Given:
Required:
We need to find the points are included in the solution
Explanation:
Recall that the solution of the system of inequalities is the intersection region of all the solutions in the system.
The points G and F lie inside the intersection.
The points in the solution are G and F.
Final answer:
Points F and G.
A shoe salesman earns a commission of 30%
of all shoe sales made.
Yesterday he sold 3 pairs of shoes for $70 each and 2 pairs of shoes for $80
each. How much did he earn in commission yesterday?
Answer: $111 is earn by shoe salesman as commission .
Step-by-step explanation:
As given the statement in the question be as follow.
Shoes salesman sold 3 pairs of shoes for $70 each and 2 pairs of shoes for $80 each.
Total cost of the pair of shoes = 3 × 70 + 2 × 80
= 210 + 160
= $ 370
As given
shoe salesman earns a commission of 30% of all shoe sales made.
30% is written is decimal form
= 0.30
Commission earns = 0.30 × Total cost of the pair of shoes .
= 0.30 × 370
= $ 111
Therefore $111 is earn by shoe salesman as commission .
what property tells us that m
Reflexive Property
1) For this assertion m∠GHK ≅ m∠GHK we have the Reflexive Property, which states that the same segment or geometric entity has the same measure.
"A quantity is congruent to itself"
m∠GHK ≅ m∠GHK
a =a
In triangle ABC, angle A is 44 degrees and angle B is 76 degrees. What is the measure of the third angle?
Answer:
60 degrees
Step-by-step explanation:
Total of angles of a triangle is 180
180-76-44= 60
Julie is 6 feet tall if she stands 15 feet from the flagpole and holds a cardboard square the edges of the square light up with the top and bottom of the flagpole approximate the height of the flagpole
Using tangent function:
[tex]\begin{gathered} \tan (\theta)=\frac{opposite}{adjacent} \\ \frac{6}{15}=\frac{15}{x-6} \\ solve_{\text{ }}for_{\text{ }}x\colon \\ 6(x-6)=15^2 \\ 6x-36=225 \\ 6x=225+36 \\ 6x=261 \\ x=\frac{261}{6} \\ x=43.5ft \end{gathered}[/tex]What is the length of the side adjacent to angle 0?
To answer this question, we always need to take into account the reference angle in a right triangle. The reference angle here is theta, Θ, and we have that:
Then, the length of the side adjacent to theta is equal to 15.
In summary, we have that the length of the side adjacent to the angle Θ is equal to 15.
Rick's average score on his first three tests in math is 80. What must he score on his next test to raise his average to 84?
SOLUTION
Now, we don't know the scores for his first three tests. But we are told that the average score for the first three tests was 80.
So, let the scores of the first three tests be a, b, and c. That means
[tex]\frac{a+b+c}{3}=80[/tex]Also, let's assume the total score for his first three tests was x, This means that
[tex]\begin{gathered} a+b+c=x \\ or \\ x=a+b+c \end{gathered}[/tex]Comparing with the first equation it means that
[tex]\begin{gathered} \frac{a+b+c}{3}=80 \\ \frac{x}{3}=80 \\ x=3\times80 \\ x=240 \end{gathered}[/tex]Now we are asked "What must he score on his next test to raise his average to 84?"
So this means the total tests becomes 4. Hence
[tex]\begin{gathered} \frac{a+b+c+d}{4}=84 \\ \frac{x+d}{4}=84 \\ \frac{240+d}{4}=84 \\ 240+d=84\times4 \\ 240+d=336 \\ d=336-240 \\ d=96 \end{gathered}[/tex]So he must score 96 to raise his average score to 84.
Hence, the answer is 96
5|x +1| + 7 = -38
Solve for x
Answer: No solutions
Step-by-step explanation:
[tex]5|x+1|+7=-38\\\\5|x+1|=-45\\\\|x+1|=-9[/tex]
However, as absolute value is non-negative, there are no solutions.
4. Which of the following rules is the composition of a dilation of scale factor 2 following (after) a translation of 3 units to the right?
ANSWER
A. (2x + 3, 2y)
EXPLANATION
Let the original coordinates be (x, y)
First, there was a dilation of scale factor 2.
This means that the coordinates become:
2 * (x, y) => (2x, 2y)
Then, there was a translation of 3 units to the right. That is a translation of 3 units on the horizontal (or x axis).
That is:
(2x + 3, 2y)
So, the answer is option A.
Segment AB and segment CD intersect at point E. Segment AC and segment DB are parallel.
To begin we shall sketch a diagram of the line segments as given in the question
As depicted in the diagram, line segment AC is parallel to line segment DB.
This means angle A and angle B are alternate angles. Hence, angle B equals 41 degrees. Similarly, angle C and angle D are alternate angles, which means angle C equals 56.
Therefore, in triangle EAC,
[tex]\begin{gathered} \angle A+\angle C+\angle AEC=180\text{ (angles in a triangle sum up to 180)} \\ 41+56+\angle AEC=180 \\ \angle AEC=180-41-56 \\ \angle AEC=83 \end{gathered}[/tex]The measure of angle AEC is 83 degrees
Create a polynomial of degree 6 that has no real roots. Explain why it has no real roots. Is it possible to have a polynomial with an odd degree that has no real roots? Explain.
Create a polynomial of degree 6 that has no real roots.
y = ( x^2 + 4) ( x^2 +7 ) ( x^2+5)
Multiplying all the terms together
y =x ^6 + 16 x^4 + 83 x^2 + 140
Using the zero product property
0= x^2 +4 x^2+7 =0 x^2 + 5 =0 will each give a complex solution
x^2 = -4 x^2 = -7 x^2 = -5
This means x = 2i or -2i x = i sqrt(7) or -i sqrt(7) x = i sqrt (5) or - i sqrt(5)
These solutions can be in the form a+bi
Therefor it will have no real roots
y = x^6 + 16 x^4 + 83 x^2 + 140 has no real solutions
Complex solutions come in pairs, so an odd degree must have a real solution
Which of the following is NOT an equation?1. 5(2x+1)=10x+52. 4x-13. 5+3=104. x/2+1=7
By definition, an equation is a statement that two mathematical expressions are equal.
Equations always contain the equal sign "="
Out of the 4 expressions listed, number 2. does not contain the equal sign, which means that this expression is not an equation.
All other expressions contain the equal sign, they can be considered equations.
When point A'(-2,4) is reflected over they-axis, where is the image A"?(2,-4)(2,4)(4,-2)
Answer:
Explanation
Given a coordinate (x, y), If this coordinate reflected over y axis, the resulting coordinate will be expressed as (-x, y). Note that only the sign of the x coordinate axis was
In a nearby park, a field has been marked off for the neighborhood Pop Warner football team. If the field has a perimeter of 310 yd and an area of 4950 yd', what are the dimensions of the field?
Answer:
The dimension of the field is ( 110 x 45)
Exolanations:
Perimeter of the field, P = 310 yd
Area of the field, A = 4950 yd²
Note that the shape of a field is rectangular:
Perimeter of a rectangle, P = 2(L + B)
Area of a rectangle, A = L x B
Substituting the values of the perimeter, P, and the Area, A into the formulae above:
310 = 2(L + B)
310 / 2 = L + B
155 = L + B
L + B = 155...............................................(1)
4950 = L x B...............(2)
From equation (1), make L the subject of the formula:
L = 155 - B...................(3)
Substitute equation (3) into equation (2)
4950 = (155 - B) B
4950 = 155B - B²
B² - 155B + 4950 = 0
Solving the quadratic equation above:
B² - 110B - 45B + 4950 = 0
B (B - 110) - 45(B - 110) = 0
(B - 110) ( B - 45) = 0
B - 110 = 0
B = 110
B - 45 = 0
B = 45
Substitute the value of B into equation (3)
L = 155 - B
L = 155 - 45
L = 110
The dimension of the field is ( 110 x 45)
pls help me i’ll give ty brainlist
[tex]\sqrt{25*7x^{4}*x^{1} } \\=\sqrt{25*x^{4}*7 *x^{1} } \\=\sqrt{25x^{4} } *\sqrt{7x}\\=5x^{2} \sqrt{7x}[/tex]
Option B is the answer.
Answer: C
i hope you can see my handwriting
Express the repeating decimal 0.2 as a fraction
Answer:
The fraction form of the repeating decimal is;
[tex]\frac{2}{9}[/tex]Explanation:
We want to express the repeating decimal 0.2 (2 repeating) as a fraction.
let x represent the fraction;
[tex]\begin{gathered} x=0.2222\ldots \\ 10x=2.222\ldots \end{gathered}[/tex]Then subtract x from 10x;
[tex]\begin{gathered} 10x-x=2.222\ldots-0.222\ldots \\ 9x=2.0 \end{gathered}[/tex]Then we can divide both sides by the coefficient of x;
[tex]\begin{gathered} \frac{9x}{9}=\frac{2}{9} \\ x=\frac{2}{9} \end{gathered}[/tex]Therefore, the fraction form of the repeating decimal is;
[tex]\frac{2}{9}[/tex]Find the first four terms of the sequence given by the following
1) In this question, we need to resort to that Explicit formula, with the first term so that we can find the terms:
[tex]\begin{gathered} a_n=54+8(n-1) \\ a_1=54+8(1-1) \\ a_1=54 \\ \\ a_2=54+8(2-1) \\ a_2=54+8 \\ a_2=62 \\ \\ a_3=54+8(3-1) \\ a_3=54+8(2) \\ a_3=54+16 \\ a_3=70 \\ \\ a_4=54+8(4-1) \\ a_4=54+8(3) \\ a_4=78 \\ \end{gathered}[/tex]2) As we can see, this is an Arithmetic sequence. And the answer is:
[tex]54,62,70,78[/tex]Mary used the quadratic formula to find the zeros of the equation below. Select the correct zeros of the equation:3x^2 - 9x + 2 = 0Answer choices include:x = fraction numerator 9 plus-or-minus square root of 57 over denominator 6 end fractionx equals fraction numerator negative 9 plus-or-minus square root of 57 over denominator 2 end fractionx equals fraction numerator 9 plus-or-minus square root of 105 over denominator 2 end fractionx equals fraction numerator negative 9 plus-or-minus square root of 105 over denominator 6 end fraction
We have the next quadratic function given:
[tex]3x^2-9x+2=0[/tex]Mary used the next quadratic formula:
[tex]x=\frac{-b^\pm\sqrt{b^2-4ac}}{2a}[/tex]Replace using the form ax²+bx+c
Where a= 3
b=-9
c=2
Then:
[tex]\begin{gathered} x=\frac{-\lparen-9)\pm\sqrt{\left(-9\right)^2-4\left(3\right)\left(2\right)}}{2\left(3\right)} \\ x=\frac{9\pm\sqrt[]{57}}{6}\frac{}{} \end{gathered}[/tex]Therefore, the correct answer is "x = fraction numerator 9 plus-or-minus square root of 57 over denominator 6 end fraction"
find the slope of the line passing through the points (-5,4) and (3,-3)
P1 = (-5, 4)
P2 = (3, -3)
Formula
[tex]\text{slope = }\frac{(y2\text{ - y1)}}{(x2\text{ - x1)}}[/tex]Substitution
[tex]\begin{gathered} \text{ slope = }\frac{(-3-4)}{(3\text{ + 5)}} \\ \text{ slope = }\frac{-7}{8} \end{gathered}[/tex]Result
[tex]\text{ slope = }\frac{-7}{8}[/tex]Given f(x), find g(x) and h(x) such that f(x)= g(h(x)) and neither g(x) nor h(x) is solely x.
Given:
[tex]\begin{gathered} f(x)=g(h(x)) \\ f(x)=\sqrt[]{-4x^2-3}+2 \end{gathered}[/tex]Solve :
[tex]g(h(x)=\sqrt[]{-4x^2-3}+2[/tex]The function g(x) convert then x is equal to h(x) then:
[tex]\begin{gathered} h(x)=-4x^2 \\ g(x)=\sqrt[]{x-3}+2 \end{gathered}[/tex]If a,b ,and c represent the set of all values of x that satisly the equation below, what is the value(A+ b+ c) + (abc)?X^3-20x = x^2(A) -1(B) 0(C) 1(D) 9
First, we need to find the solutions a, b, and c of the equation:
[tex]x^3-20x=x^2[/tex]We can rewrite it as:
[tex]\begin{gathered} x^3-x^{2}-20x=0 \\ \\ x(x^{2}-x-20)=0 \\ \\ x=0\text{ or }x^{2}-x-20=0 \end{gathered}[/tex]Thus, one of the solutions is a = 0.
To find the other solutions, we can use the quadratic formula. We obtain:
[tex]\begin{gathered} x=\frac{-(-1)\pm\sqrt[]{(-1)^{2}-4(1)(-20)}}{2(1)} \\ \\ x=\frac{1\pm\sqrt[]{1+80}}{2} \\ \\ x=\frac{1\pm\sqrt[]{81}}{2} \\ \\ x=\frac{1\pm9}{2} \\ \\ b=\frac{1-9}{2}=-4 \\ \\ c=\frac{1+9}{2}=5 \end{gathered}[/tex]Now, we need to find the value of the expression:
[tex]\mleft(a+b+c\mright)+abc[/tex]Using the previous solutions, we obtain:
[tex]\mleft(0-4+5\mright)+0(-4)(5)=1+0=1[/tex]Therefore, the answer is 1.
find the slope. A. y= -1/2x - 19/2.
The equation of the line follows the following general structure:
[tex]y=mx+b[/tex]Where m is the slope of the line and b is the y intercept.
Find the corresponding values in the given formula, this way:
In the given equation, m has a value of -1/2, it means the slope is -1/2.
Find the simple interest owed for the following loan. Principal = 2775 Rate = 7.5% Time = 5 1/2 years
We would apply the simple interest formula which is xpressed as
I = PRT/100
Where
I represents interest
P represents principal or amount borrowed
T represents time in years
R represents rate.
From the information given,
P = 2775
R = 7.5
T = 5 1/2 = 5.5
I = (2775 * 7.5 * 5.5)/100
I = 1144.6875
Rounding to the nearest cent,
I = 1144.69
xyx2xy1645256720 2258 484 1,2762873 7842,044 3294 1,0243,008 45141 2,025 6,345 ∑x=143 ∑y=411 ∑x2=4,573 ∑xy=13,393 Which regression equation correctly models the data?y = 2.87x + 0.12y = 2.87x + 11.85y = 3.39x – 14.75y = 3.39x – 9.24
We are asked to identify the correct regression equation.
The regression equation is given by
[tex]y=bx+a[/tex]Where the coefficients a and b are
[tex]a=\frac{\sum y\cdot\sum x^2-\sum x\cdot\sum xy}{n\cdot\sum x^2-(\sum x)^2}[/tex][tex]b=\frac{n\cdot\sum xy-\sum x\cdot\sum y}{n\cdot\sum x^2-(\sum x)^2}[/tex]Where n is the number of observations that is 5.
Let us substitute the following into the above formula.
∑x=143
∑y=411
∑x^2=4,573
∑xy=13,393
[tex]a=\frac{411\cdot4573-143\cdot13393}{5\cdot4573-(143)^2}=-14.75[/tex][tex]b=\frac{5\cdot13393-143\cdot411}{5\cdot4573-(143)^2}=3.39[/tex]So, the coefficients are
a = -14.75
b = 3.39
Therefore, the correct regression equation is
[tex]y=3.39x-14.75[/tex]14 pointsWhich are the coefficients of the terms in the algebraic expression, x2 - 3x?O and -31 and -3O and 351 and 36
Answer:
The coefficients of the terms in the algebraic expression are 1 and -3
[tex]1\text{ }and-3[/tex]Explanation:
The coefficients are the number that multiplies an algebraic term in an algebraic expression.
for example; the coefficient of 3x is 3.
[tex]3x=3\times x[/tex]For the question;
given the expression;
[tex]x^2-3x[/tex]The coefficient of x^2 is 1
[tex]x^2=1\times x^2[/tex]while the coefficient of x is -3
[tex]-3x=-3\times x[/tex]Therefore, the coefficients of the terms in the algebraic expression are 1 and -3
[tex]1\text{ }and-3[/tex]3/5 ÷ 1/3 = ?????????
Change the division sign to multiplication and then invert 1/3
That is;
[tex]\frac{3}{5}\times3[/tex][tex]=\frac{9}{5}\text{ =1}\frac{4}{5}[/tex]Graph two or more functions in the same family for which the parameter being changed is the slope, m. and is less than 0.Refer to the graph of f(x) = x + 2
We have the expression:
[tex]f(x)=x+2[/tex]If the slope is changing being less than 0, that is:
please help me work through this homework problem! thank you!
Given:
Given the function
[tex]y=3+\frac{3}{x}+\frac{2}{x^2}[/tex]and a point x = 3.
Required: Equation of the line tangent to y at x = 3.
Explanation:
The derivative of a function is he slope of the tangent line of the function at a given point. So, finding the derivative gives the slope of the tangent line.
[tex]y^{\prime}=-\frac{3}{x^2}-\frac{4}{x^3}[/tex]Substitute 3 for x into the derivative.
[tex]\begin{gathered} y^{\prime}|_{x=3}=-\frac{3}{3^2}-\frac{4}{3^3} \\ =-\frac{31}{27} \end{gathered}[/tex]Therefore, the slope of the tangent line is -31/27.
Substitute 3 for x into y.
[tex]\begin{gathered} y|_{x=3}=3+\frac{3}{3}+\frac{2}{3^2} \\ =3+1+\frac{2}{9} \\ =4+\frac{2}{9} \\ =\frac{38}{9} \end{gathered}[/tex](3, 38/9) is the only point on the tangent line where it intersects the original graph.
Plug these coordinates along with slope into the general point-slope form to find the equation.
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-\frac{38}{9}=-\frac{31}{27}(x-3) \end{gathered}[/tex]Solving for y will give the equation in slope-intercept form.
[tex]\begin{gathered} y=-\frac{31}{27}(x-3)+\frac{38}{9} \\ =-\frac{31}{27}x+\frac{69}{9} \end{gathered}[/tex]Final Answer: The equation of the tangent line is
[tex]y=-\frac{31}{27}x+\frac{69}{9}[/tex]
Solve for the unknown: 6(B+2) = 30
The unknown is B
[tex]6(B+2)=30[/tex][tex]\begin{gathered} 6B+12=30 \\ 6B+12-12=30-12 \\ 6B=18 \\ B=\frac{18}{6} \\ B=3 \end{gathered}[/tex]4x squared- 5x +4-(9x squared +3x -1)
hello
the question here requires the subtraction of polynomials
[tex]\begin{gathered} 4x^2-5x+4 \\ - \\ 9x^2+3x-1 \end{gathered}[/tex]if we are to do this, we have to subtract the polynomials based on their degree
this would be equal to
[tex]-5x^2-8x+5[/tex]the above polynomial is the result after subtraction, but we can as well, decide to multiply through by -1, to make or eilimate the negative sign on the second degree polynomal
[tex]\begin{gathered} (-5x^2-8x+5)\times-1 \\ = \\ 5x^2+8x-5 \end{gathered}[/tex]I List two types of angle pairs: 14) 15)
Let's recall that a type of angle pairs are complementary angles. They're complementary if the sum of their degree measurements equals 90 degrees or the right angle.
Example:
[tex]\angle ABZ\text{ and }\angle ZBC\text{ are complementary angles }[/tex]Let's recall that a second type of angle pairs are suplementary angles. In this case, the angles add up to 180 degrees.
Example:
[tex]\angle ABF\text{ and }\angle FBC\text{ are suplementary angles}[/tex]