ray OJ bisects angle IOK, if angle IOJ= 2x-5 and measure of angle JOK=x+11, then find measure of angle IOJ
Given ray OJ bisects angle IOK, we can set an equation for angles IOJ and JOK, solve for x and substitute x into the given measure of angle IOJ: 2*(16)-5 equals 27 degrees.
Explanation:Since ray OJ bisects angle IOK, it means that angle IOJ and angle JOK are equal in measurement. Given in the problem, angle IOJ= 2x-5 and angle JOK= x+11. Because they are equal, we can set up the equation 2x-5 = x+11.
To solve for x, we subtract x from both sides of the equation, resulting in x-5 =11. And then, if we add 5 to both sides of the equation, the end result is x = 16.
Substituting x into the given measurement for angle IOJ, which is 2x-5, we would get the measure of angle IOJ as 2*(16)-5 = 27 degrees.
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Evaluate the expression for a = 2 and b = 5. 100 20 50 625
Micheala mixed peanuts and almonds. · Micheala can buy 3 pounds of peanuts for a total cost of $8.70. · The cost per pound for almonds is 80% more than the cost per pound for peanuts. · Micheala bought enough peanuts that, when he mixed them with the almonds, the mixture had a value of $3.94 per pound. What is the approximate percent by weight of almonds?
An isosceles triangle’s altitude will bisect its base. Which expression could be used to find the area of the isosceles triangle above?
The expression could be used to find the area of the isosceles triangle above is [tex]\sqrt{40} \cdot \sqrt{40} /2[/tex]
The length of the base is the distance between the points 4+2i and 10+4i, so
Base= |10+ 4i (4+2i)| = |10+4i-4-2i|= |6 + 2i| = [tex]\sqrt{6^2 +2^2}[/tex] = [tex]\sqrt{36+4}[/tex]= √40
The middle point of the base is placed at point
4+2i+ 10 + 4i/2 = 6i +14/2 = 7+ 3i
The length of the height is the distance between the points 5+9i and 7+3i
Height = 5 +91 (7+3i)| =|5+ 9i −7 - 3i| = |−2+6i| = [tex]\sqrt{(-2)^2 + 6^2} = \sqrt {4+36}[/tex] = √40
So, the area of the triangle is
[tex]A= 1/2 \cdot Base \cdot Height= \sqrt{40} \cdot \sqrt{40} /2[/tex]
Therefore, The expression could be used to find the area of the isosceles triangle above is [tex]\sqrt{40} \cdot \sqrt{40} /2[/tex]
The probable question may be:
An isosceles triangle’s altitude will bisect its base. Which expression could be used to find the area of the isosceles triangle above?
Points on the graph of the triangle are (5+9i), (10+4i), and (4+2i).
A. \sqrt{40} \cdot \sqrt{40} /2
B. \sqrt{40} \cdot \sqrt{68} /2
C. \sqrt{232} \cdot \sqrt{288} /2
D. \sqrt{232} \cdot \sqrt{164} /2
Final answer:
To find the area of an isosceles triangle when given the length of the base and the altitude, you can use the formula A = ½ × base × height. The altitude of an isosceles triangle will bisect its base, so you can divide the base in half to find the length of the base above the altitude.
Explanation:
An isosceles triangle's altitude will bisect its base. To find the area of the isosceles triangle, we can use the formula A = ½ × base × height. Since the altitude bisects the base, we can divide the base in half to find the length of the base above the altitude. Let's say the length of the base is 2x and the length of the altitude is h. So, the expression to find the area of the isosceles triangle is A = ½ × (2x) × h = xh.
Lonnie ordered 12$ copies of the same book.h3 books cost 19$ each and the order has a 15$ charge.what is the total cost of Lonnie's order?
HELP PLEASE SOS :))
PLease PLease help with the questions below
Multiply. (7x+3)(4x−5) Enter your answer, in standard form, in the box.
(7x+3) x (4x-5)
7x*4x = 28x^2
7x*-5 = -35x
3+4x = 12x
3*-5 = -15
so you get 28x^2 -23x -15
Answer: The required multiplied expression is [tex]28x^2-23x-15.[/tex]
Step-by-step explanation: We are given to multiply the following linear factors and write the answer is standard form :
[tex]M=(7x+3)(4x-5)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We will be using the following distributive property :
[tex](a+b)(c+d)=a(c+d)+b(c+d).[/tex]
From (i), we get
[tex]M\\\\=(7x+3)(4x-5)\\\\=7x(4x-5)+3(4x-5)\\\\=28x^2-35x+12x-15\\\\=28x^2-23x-15.[/tex]
Thus, the required multiplied expression is [tex]28x^2-23x-15.[/tex]
HELP ME PLEASE ! I NEED TO DO THIS QUICKLY !!!!! Translate each sentence into an equation!!!!!!!!!!!! 7 berries are 5 less than twice them number of berries Mickey had for lunch! AND also another one Negative 4 times the difference of a number and 7 is 12
A professional basketball player makes 80 % of the free throws he tries. assuming this percentage will hold true for future attempts, find the probability that in the next 8 tries, the number of free throws he will make is exactly 8.
need help from a math wiz
Factor the trinomial below. x2 – 2x – 35 A. (x – 5)(x + 7) B. (x – 5)(x – 7) C. (x + 5)(x – 7) D. (x + 5)(x + 7)
What is two times a number n is three times the sum of n and nine
Please help! Use the quadratic function to predict y if x equals 6. Y=2x^2-2x-2
A) y= -58
B) y= 58
C) y=-60
D) y= 60
Answer:
The correct option is B.
Step-by-step explanation:
The given quadratic function is
[tex]y=2x^2-2x-2[/tex]
We have to find the value of the function y at x=6.
Substitute x=6 in the given function.
[tex]y=2(6)^2-2(6)-2[/tex]
[tex]y=2(36)-12-2[/tex]
[tex]y=72-14[/tex]
[tex]y=58[/tex]
The value of the function is 58 at x=6 is 58.
Therefore the correct option is B.
Se the change of variables s=xy, t=xy^2 to compute \iint_r xy^2\,da, where r is the region bounded by xy=3,\ xy=7,\ xy^2=3,\ xy^2=7.
The value of [tex]\int\limits^._R {xy^2} \, dA[/tex] is 16.
What is integration?The calculation of an integral is called integration. In math, many useful quantities like areas, volumes, displacement, and so on can be found using integrals. When we discuss integrals, we typically refer to definite integrals. Antiderivatives make use of the indefinite integrals. Apart from differentiation, one of the two major calculus topics in mathematics is integration.
Given xy =3, xy = 7,
xy² = 3, xy² = 7
s = xy and t = xy²
dividing t by s we get
t/s = xy²/xy
y = t/s
and x = s/y = s²/t
now differentiate x and y partially with respect to s and t
∂x/∂s = 2s/t
∂x/∂t = -s²/t²
∂y/∂s = -t/s²
∂y/∂t = 1/s
The Jacobian is
∂(x, y)/∂(s, t) = [tex]\left|\begin{array}{ccc}dx/ds&dx/dt\\dy/ds&dy/dt\end{array}\right|[/tex]
∂(x, y)/∂(s, t) =[tex]\left|\begin{array}{ccc}2s/t&-s^{2}/t^{2} \\-t/s^{2} &1/s\end{array}\right|[/tex]
∂(x, y)/∂(s, t) =2/t - 1/t
∂(x, y)/∂(s, t) = 1/t
so for [tex]\int\limits^._R {xy^2} \, dA[/tex] = [tex]\int\limits^a_b \int\limits^a_b {t}\frac{d(x, y)}{d(s, t) } \, dsdt[/tex]
for (s, t) (3 ≤ s ≤7, 3 ≤ t ≤ 7)
= [tex]\int\limits^7_3 \int\limits^7_3 {t}*1/t } \, dsdt[/tex]
= [tex]\int\limits^7_3ds \int\limits^7_3 dt[/tex]
=[s]₃⁷ [t]₃⁷
= (7 - 3)(7 - 3)
= 16
Hence the value is 16.
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Which undefined geometric term is described as a location on a coordinate plane that is designated by an ordered pair, (x, y)?
Which is the directrix of a parabola with equation x^2=y
Which statement about the equation is true? 3y – 1 = 1/3 – 4y
The equation has one solution.
The equation has no solution.
The equation has a few solutions.
The equation has many solutions.
A donkey weighs 570 pounds and an elephant weighs 5 tons. How much more does the elephant weigh?
The correct answer is:
5 tons = 10,000 lb
10,000 lb − 570 lb = 9,430 lb = 4 tons and 1,430 lb
This answer is if you want it in tons and pounds.
Find the area of the parallelogram when base = 10yd and height = 5yd. A. 15yd B. 15yd2 C. 50yd D. 50yd2
area = l x w
10 * 5 = 50
answer is D. 50 yd^2
At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not exist, enter dne.) (x, y, z) =
Final answer:
The point on the paraboloid y = x^2 + z^2 where the tangent plane is parallel to the plane 3x + 2y + 7z = 2 is found by setting the normal vectors proportional. Solving the equations, the point is (3/4, 29/8, 7/4).
Explanation:
To find the point on the paraboloid y = x2 + z2 where the tangent plane is parallel to the plane 3x + 2y + 7z = 2, we first need to determine the normal vector of the given plane. The normal vector of the plane is defined by its coefficients, which are (3, 2, 7). For the paraboloid, we can find the normal vector at any point by taking the gradient of the function y.
The gradient of y with respect to x and z is (2x, 1, 2z). A tangent plane to the paraboloid at point (x, y, z) will have this gradient as its normal vector. To find the point where this tangent plane is parallel to the given plane, we set the gradients equal to each other up to a constant factor because parallel planes have proportional normal vectors.
Therefore, we solve the equations:
2x = 3k
1 = 2k
2z = 7k
From the second equation, k = 1/2. Substituting k into the other equations, we find x = 3/4 and z = 7/4. Now we can substitute x and z into the equation of the paraboloid to find y:
y = (3/4)2 + (7/4)2 = 9/16 + 49/16 = 58/16 = 29/8.
The point on the paraboloid where the tangent plane is parallel to the given plane is (3/4, 29/8, 7/4).
Is it possible to draw a graph of this system? If yes, draw it.
[tex] \left \{ {{3x-1=0} \atop {4x+2y=0}} \right. [/tex]
Thanks in adavance!
How many license plates can be made using 3 digits and 4 letters if repeated digits and letters are not allowed?
Liam earns $7.50 an hour. His benefits package is equal to 25 percent of his hourly wages. When you include the value of his benefits, how much does Liam earn per hour?
translate the word phrase nto a variable expression the quotient of a number and 3 is increased by 1
25 POINTS PLEASE HELP SOMEONE WHO CAN
Solve the system by substitution
-x-y-z=-8
-4x+4y+5z=7
2x+2z=4
1. Which domain restrictions apply to the rational expression?
x^2+5x+6
--------------
x^2-9
Select each correct answer.
x≠2
x≠1
x≠−3
x≠3
x≠−2
x≠−1
(4x)/(x-3)+(2)/(x^(2)-9)= (1)/(x+3)
A. 19/3
B. 13/3
C. -5
D. -17/3
9. Which statement best reflects the solution(s) of the equation?
1/x + 1/x-3 = x-2 / x-3
A. There is only one solution: x = 1.
The solution x = 0 is an extraneous solution.
B. There is only one solution: x = 1.
The solution x = 3 is an extraneous solution.
C. There is only one solution: x = 3.
The solution x = 1 is an extraneous solution.
D. There are two solutions: x = 1 and x = 3.
10. (√3x+1)−x+3=0
A. There is only one solution: x = 1.
The solution x = 8 is an extraneous solution.
B. There is only one solution: x = 8.
The solution x = 1 is an extraneous solution.
C. There is only one solution: x = 8.
The solution x = 0 is an extraneous solution.
D. There are two solutions: x = 1 and x = 8.
Answer with explanation:
1. The given rational expression is
[tex]\rightarrow\frac{x^2+5 x+6}{x^2-9}\\\\\rightarrow\frac{x^2+5 x+6}{(x-3)(x+3)}[/tex]
The function is not defined ,when
→(x-3)(x+3)=0
→x-3≠0 ∧ x+3≠0
→x≠3, ∧ x ≠ -3
⇒Option C and D
→x≠−3
→x≠3
2.
[tex]\rightarrow\frac{4 x}{x-3}+\frac{2}{x^2-9}=\frac{1}{x+3}\\\\\rightarrow\frac{4 x}{x-3}+\frac{2}{(x-3)(x+3)}=\frac{1}{x+3}\\\\\rightarrow\frac{4 x(x+3)+2}{(x-3)(x+3)}=\frac{1}{x+3}\\\\\rightarrow4 x(x+3)+2=\frac{(x-3)(x+3)}{x+3}\\\\\rightarrow4x^2+12 x+2=x-3\\\\\rightarrow4x^2+11x+5=0\\\\ \text{Using Discriminant method for a quadratic equation}\\\\ax^2+bx +c=0\\\\x=\frac{-b\pm\sqrt{D}}{2 a}\\\\D=b^2-4 ac\\\\x=\frac{-11\pm\sqrt{121-80}}{2 \times 4}\\\\x=\frac{-11\pm\sqrt{41}}{8}[/tex]
None of the option
3.
[tex]\rightarrow \frac{1}{x}+\frac{1}{x-3}=\frac{x-2}{x-3}\\\\\rightarrow\frac{x-3+x}{x(x-3)}=\frac{x-2}{x-3}\\\\\rightarrow2x-3=\frac{x(x-3)(x-2)}{x-3}\\\\\rightarrow 2 x-3=x^2-2 x\\\\\rightarrow x^2-4x+3=0\\\\\rightarrow (x-1)(x-3)=0\\\\x=1,3[/tex]
For, x=3 , the equation is not defined.
So, there is single solution which is , x=1.
Option B:→ There is only one solution: x = 1.
The solution x = 3 is an extraneous solution.
4.
[tex]\rightarrow \sqrt{3}x+1-x+3=0\\\\\rightarrow \sqrt{3}x -x=-4\\\\\rightarrow x(\sqrt{3}-1)=-4\\\\\rightarrow x=\frac{-4}{\sqrt{3}-1}\\\\\rightarrow x=\frac{-4\times(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\\\x=\frac{-4\times(\sqrt{3}+1)}{2}\\\\x=-2(\sqrt{3}+1)[/tex]
None of the option
1) Options c) and d) are correct.
2) None of the options are correct.
3) Option B) is correct.
4) None of the options are correct.
Step-by-step explanation:
1) Given : [tex]4x^2+12x+2=x-3[/tex]
Expression -- [tex]\dfrac{x^2+5x+6}{x^2-9}[/tex]
Solution :
[tex]\dfrac{x^2+5x+6}{x^2-9}=\dfrac{x^2+3x+2x+6}{(x+3)(x-3)}[/tex]
[tex]\dfrac{x^2+5x+6}{x^2-9}=\dfrac{(x+3)(x+2)}{(x+3)(x-3)}[/tex]
Therefore, [tex]\rm x\neq 3 \; and\;x\neq -3[/tex] ,option c) and d) is correct.
2) Given :
Expression - [tex]\dfrac{4x}{x-3}+\dfrac{2}{x^2-9}=\dfrac{1}{x+3}[/tex]
Solution :
[tex]\dfrac{4x}{(x-3)}+\dfrac{2}{(x+3)(x-3)}=\dfrac{1}{(x+3)}[/tex]
[tex]\dfrac{4x(x+3)+2}{(x-3)(x+3)}=\dfrac{1}{(x+3)}[/tex]
[tex]4x^2+12x+2=x-3[/tex]
[tex]4x^2+11x+5=0[/tex]
[tex]x=\dfrac{-11\pm\sqrt{121-80} }{8}[/tex]
[tex]x = \dfrac{-11\pm\sqrt{41} }{8}[/tex]
None of the options are correct.
3) Given :
Expression - [tex]\dfrac{1}{x}+\dfrac{1}{x-3}=\dfrac{x-2}{x-3}[/tex] ----- (1)
Solution :
[tex]\dfrac{x-3+x}{(x)(x-3)}=\dfrac{x-2}{x-3}[/tex]
[tex]2x-3=x(x-2)[/tex]
[tex]x^2-4x+3=0[/tex]
[tex]x^2-3x-x+3=0[/tex]
[tex](x-3)(x-1)=0[/tex]
At x = 3 equation (1) is not define. Therefore, the correct answer is option
B) There is only one solution: x = 1. The solution x = 3 is an extraneous solution.
4) Given :
Exprression - [tex](\sqrt{3}x +1)-x+3=0[/tex]
Solution :
[tex]x(\sqrt{3}-1 )=-4[/tex]
[tex]x=\dfrac{-4}{\sqrt{3}-1 }[/tex]
None of the options are correct.
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Mr. lance keeps a jar of candy on his desk for his students to pick from when they do well on a test. The jar contain 5 snickers, 2 butterfingers, 4 almond joys and 3 milky ways. If two students get to pick candy from the jar, what is the probability that the first student picks a snickers and then a second student also picks a snickers?
The probability that the first student picks a Snickers and then the second student also picks a Snickers is 10/91.
Explanation:The question is asking for the probability that two students will each pick a Snickers from a jar containing various types of candy. We start with a total of 14 candies: 5 Snickers, 2 Butterfingers, 4 Almond Joys, and 3 Milky Ways. When the first student picks a Snickers, there are 5 possible Snickers they can pick out of the 14 total candies.
The probability for the first student is therefore rac{5}{14}. Assuming they pick a Snickers, there are now 13 candies left in total, with 4 being Snickers. So, the second student now has a probability of rac{4}{13} to pick a Snickers.
To find the overall probability of both events happening, we multiply the probabilities of each individual event occurring. The calculation is:
rac{5}{14} \times rac{4}{13} = rac{20}{182} = rac{10}{91}
The probability that the first student picks a Snickers and then the second student also picks a Snickers is rac{10}{91}.
2 sides of an isoceles triangle are 2 and 12. find the length of the third side
The third side of the isosceles triangle with the given sides of 2 and 12 is 12 units.
To find the length of the third side of an isosceles triangle where two sides are given as 2 and 12, we must consider two cases.
In an isosceles triangle, two sides are equal in length and the two angles opposite these sides are also equal. This means either the two given sides are the equal sides, which is not possible since they are different lengths, or one of the given lengths will be the base and the equal sides are yet to be determined.
Case 1: If the side of length 2 is the base, the length of the equal longer sides will both be 12.
Case 2: If the side of length 12 is the base, then the length of the two equal sides must both be 2.
However, we must recognize that the first case is incorrect because it does not form a triangle (a triangle cannot have two sides of length 12 and one side of length 2, as the two longer sides would be parallel and never meet to form a triangle).
Therefore, the correct answer is given by Case 2: the third side of the triangle, which is the base, is 12 units long, and the two equal sides are 2 units long each.
Find the polynomial.
{-1/3, 4} is the solution set of?
A. 3x^2 - 11x + 4 = 0
B. 3x^2 - 11x - 4 = 0
C. 1/3x^2 - 11x - 4 = 0
D.-1/x^2 - 11x - 4 = 0
Answer:
option B
Step-by-step explanation:
A. [tex]3x^2 - 11x + 4 = 0[/tex]
3*4 = 12
We find out two factors whose sum is -11 and product is 12
1 times 12 = 12
To get sum -11 , then one factor should be negative. So, factoring is not possible.
B . [tex]3x^2 - 11x - 4 = 0[/tex]
3*(-4) = -12
We find out two factors whose sum is -11 and product is -12
1 times (-12) = -12
1 + (-12) = -11
So two factors are 1 and -12
Split the middle term -11x using factors 1 and -12
So equation becomes
[tex]3x^2 + 1x - 12x - 4 = 0[/tex]
Now group first two terms and last two terms
[tex](3x^2 + 1x)+ (- 12x - 4) = 0[/tex]
[tex]x(3x+ 1)-4(3x +1)=0[/tex]
(3x+1)(x-4)=0
Now we set each parenthesis =0 and solve for x
3x+1 =0 , subtract 1 on both sides
3x = -1 ( divide both sides by 3)
x= -1/3
Now we set x-4=0
add 4 on both sides
so x=4
Option B is correct
A line goes through the points (8,9) and (-2,4) .
(a) What is the slope of the line? Show your work
(b) Write the equation of the line in point-slope form. Show your work
(c) Write the equation of the line in slope-intercept form.
show steps
The equation of the line in slope-intercept form would be; y=1/2x+5.
What is the slope?The slope is the ratio of the vertical changes to the horizontal changes between two points of the line.
(a) The slope of line passing through two points is given by;
[tex]m = \dfrac{y-q}{x-p}[/tex]
Here m = 4 - 9/ -2 - 8
m = 5/10
m = 1/2
(b) The equation of line passing through a point and having slope is given by ;
y-y₁ = m(x-x₁)
Here, y-4 = 1/2(x+2)
(c) The slope intercept form;
y-4=1/2(x+2)
y-4=1/2x+1
y=1/2x+5
Hence, the equation of the line in slope-intercept form would be; y=1/2x+5.
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