ANSWER :
c. None
d. 0
e. (-1, 0)
EXPLANATION :
c. Values of x in which f(x) = -2
From the illustration, the graph does not pass through the y = -2
So there's NO values of x that will give f(x) = -2
d. Values of x in which f(x) = -3
From the illustration, when x = 0, f(x) will be -3
So the value of x is 0
e. x-intercepts are the points in which the graph intersects the x-axis.
In this case, the graph intersects at point (-1, 0)
Suppose you are measuring a moving box to see if it has enough room in it. The moving box is a cube, and the length of one side is 2 ft. long. What is the volume of the box?
Solution:
Suppose you are measuring a moving box to see if it has enough room in it.
The moving box is a cube.
Given that the length of one side is 2 ft. long, i.e.
[tex]l=2\text{ ft}[/tex]To find the volume of a cube, the formula is
[tex]V=l^3[/tex]Substitute for l into the formula above
[tex]\begin{gathered} V=l^3=2^3=8\text{ ft}^3 \\ V=8\text{ ft}^3 \end{gathered}[/tex]Hence, the volume of the box is 8 ft³
6x - 5y = - 4Direct variationХ5?k=Not direct variation2y = 14xDirect variationk ==Not direct variation
Direct Variation is of the form (directly proportional);
[tex]\begin{gathered} y\propto x \\ y=kx \end{gathered}[/tex]While indirect variation (inversely proportional) is of the form;
[tex]\begin{gathered} y\propto\frac{1}{x} \\ y=\frac{k}{x} \end{gathered}[/tex]So, for each of the given question we want to determine if they are direct or indirect variation;
1.
[tex]\begin{gathered} 6x-5y=-4 \\ -5y=-4-6x \\ -5y=-6x-4 \\ y=\frac{-6x-4}{-5} \\ y=\frac{6}{5}x+\frac{4}{5} \end{gathered}[/tex]Therefore, this is a Direct variation with proportionality constant;
[tex]k=\frac{6}{5}[/tex]2.
[tex]\begin{gathered} 2y=14x \\ y=\frac{14x}{2} \\ y=7x \end{gathered}[/tex]Therefore, this is a Direct variation with proportionality constant;
[tex]k=7[/tex]The perimeter, P, of a rectangle is the sum of twice the length and twice the width. P= 21+ 2w units P= 2([+w) units P= 2(x+3) units P= 2(5)-2(9) units P= 4 x units
We can see the problem states that P = 2(x+3) and also states that P=4x
Those equations lead to the expression
2(x+3)=4x
Operating
2x+6=4x
Subtracting 2x
6 = 2x
Solving for x
x = 6/2 = 3
Thus, the perimeter is
P = 2(3+3) = 12 units
Find the value of x, if m<3 is 13x-13 and m<4 is 8x+67.2 points43Your answer
Answer
x = 6
Explanation:
m<3 and m<4 are supplementary angles
supplementarrh angles is 180 degrees
m<3 + m<4 = 180
13x - 13 + 8x + 67 = 180
collect the like terms
13x + 8x - 13 + 67 = 180
21x + 54= 180
Isolate 21x
21x = 180 - 54
21x = 126
divide both sides by 21
21x/21 = 126/21
x = 6.
Therefore, x = 6
Compute.
\[ \left(\dfrac 8 3\right)^{-2} \cdot \left(\dfrac 3 4\right)^{-3}\]
[tex]\left(\cfrac{8}{3}\right)^{-2} \left(\cfrac{3}{4}\right)^{-3}\implies \left(\cfrac{3}{8}\right)^{+2} \left(\cfrac{4}{3}\right)^{+3}\implies \cfrac{3^2}{8^2}\cdot \cfrac{4^3}{3^3}\implies \cfrac{4^3}{8^2}\cdot \cfrac{3^2}{3^3} \\\\\\ \cfrac{64}{64}\cdot \cfrac{1}{3}\implies 1\cdot \cfrac{1}{3}\implies \cfrac{1}{3}[/tex]
TRIGONOMETRY if 0 is in the first quadrant and cos 0=3/5 what is sin (1/20)?Where 0 is theta
Given:
[tex]\cos \text{ }\theta\text{ = }\frac{3}{5}[/tex]Using the trigonometric identity:
[tex]undefined[/tex]6 points 3 The coordinates of the vertices of the triangle shown are P (2,13), Q (7,1), and R (2, 1). 14 13 12 11 10 9 8 6 5 3 2. 1 R Q 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 2 7 8 What is the length of segment PQ in units?
We have the following:
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]replacing:
P (2,13) = (x1,y1)
Q(7,1) = (x2,y2)
[tex]\begin{gathered} d=\sqrt[]{(7-2)^2+(1-13)^2} \\ d=\sqrt[]{5^2+12^2} \\ d=\sqrt[]{25+144} \\ d=\sqrt[]{169} \\ d=13 \end{gathered}[/tex]The answer is 13 units
so i have to factor x(3x+10)=77
SOLUTION
Given the equation as seen below, we can use the following steps to get the factors
[tex]x(3x+10)=77[/tex]Step 1: Remove the bracket by multiplying the value outside the bracket with the one inside the bracket using the distributive law. We have:
[tex]\begin{gathered} x(3x)+x(10)=77 \\ 3x^2_{}+10x=77 \\ 3x^2+10x-77=0 \end{gathered}[/tex]Step 2: Now that we have a quadratic equation, we solve for x using the quadratic formula:
[tex]\begin{gathered} 3x^2+10x-77=0 \\ u\sin g\text{ the form }ax^2+bx+c=0 \\ a=3,b=10,c=-77 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ =\frac{-10\pm\sqrt[]{10^2-4(3)(-77)}}{2(3)} \\ =\frac{-10\pm\sqrt[]{100+924}}{6} \\ =\frac{-10\pm\sqrt[]{1024}}{6} \\ =\frac{-10+32}{6}\text{ or }\frac{-10-32}{6} \\ \frac{22}{6}\text{ or -}\frac{42}{6} \\ =\frac{11}{3}\text{ or }-7 \end{gathered}[/tex]Hence, it can be seen from above that the factors will be -7 or 11/3.
solve the equation 3.75x+3.7=1.7+1.75xA 10 B -1 C 1D .1
The question is given to be:
[tex]3.75x+3.7=1.7+1.75x[/tex]Step 1
Multiply every number by 100:
[tex]\begin{gathered} 3.75x\cdot100+3.7\cdot100=1.7\cdot100+1.75x\cdot100 \\ 375x+370=170+175x \end{gathered}[/tex]Step 2
Subtract 370 from both sides:
[tex]\begin{gathered} 375x+370-370=170+175x-370 \\ 375x=175x-200 \end{gathered}[/tex]Step 3
Subtract 175x from both sides:
[tex]\begin{gathered} 375x-175x=175x-200-175x \\ 200x=-200 \end{gathered}[/tex]Step 4
Divide both sides by 200:
[tex]\begin{gathered} \frac{200x}{200}=-\frac{200}{200} \\ x=-1 \end{gathered}[/tex]ANSWER
The correct option is OPTION B, the SECOND OPTION.
Find the intersection if possibleExpress your answer in interval notation
Solution:
The first set given as;
[tex][-9,-1)[/tex]Then in list form, the set is;
[tex]\lbrace-9,-8,-7,-6,-5,-4,-3,-2\rbrace[/tex][tex]\lbrace-9,-8,-7,-6,-5,-4,-3,-2\rbrace[/tex]Also, the second set given as;
[tex](-3,4)[/tex]Then, in list form, the set is;
[tex]\lbrace-2,-1,0,1,2,3\rbrace[/tex]The intersection of the two sets is;
[tex]\begin{gathered} \lbrace-9,-8,-7,-6,-5,-4,-3,-2\rbrace\cap\lbrace-2,-1,0,1,2,3\rbrace=\lbrace-2\rbrace \\ \\ \text{ Note that there are some real numbers on the number line} \end{gathered}[/tex]Thus, the solution in interval notation is;
[tex](-3,-1)[/tex]ANSWER: (-3,-1)
if you receive a 175.84 cents on 314 invested at a rate of 7% for how long did yo invest the principle
Answer:
The number of years you should invest the principal is;
[tex]8\text{ years}[/tex]Explanation:
Given;
[tex]\begin{gathered} \text{Interest i = \$175.84} \\ \text{ Principal P = \$}314 \\ \text{Rate r = 7\% =0.07} \end{gathered}[/tex]Recall that the formula for simple interest is;
[tex]\begin{gathered} i=P\times r\times t \\ t=\frac{i}{Pr} \\ \text{where;} \\ t=\text{time of investment} \end{gathered}[/tex]substituting the given values;
[tex]\begin{gathered} t=\frac{i}{Pr} \\ t=\frac{175.84}{314\times0.07} \\ t=\frac{175.84}{21.98} \\ t=8 \end{gathered}[/tex]Therefore, the number of years you should invest the principal is;
[tex]8\text{ years}[/tex]We can also solve as;
[tex]\begin{gathered} i=P\times r\times t \\ 175.84=314\times0.07\times t \\ 175.84=21.98t \end{gathered}[/tex]then we can divide both sides by 21.98;
[tex]\begin{gathered} \frac{175.84}{21.97}=\frac{21.98t}{21.98} \\ 8=t \\ t=8\text{ years} \end{gathered}[/tex]57 es 95% de que número
60
1) Considerando que 57 es 95% de algun numero, vamos escribir esta ecuacion
0.95x = 57 Escribindo 95% como 0.95, dividir los dos lados por 0.95
x =57/0.95
x=60
2) Asi 57 es 95% de 60.
find the absolute extrema for the function on the given inveral
In order to find the minimum and maximum value in the given interval, first let's find the vertex coordinates:
[tex]\begin{gathered} f(x)=3x^2-24x \\ a=3,b=-24,c=0 \\ \\ x_v=\frac{-b}{2a}=\frac{24}{6}=4 \\ y_v=3\cdot4^2-24\cdot4=3\cdot16-96=-48 \end{gathered}[/tex]Since the coefficient a is positive, so the y-coordinate of the vertex is a minimum point, therefore the absolute minimum is (4,-48).
Then, to find the maximum, we need the x-coordinate that is further away from the vertex.
Since 0 is further away from 4 than 7, let's use x = 0:
[tex]f(0)=3\cdot0-24\cdot0=0[/tex]Therefore the absolute maximum is (0,0).
what is the difference between solving literal equations(with only variables)and solving multistep equations(woth numbers and a variables)
To solve a literal equation means to express one variable with respect to the other variables in the equation. The most important part of a literal equation is to isolate or keep by itself a certain variable on one side of the variable (either left or right) and the rest on the other side
Solving multistep equations takes more time and more operations compared to solving a literal equation.
Find the value of x that makes A || B.AB5423142 3x10 and 23 = x + 30X=[? ]
∠2 and ∠3 are alternate interior angles. In order to A II B, the alternate interior angles must be equal.
Then,
[tex]\begin{gathered} \angle2=\operatorname{\angle}3 \\ 3x-10=x+30 \end{gathered}[/tex]To find x, subtract x from both sides of the equation:
[tex]\begin{gathered} 3x-10-x=x+30-x \\ 3x-x-10=x-x+30 \\ 2x-10=30 \end{gathered}[/tex]Now, add 10 to both sides of the equation:
[tex]\begin{gathered} 2x-10+10=30+10 \\ 2x=40 \end{gathered}[/tex]Finally, divide both sides by 2:
[tex]\begin{gathered} \frac{2x}{2}=\frac{40}{2} \\ x=20 \end{gathered}[/tex]Answer: x = 20.
evaluate each using the values given y+y-(y-x); use x = 1, and y = 4Options12154
The given expression is
y + y - (y - x)
We would substitute x = 1 and y = 4 into the expression. it becomes
4 + 4 - (4 - 1)
8 - 3
= 5
the correct answer is 5
two rectangles are similar. The length of small rectangle is 4 and the length of the big rectangle is 12. If the perimeter of the smaller rectangle is 28, and what is the perimeter of the larger rectangle?
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data
small rectangle
length = 4
perimeter = 28
big rectangle
length = 12
perimeter = ?
Step 02:
small rectangle
perimeter = 2l + 2w
28 = 2 * 4 + 2 w
28 - 8 = 2w
20 / 2 = w
10 = w
big rectangle
[tex]\frac{4}{12}\text{ = }\frac{10}{w}[/tex]4 w = 10 * 12
w = 120 / 4 = 30
Perimeter = 2*12 + 2*30
= 24 + 60 = 84
The answer is:
The perimeter of the big rectangle is 84.
2.Solve the inequality for x. Show each step to the solution.12x > 3(5x – 2) – 15
we have the inequality
12x > 3(5x-2)-15
step 1
apply distributive property right side
12x > 15x-6-15
combine like terms right side
12x > 15x-21
step 2
Adds both sides 21
12x+21 > 15x-21+21
simplify
12x+21 > 15x
step 3
subtract 12x both sides
12x+21-12x > 15x-12x
21 > 3x
step 4
Divide by 3 both sides
21/3 > 3x/3
7 > x
Rewrite
x < 7An architect designs a rectangular flower garden such that the width is exactly two-thirds of the length. If 240 feet of antique picket fencing are to be used to enclose the garden, find the dimensions of the garden. What is the length of the garden? The length of the garden is What is the width of the garden? The width of the garden is
STEP 1:
We'll derive an expression for the width and the length
[tex]\begin{gathered} w=\frac{2l}{3}\text{ where} \\ w\text{ = width} \\ l=\text{ length} \end{gathered}[/tex]STEP 2:
Next, We then derive an expression for the perimeter substituting w as a function of l
[tex]\begin{gathered} \text{Perimeter = 2(l+w)} \\ 240=2(l+\frac{2l}{3}) \end{gathered}[/tex]STEP 3:
Solve for l and subsequently w
[tex]\begin{gathered} \text{Perimeter}=\text{ 240 = 2(}\frac{2l+3l}{3})=2(\frac{5l}{3}) \\ 240=\frac{10l}{3} \\ \text{Cross multiplying gives 240}\times3=5l \\ l=\frac{240\times3}{10}=72ft \\ w=\frac{2l}{3}=\frac{2\times72}{3}=48ft \end{gathered}[/tex]Therefore, length = 72 ft and width = 48ft
Exactly 25% of the marbles in a bag are black. If there are 8 marbles in the bag, how many are black?
Let the total number of marbles in the bag be 'x'.
Given that exactly 25% of the total marbles are black,
[tex]\begin{gathered} \text{ No. of black marbles}=25\text{ percent of total marbles} \\ \text{ No. of black marbles}=25\text{ percent of x} \\ \text{ No. of black marbles}=\frac{25}{100}\cdot x \\ \text{ No. of black marbles}=0.25x \end{gathered}[/tex]Also, given that there are total 8 marbles in the bag,
[tex]x=8[/tex]Then the number of black marbles will be obtained by substituting x=8,
[tex]\begin{gathered} \text{ No. of black marbles}=0.25(8) \\ \text{ No. of black marbles}=2 \end{gathered}[/tex]Thus, there are 2 black marbles in the bag.
4.A pet store sells cats for $50 and dogs for $100. If one day it sells a total of 4pets and makes $300, find out how many cats and dogs it sold by writing asystem of equations and graphing to solve it.Representations:Equations:
Answer:
2 cats and 2 dogs
Explanation:
Representations:
x = number of cats sold
y = number of dogs sold
Equations:
We know that it sells a total of 4 pets, so the sum of the number of cats and dogs is 4. So:
x + y = 4
On the other hand, they make $300, and they make $50 for each cat and $100 for each dog, so:
$50x + $100y = $300
So, the system of equation is:
x + y = 4
50x + 100y = 300
Graph:
Now, we need to graph the equations, so we need to identify two points for each equation:
For x + y = 4
If x = 0, then:
0 + y = 4
y = 4
If x = 4, then:
4 + y = 4
4 + y - 4 = 4 - 4
y = 0
For 50x + 100y = 300
If x = 0, then:
50(0) + 100y = 300
100y = 300
100y/100 = 300/100
y = 3
If x = 4, then:
50(4) + 100y = 300
200 + 100y = 300
200 + 100y - 200 = 300 - 200
100y = 100
100y/100 = 100/100
y = 1
Therefore, we have the points (0, 4) and (4, 0) to graph the line of the first equation and the points (0, 3) and (4, 1) to graph the line of the second equation.
So, the graph of the system is:
Therefore, the solution is the intersection point (2, 2), so they sold 2 cats and 2 dogs that day.
Determine a series of transformations that would map Figure 1 onto Figure J. y 11 Figure J NOW ona 00 05 15 1 -12-11-10-9-8-7-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 -2 Šť b bo v och t co is with Figure I -11 -12 A followed by a o
EXPLANATION
The transformations that would map Figure 1 onto Figure J are:
A rotation followed by a translation
The area of A triangle with base b and height h is given by A 1/2bh. Find the area when b=24 m and h=30
Given:
Base of a triangle 24m and height = 30m
Required:
Find the area of a triangle.
Explanation:
We have formula for area of triangle
[tex]A=\frac{1}{2}\times b(base)\times height(h)[/tex]Now,
[tex]\begin{gathered} A=\frac{1}{2}\times24\times30 \\ A=360m^2 \end{gathered}[/tex]Answer:
The area of triangle is 360 meter square.
Morgan wants to order at least $45 worth of merchandise, so she will get free shipping. If Morgan has picked out a key chain for $8 and a bag for $19, which inequality represents the amount of money, m, she needs to spend to get free shipping?
Morgan wants to order at least $45 worth of merchandise.
She spent $8 and $19. m represents the amount of money she needs to spend to get free shipping. The inequality is:
8 + 19 + m ≥ 45
m ≥ 45 - 8 - 19
m ≥ 18
Can someone please help me find the valu of X?
Answer:
x = 10
Explanation:
Because the transverse lines are parralell, the following must be true
[tex]\frac{x+8}{x+2}=\frac{3}{2}[/tex]cross multipication gives
[tex]\begin{gathered} 2(x+8)=3(x+2) \\ \end{gathered}[/tex]which simplifies to give
[tex]\begin{gathered} 2x+16=3x+6 \\ \end{gathered}[/tex]subtracting 2x from both sides gives
[tex]16=x+6[/tex]subtracting 6 from both sides gives
[tex]10=x[/tex]Hence the value of x is 10.
I’m sorry to keep bothering you guys but you’re the third person that I’m trying the last two their answers went partway and then stopped I just need to see how this is worked out
In this case, we'll have to carry out several steps to find the solution.
Step 01
Data:
Graph:
Height of Golf Ball
Step 02:
functions:
We must analyze the graph to find the solution.
Function:
Non linear
intercepts:
x-intercepts: 0 and 120
y-intercept: 0
symmetry:
x = 50
positive:
domain: (0 , 120)
negative:
there are no negative values
increasing:
interval on x: (0 , 50)
decreasing:
interval on x: (50, 120)
That is the full solution.
For the function f(x) = x2 + 2x - 15 solve the following.f(x) = 0
Given:
[tex]f\mleft(x\mright)=x^2+2x-15[/tex]To find: The value of x when
[tex]f(x)=0[/tex]Explanation:
Since,
[tex]f(x)=0[/tex]We can write it as,
[tex]\begin{gathered} x^2+2x-15=0 \\ x^2+5x-3x-15=0 \\ x(x^{}+5)-3(x-5)=0 \\ (x+5)(x-3)=0 \\ x=-5,3 \end{gathered}[/tex]Hence, the solution is x = -5, and 3.
Final answer: The solution is,
[tex]\mleft\lbrace-5,3\mright\rbrace[/tex]You might need: Calculator h(r) = 72 +11r - 26 1) What are the zeros of the function? Write the smaller r first, and the larger r second.
1) Notice that:
[tex]r^2+11r-26=r^2+13r-2r-2(13).[/tex]Grouping like terms we get:
[tex]\begin{gathered} r^2+13r-2r-2(13)=r(r+13)-2(r+13) \\ =(r-2)(r+13). \end{gathered}[/tex]Therefore:
[tex]h(r)=(r-2)(r+13).[/tex]Then the zeros of h(r) are:
[tex]r=2\text{ and }r=-13.[/tex]2) Notice that:
[tex]\begin{gathered} h(r)=r^2+11r-26=r^2+11r+(\frac{11}{2})^2-(\frac{11}{2})^2-26 \\ =(r+\frac{11}{2})^2-\frac{121}{4}-26=(r+\frac{11}{2})^2-\frac{225}{4}. \end{gathered}[/tex]Therefore the vertex of the given parabola is:
[tex](-\frac{11}{2},-\frac{225}{4}).[/tex]Answer:
1)
[tex]\begin{gathered} smaller\text{ r=-13,} \\ larger\text{ r=2.} \end{gathered}[/tex]2) Vertex:
[tex](-\frac{11}{2},-\frac{225}{4}).[/tex]
If trapezoid ABCD was dilated by a scale factor of 2\3 to form trapezoid A'B'C'D,what is the area of trapezoid ABCD?The area of trapezoid A'B'C'D is 12 units^2
As a general rule, we know that the area of a dilated figure is the area of the original figure multiplied by the square of the scale factor. We can see this in the following formula:
[tex]A=A^{\prime}\cdot k^2[/tex]where A is the area of the original figure, A' is the area of the dilated figure and k is the scale factor.
In this case, we have that the area of the dilated figure (trapezoid A'B'C') is 12 square units, and the scale factor is k = 2/3. Then, using the equation we get the following:
[tex]\begin{gathered} A^{\prime}=12 \\ k=\frac{2}{3} \\ \Rightarrow A=12\cdot(\frac{2}{3})^2=12\cdot(\frac{4}{9})=\frac{12\cdot4}{9}=\frac{48}{9}=5\frac{1}{3} \\ A=5\frac{1}{3}u^2 \end{gathered}[/tex]therefore, the area of trapezoid ABCD is 5 1/3 square units
Just need help with number 3. Please. Thankyou! Been stuck on this one for a while now.
A cosine function is given in the form:
[tex]y=A\cos(x-h)+k[/tex]where |A| is the amplitude, h is the horizontal shift (also called a phase shift) and k is the vertical shift.
The function is given to be:
[tex]\begin{gathered} y=8\cos(\frac{1}{4}x) \\ A=8 \end{gathered}[/tex]Therefore, the amplitude is 8.