The coordinates given are (-8,-55) , (10,89)
The slope of the line is
[tex]m=\frac{89+55}{10+8}=\frac{144}{18}=8[/tex]For line 2 =
The coordinates given are (9,-44) , (4,-14)
The slope of the line is
[tex]m=\frac{-14+44}{4-9}=\frac{30}{-5}=-6[/tex]
The lines are not perpendicular or parallel because the slope of the lines does not satisfy the condition of perpendicular or parallel slopes.
Hence the answer is neither.
If you apply the changes below to the absolute value parent function, f(x) = [X,which of these is the equation of the new function?• Shift 2 units to the left.• Shift 3 units down.O A. g(x) = 5x + 21 - 3O B. g(x) = (x - 21 - 3O c. g(x) = 5x + 31 - 2O D. g(x) = \x - 31 - 2
h is the translation left or right
h > 0 the function shifts to the right h units
h < 0 the function shifts to the left h units
k is the translation up or down
k > 0 the function shifts up k units
k < 0 the function shifts down k units
For the given transformations:
Shift 2 units to the left. h: -3
Shift 3 units down. k: -3
Then, the function g(x) is:
[tex]\begin{gathered} g(x)=\lvert x-(-2)\rvert-3 \\ \\ g(x)=\lvert x+2\rvert-3 \end{gathered}[/tex]solve[tex]3 - \frac{x}{2} \geqslant 15[/tex]the equation
GIVEN:
We are given the following inequality;
[tex]3-\frac{x}{2}\ge15[/tex]Required;
To solve the inequality for x.
Step-by-step solution;
We begin by collecting like terms. Subtract 3 from both sides;
[tex]\begin{gathered} 3-3-\frac{x}{2}\ge15-3 \\ \\ -\frac{x}{2}\ge12 \end{gathered}[/tex]Now we cross multiply;
[tex]-x\ge24[/tex]We now multiply both sides of the inequality by -1.
Note that when an inequality is multiplied or divided by a negative value, then the inequality sign flips over.
Therefore;
[tex]\begin{gathered} -x\ge24 \\ \\ -x(-1)\ge24(-1) \\ \\ x\leq-24 \end{gathered}[/tex]ANSWER:
[tex]x\leq-24[/tex]Ahmed takes out a loan charging 6.7% simple interest for 10 years.
At the end of 10 years Ahmed pays back $1278 in just interest. Round your answer
to the nearest penny. The original amount of the loan (principal) was
A/
Round your answer to the nearest penny..
Answer:
$1907.46
Step-by-step explanation:
You want the principal amount of a 10-year loan that earns $1278 in simple interest at the annual rate of 6.7%.
Simple InterestThe interest is given by the formula ...
I = Prt
Solving for P gives ...
P = I/(rt)
P = $1278/(0.067·10) ≈ $1907.46
The amount of the loan was $1907.46.
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Drag each label to the correct location on the table. Each label can be used more than once, but not all labels will be used. Simplify the given polynomials. Then, classify each polynomial by its degree and number of terms.polynormial 1:[tex](x - \frac{1}{2})(6x + 2)[/tex]polynormial 2:[tex](7 {x}^{2} + 3x) - \frac{1}{3} (21 { x}^{2} - 12)[/tex]polynormial 3:[tex]4(5 {x}^{2} - 9x + 7) + 2( - 10 {x}^{2} + 18x - 03) [/tex]
Given the polynomials, let's simplify the polynomials and label them.
Polynomial 1:
[tex]\begin{gathered} (x-\frac{1}{2})(6x+2) \\ \text{Simplify:} \\ 6x(x)+2x+6x(-\frac{1}{2})+2(-\frac{1}{2}) \\ \\ =6x^2+2x-3x-1 \\ \\ =6x^2-x-1 \end{gathered}[/tex]After simplifying, we have the simplified form:
[tex]6x^2-x-1[/tex]Since the highest degree is 2, this is a quadratic polynomial.
It has 3 terms, therefore by number of terms it is a trinomial.
Polynomial 2:
[tex]\begin{gathered} (7x^2+3x)-\frac{1}{3}(21x^2-12) \\ \\ \text{Simplify:} \\ (7x^2+3x)-7x^2+4 \\ \\ =7x^2+3x-7x^2+4 \\ \\ \text{Combine like terms:} \\ 7x^2-7x^2+3x+4 \\ \\ 3x+4 \end{gathered}[/tex]Simplified form:
[tex]3x+4[/tex]The highest degree is 1, therefore it is linear
It has 2 terms, therefore by number of terms it is a binomial
Polynomial 3:
[tex]\begin{gathered} 4(5x^2-9x+7)+2(-10x^2+18x-13) \\ \\ \text{Simplify:} \\ 20x^2-36x+28-20x^2+36x-26 \\ \\ \text{Combine like terms:} \\ 20x^2-20x^2-36x+36x+28-26 \\ \\ =2 \end{gathered}[/tex]Simplified form:
[tex]2[/tex]The highest degree is 0 since it has no variable, therefore it is a constant.
It has 1 term, by number of terms it is a monomial.
ANSWER:
Polynomial Simplified form Name by degree Name by nos. of ter
1 6x²-x-1 quadratic Trinomial
2 3x + 4 Linear Binomial
3 2 Constant Monomial
Tim bought a new car for 25,000 one year later the value of the car decrease to 20,000 what is the percentage of the decrease of the car
Answer:
The percentage decrease in the value of the car is;
[tex]20\text{\%}[/tex]Explanation:
Given that the initial price of the car is;
[tex]25,000[/tex]And after one year the price decreased to;
[tex]20,000[/tex]The percentage change in the price will be;
[tex]\begin{gathered} \text{ \%P }=\frac{25000-20000}{25000}\times100\text{\%} \\ \text{ \%P }=\frac{5000}{25000}\times100\text{\%} \\ \text{ \%P }=0.2\times100\text{\%} \\ \text{ \%P }=20\text{\%} \end{gathered}[/tex]Therefore, the percentage decrease in the value of the car is;
[tex]20\text{\%}[/tex]Debra has ridden 6 miles of a bike course. The course is 15miles long. What percentage of the course has Debra ridden so far
From the question
Miles covered for bike course = 6 miles
Total miles of course = 15 miles
In percentage, this becomes
Let z = percentage of miles covered
Hence
[tex]z=\frac{\text{miles covered}}{Total\text{ miles}}\times100\text{\%}[/tex]Substitute in the values to get
[tex]z=\frac{6}{15}\times100\text{\%}[/tex]Simplify:
[tex]\begin{gathered} z=\frac{2}{5}\times100\text{\%} \\ z=2\times20\text{\%} \\ z=40\text{\%} \end{gathered}[/tex]Therefore, the percentage of the course Debra has ridden so far is 40%
9 x+3=9 3x=9 3+x=9 x=9-3 x=9=3 < those are the answers
Based on the diagram of the figure, the equation is:
x + 3 = 9
Trapezoid A'B'C'D was formed after a translation and reflection.YA61AB41D2The amount of unitstrapezoid ABCD wastranslated is the nextnumber of yourcombinationReturn-621D4Α'В"to the-6Dungeon
The trapezoid on the left side of the y-axis, is first of all moved 6 units down (negative 6 units on the y axis).
Next its moved 6 units to the right (positive 6 units on the x-axis).
Then it rotates 180 degrees clockwise, and the transformation from trapezoid ABCD to A'B'C'D' is complete.
Select ALL the expressions that have the same value as 9s
1. 9+s
2. 9*s
3. 3s + 6s
4. 3(3s)
5. s(5+4)
6. 3s * 3s
7. s+s+s+s+s+s+s+s+s2. 3s+2t=-4-6s-10t=-7What’s the value of s ?
ANSWER
2, 3, 4, 5, and 7 are the correct expressions that have the same values as 9s
STEP-BY-STEP EXPLANATION
Using Simultaneous equation
3s + 2t = -4 .............................................(1)
-6s - 10t = -7 .............................................(2)
multiply equation 1 by 2 and multiply equation 2 by 1:
equ 1 x 2: 6s + 4t = -8 ...............................(3)
equ 2 x 1: -6s -10t = -7 ................................(4)
Add equation 3 and 4 together
0s -6t = - 15
Divide through by -6:
[tex]\begin{gathered} t\text{ = }\frac{-15}{-6} \\ t\text{ = }\frac{5}{2} \end{gathered}[/tex]substitute the value of t into equation 1:
[tex]\begin{gathered} 3s\text{ + 2t = -4} \\ 3s\text{ + 2(}\frac{5}{2})\text{ = -4} \\ 3s\text{ + 5 = -4} \\ 3s\text{ = -4 -5} \\ 3s\text{ = -9 } \\ s\text{ = }\frac{-9}{3} \\ s\text{ = -3} \\ \end{gathered}[/tex]Now solving for the expression that has the same value as 9s:
Note: 9s = 9(-3) = -27
1. 9 + s = 9 - 3 = 6
2. 9 * s = 9 * -3 = -27
3. 3s + 6s = 3(-3) + 6(-3) = -9 -18 = -27
4. 3(3s) = 9s = 9(-3) = -27
5. s(5 + 4) = s(9) = -3(9) = -27
6. 3s * 3s = 3(-3) * 3(-3) = -9 * -9 = 81
7. s+s+s+s+s+s+s+s+s = -3-3-3-3-3-3-3-3-3 = -27
Hence, 2, 3, 4, 5, and 7 are the correct expressions that have the same values as 9s.
If: x+y+z=2-x+3y+2z=84x+y=4Find the value of x, y and z
We have
[tex]\begin{gathered} x+y+z=2 \\ -x+3y+2z=8 \\ 4x+y=4 \end{gathered}[/tex]We have with the third equation
[tex]y=4-4x[/tex]We substitute in the first and second equation
[tex]\begin{gathered} x+4-4x+z=2 \\ -3x+z=-2 \end{gathered}[/tex][tex]\begin{gathered} -x+3(4-4x)+2z=8 \\ -x+12-12x+2z=8 \\ -13x+2z=8-12 \\ -13x+2z=-4 \end{gathered}[/tex]Then we have
[tex]z=-2+3x[/tex]We substitute
[tex]\begin{gathered} -13x+2(-2+3x)=-4 \\ -13x-4-6x=-4 \\ -19x=0 \\ x=0 \end{gathered}[/tex]if x=0
[tex]z=-2[/tex]and if x=0
[tex]y=4[/tex]ANSWER
x=0
y=4
z=-2
The graph below shows an office worker's annual salary:What are the domain and range of the function? Why are the x-values nonnegative?
Okay, here we have this:
Considering that the domain refers to the possible values of x that can be substituted in the correspondence rule of a function. We can see in the function that the domain is [0, +∞) (because there is an arrow that indicates that it continues to increase).
And as the range is the set of numbers that depend on the substitution (tabulation) of the values that "x" can take. We can see that the range is [5000, +∞).
And the number of years is non-negative because the years that elapse are always counted forward, that is, they begin to count from zero.
Simplify using the distributive property.8(y + 12)8 y + 9620 + y8 y + 1220 y
Solution:
Concept:
The distributive property of multiplication states that when a number is multiplied by the sum of two numbers, the first number can be distributed to both of those numbers and multiplied by each of them separately, then adding the two products together for the same result as multiplying the first number by the sum.
The expression is given below as
[tex]\begin{gathered} 8(y+12) \\ =8\times y+8\times12 \\ =8y+96 \end{gathered}[/tex]Hence,
The final answer is
[tex]\Rightarrow8y+96[/tex]Which ratio of cups of banana to cups of apple juice is also equivalent to ¼:⅓?• 4/4 : 3/3• 3/3 : 3/4• 3/4 : 4/3• 3/4 : 3/3
ANSWER
3/4 : 3/3
EXPLANATION
We have the ratio of cups of banana to cups of apple juice to be 1/4 : 1/3
Equivalent ratios can be gotten by multiplying the ratio with a common integer.
This means we can multiply bot sides of the ratio with the same integer e.g. 2, 4, 7...
Since from the diagram, the next equivalent ratio is given (2/4 : 2/3), we can obtain the next equivalent ratio by multiplying the ratio by 3. That is:
3/4 : 3/3
Therefore, the correct option is 3/4 : 3/3
You TryWrite an equation for each of the following,then solve for the variable.20 is the same as the sum of 4 and g.
Given statement:
20 is the same as the sum of 4 and g
Let us break down the statement into parts and then write the equation
the sum of 4 and g:
[tex]\text{= 4 + g}[/tex]This sum is equal to 20:
[tex]4\text{ + g = 20}[/tex]Hence, the equation is:
[tex]4\text{ + g = 20}[/tex]Solving for the variable:
[tex]\begin{gathered} \text{Collect like terms} \\ g\text{ = 20 -4} \\ g\text{ = 16} \end{gathered}[/tex]Answer Summary
[tex]\begin{gathered} \text{equation: 4 + g = 20} \\ g\text{ = 16} \end{gathered}[/tex]What is the probability of a person who plays equation IQ being in grade 6? Enter the answer as a percentage, round to the nearest tenth place.
ANSWER
28.3%
EXPLANATION
The total number of people that play Equation IQ is 300 - we can know this by adding either the more right column or the lowest row of the table.
From those people, 85 are in grade 6. The probability of a person that plays Equation IQ being in grade 6 is:
[tex]P(\text{grade 6})=\frac{\#\text{ people in grade 6}}{\#\text{ total number of players}}=\frac{85}{300}\approx0.2833\ldots[/tex]To write it as a percentage we just have to multiply it by 100:
[tex]0.283\times100=28.3\text{ \%}[/tex]Solve F=mv^2/R for V
SOLUTION
We want to solve for v in
[tex]F=\frac{mv^2}{R}[/tex]This means we should make v the subject, that is make it stand alone. This becomes
[tex]\begin{gathered} F=\frac{mv^2}{R} \\ m\text{ultiply both sides by }R,\text{ we have } \\ F\times R=\frac{mv^2}{R}\times R \\ R\text{ cancels R in the right hand side of the equation we have } \\ FR=mv^2 \end{gathered}[/tex]Next, we divide both sides by m, we have
[tex]\begin{gathered} FR=mv^2 \\ \frac{FR}{m}=\frac{mv^2}{m} \\ m\text{ cancels m, we have } \\ \frac{FR}{m}=v^2 \\ v^2=\frac{FR}{m} \end{gathered}[/tex]Lastly, we square root both sides we have
[tex]\begin{gathered} v^2=\frac{FR}{m} \\ \sqrt[]{v^2}=\sqrt[]{\frac{FR}{m}} \\ \text{square cancels square root, we have } \\ v=\sqrt[]{\frac{FR}{m}} \end{gathered}[/tex]Hence the answer is
[tex]v=\sqrt[]{\frac{FR}{m}}[/tex]Help me with number 4 please Identify the 17th term of a geometric sequence where a1 = 16 and a5 = 150.06 Round the common ratio and 17th term to the nearest hundredth.
Common ratio = 1.75
17th term = 123,802.31
Explanations:Given the following parameters:
[tex]\begin{gathered} a_1=16 \\ a_5=150.06 \end{gathered}[/tex]Since the sequence is geometric, the nth term of the sequence is given as;
[tex]a_n_{}=a_{}r^{n-1}[/tex]a is the first term
r is the common ratio
n is the number of terms
If the first term a1 = 16, then;
[tex]\begin{gathered} a_1=ar^{1-1}_{} \\ 16=ar^0 \\ a=16 \end{gathered}[/tex]Similarly, if the fifth term a5 = 150.06, then;
[tex]\begin{gathered} a_5=ar^{5-1} \\ a_5=ar^4 \\ 150.06=16r^4 \\ r^4=\frac{150.06}{16} \\ r^4=9.37875 \\ r=1.74999271132 \\ r\approx1.75 \end{gathered}[/tex]Hence the common ratio to the nearest hundredth is 1.75
Next is to get the 17th term as shown;
[tex]\begin{gathered} a_{17}=ar^{16} \\ a_{17}=16(1.75)^{16} \\ a_{17}=16(7,737.6446) \\ a_{17}\approx123,802.31 \end{gathered}[/tex]Hence the 17th term of the sequence to the nearest hundredth is 123,802.31
someone help me
please and thank you
Answer: Answer C
Step-by-step explanation:
Because it is a reflection of the Y-axis, the X-coordinates would remain the same but the Y-coordinates would change.
how do I solve them to know what is the correct answer
For:
[tex]8x^3+16x^2[/tex]Factor 8x² out of the expression:
[tex]8x^2(x+2)[/tex]-------------------------------------------------------------------
For:
[tex]\begin{gathered} 2x^2-x+8x-4 \\ \end{gathered}[/tex]Add like terms:
[tex]2x^2+7x-4[/tex]The coefficient of x² is 2 and the constant term is -4. The product of 2 and -4 is -8. The factors of -8 which sum to 7 are -1 and 8, thus:
[tex]\begin{gathered} 2x^2+7x-4=4(2x-1)+x(2x-1) \\ so\colon \\ 4(2x-1)+x(2x-1)=(2x-1)(x+4) \end{gathered}[/tex]Show that if the diagonals of a quadrilateral bisect each other at right angles then it is a rhombus.
To Show that if the diagonals of a quadrilateral bisect each other at right angles then it is a rhombus.
Proof:
Let ABCD be a quadrilateral such that the diagonals bisect each other,
Therefore,
[tex]\begin{gathered} OA=OC\ldots(1) \\ OB=OD\ldots(2) \end{gathered}[/tex]the diagonal bisect at right angle.
Hence,
[tex]\begin{gathered} \angle AOB=90^{\circ} \\ \angle BOC=90^{\circ} \\ \angle COD=90^{\circ} \\ \angle AOD=90^{\circ}\ldots(3) \end{gathered}[/tex]to prove: ABCD is rombus,
Rombus: its is a parallelogram, with all the sides equal.
so, to prove ABCD a parallelogram.
consider the triangle,
[tex]\begin{gathered} triangleAOD\text{ and triangle }COB, \\ OA=OC \\ \angle AOD=\angle COB \\ OD=OB \\ \end{gathered}[/tex]thus, traingle
[tex]\text{AOD}\cong COB[/tex]consider the sides, AD and BC
with the transversal ac,
The angles,
[tex]\angle OAD\text{ AND }\angle OCB[/tex]are alterntaive angles. they are equal.
this implies, AD is parallel BC.
similarly, AB is parallel to DC.
Hence, AD II BC and AB II DC.
In ABCD the opposite sides are parallel,
This implies, ABCD is parallelogram.
Now, to prove that ABCD is a rombus.
for that all the sides of ABCD should be equal.
now, consider the triangle AOD and COD.
[tex]\begin{gathered} OA=OC \\ \angle AOD=\angle COD \\ OD=OD\text{ common side} \end{gathered}[/tex]By SAS congruent rule,
Traingles,
[tex]AOD\cong COD[/tex]Thus, by CPCT Corresponding parts of congruent triangles ,
AD= CD
we know that,
AD=CB and CD=AB
Thus, AD=CD=CB=AB.
hence, all the sides are eqaul and ABCD is parallelogram.
So, ABCD is a rhombus.