The mean value of the sample is 98.1 F and its standard deviation is 0.7 F.
The margin of error of the mean value is given by 0.7/sqrt(751) = 0.7/27.4 = 0.026 (rounded to the nearest thousandth)
Using the Z test, we got: Z = (98.6 - 98.1)/(0.026) = 0.5/0.026 = 196
Therefore, the mean value of the sample is incompatible with 98.6 and we can claim that the mean body temperature is less than it.
a cylindrical container with an 8 in. diameter and an 2 in. height is completely filled with water. all of its contents are poured into another cylindrical container that is 32 in. tall. if the water completely fills the second container, the second container's diameter is ___ in.
Volume of a cylinder: π r^2 h
Where:
r = radius
h= height
Calculate the volume of the first cylindrical container:
radius = diameter/2
Volume = π (8/2)^2 (2) = 100.5 in3
Second container
Volume = 100.5
heigth = 32 in
100.5 = π r^2 32
Solve for r
100.5/ (π 32 ) = r^2
1 = r^2
√1 = r
r=1
Diameter = 2 r = 2 (1) = 2 in
Find the slope of the line in simplest form
Answer:
[tex]\boxed{\sf \sf Slope(m)=-\cfrac{5}{4}}[/tex]
Step-by-step explanation:
To find the slope between two points we'll use the slope formula:-
[tex]\boxed{\bf \mathrm{Slope}=\cfrac{y_2-y_1}{x_2-x_1}}[/tex]
Given points:-
(-3, 1)(1, -4)[tex]\sf \left(x_1,\:y_1\right)=\left(-3,\:1\right)[/tex]
[tex]\sf \left(x_2,\:y_2\right)=\left(1,\:-4\right)[/tex]
[tex]\sf m=\cfrac{-4-1}{1-\left(-3\right)}[/tex]
[tex]\sf m=-\cfrac{5}{4}[/tex]
Therefore, the slope of the line is -5/4!
____________________
Hope this helps!
Have a great day!
10. Explain how you would prove the following.Given: HY = LY:WH LFProve: A WHY = AFLY
It is being proved that triangle Δ WHY ≅ Δ FLY by ASA rule.
In triangle Δ WHY and Δ FLY, we have that:
HY ≅ LY ( given)
∠WHY = ∠ FLY (alternate interior angles as WH || LF)
∠WYH = ∠ FYL ( Vertically opposite angles)
We get that:
Δ WHY ≅ Δ FLY ( ASA rule)
It is proved that Δ WHY ≅ Δ FLY by ASA rule.
Therefore, we get that, it is being proved that triangle Δ WHY ≅ Δ FLY by ASA rule.
Learn more about triangle here:
https://brainly.com/question/17335144
#SPJ9
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!
Which of the following functions are linear? Select all that apply. A The function that assigns to each number r the value 9x2. 3 B The function that assigns to each positive number x the value The function that assigns to each positive number 2 the value 5V2r. D The function that assigns to each number 2 the value 78. 1 E The function that assigns to each number the value 2. F|The function that assigns to each number 2 the value x + 8.
The answers are D, E, F
Being linear has one dependent variable and one independent variable, similar to a straight line.
only D,E and F satisfy this.
Hi can someone please help me out on this drag and drop assignment? I’ll appreciate the help :)
Answer:
Step-by-step explanation:
1.
circumference of circular fence = 2πr
π = 3.14
r = radius
radius of circular fence, r= 10 feet
putting the values in the formula,
circumference = 2× 3.14 × 10
= 62.8 feet
therefore the fencing brad need will be 62.8 feet
2.
Area of the circular hot tub = πr²
π= 3.14
r = radius
as given in the question,
diameter = 80 inches
we know diameter is equal to half radius so r = 40 inches
putting the values in the formula,
area = 3.14 × 40× 40
= 5,024 inches
hence the area of the hot tub is equal to 5,025 inches
3.
Area of the circular wall clock = πr²
π= 3.14
r = radius
as per the question ,
area of 5 inches wall clock = 3.14× 5×5 = 78.5 inches
and area of 6 inches wall clock = 3.14 ×6×6 = 113.04 inches
To find the how much wall space will 6 inches wall clock takes we have to subtract both areas,
area of 6 inches wall clock - area of 5 inches wall clock = 113.04 - 78.5
= 34.54 inches
6 inches wall clock will take 34.54 inches of area more.
4.
In one rotation tire will cover whole area so to find the diameter we have to put area of circle equals to area of tire
Area of the circle = πr²
π= 3.14
r = radius
as per the question ,
area of the tire =116.18 inches
πr² = 116.18
r=6.08 inches
diameter = 2 × r= 12.16 inches
read more about circle:
https://brainly.com/question/2870743
Natalie bought a tank for her pet fish. She is measuring how much water will fill the tank.Which measurement will best help Natalie determine how much water will fill the tank?O Natalie should measure the area of the fish tank in square units.O Natalie should measure the area of the fish tank in cubic units.O Natalie should measure the volume of the fish tank in square units.O Natalie should measure the volume of the fish tank in cubic units.
Answer:
Natalie should measure the volume of the fish tank in cubic units.
Explanation:
The measure of the area is in square units and the measure of the volume is in cubic units.
The area gives you the measure of a plane surface and the volume gives you how much space a solid occupies
So, the measurement that will best help Natalie is the volume of the fish tank in cubic units.
To get around a small pond, a local electrical utility must lay two sections of underground cable that are 371 m and 440 m long. The two sections meet at an angle of 145°. How much extra cable is needed due to going around the pond?
If the cable could went through the pond there would be only one straight section conecting the two points. If we draw this new section in the picture we'll form a triangle:
In order to find how much extra cable is needed because of the pond we must find the length of the imaginary cable that connects the points through the dot, for this length we are going to use x.
The cosine rule will help as find it. Let's assume that we have a triangle with an angle A which has an opposite side with a length a and the lengths of the other two sides are b and c. Then the cosine rule states the following:
[tex]a^2=b^2+c^2-2bc\cos A[/tex]We can apply this to our triangle. The 145° angle that we know is A, its opposite side a is x and the remaining sides b and c are the two cable sections of 371 m and 440 m. Then we get:
[tex]\begin{gathered} x^2=371^2+440^2-2\cdot371\cdot440\cdot\cos145^{\circ} \\ x^2=598677.7594 \end{gathered}[/tex]Then we apply a square root to both sides of this equation:
[tex]\begin{gathered} \sqrt{x^2}=\sqrt{598677.7594} \\ x=773.74 \end{gathered}[/tex]So without the pond the length of the cable would have been of 773.74 m. In order to find the amount of extra cable needed we must take the total length of both sections and substract 773.74 m from it. Then we get:
[tex]371+440-773.74=37.26[/tex]AnswerThen the answer is 37.26m.
Can you help me answer a, b and c please?
Answer
a)
[tex]B=B_{0}(1+\frac{a}{12})^{t}[/tex]b)
[tex]B=B_{0}(1+\frac{25}{3}a)^{t}[/tex]c)
[tex]B=B_0(1+\frac{25}{3}a)^{12y}[/tex]Explanation
We're given the function:
[tex]B=B_0(1+r)^t[/tex]To represent the equation with the data given in the problem, we need to solve the three parts of this problem.
The part a ask us to write the expression in terms of annual percentage rate (APR) in decimal. If we call "a" the APR in decimal, then the monthly rate is the APR divided in 12:
[tex]r=\frac{a}{12}[/tex]Now we can rewrite the balance equation in terms of the initial investment, the number of months and the APR:
[tex]B=B_0(1+\frac{a}{12})^t[/tex]In part b, we need to write the balance equation using the APR as percentage. The APR as decimal is equal to the APR in percentage divided by 100. If we call A the APR in percentage:
[tex]a=\frac{A}{100}[/tex]Now we replace this value in the balance equation we got in part a:
[tex]B=B_0(1+\frac{100a}{12})^t[/tex]Then simplify:
[tex]B=B_0(1+\frac{25}{3}a)^t[/tex]That's the answer to b.
In part c, we need to write the balance equation with the time in years. Since 1 year has 12 months, if we call the number of months t, and the number of years y:
[tex]t=12y[/tex]Then:
[tex]B=B_0(1+\frac{25}{3}a)^{12y}[/tex]And this is the answer to c.
an object is thrown down from the top of a building. A height function for the object is given by the equation h=16(8+ t ) (5 - t) where T is the number of seconds elapsed since the object was thrown and H is the height of the object above the ground ( in feet). explain how to reason about the structure of the equation to determine when the object will hit the ground
The height is a function of the time, given by the following equation:
h(t) = 16(8+t)(5-t)
The object hits the ground when h(t) = 0. So
16(8 + t)(5 - t) = 0
This means that:
8 + t = 0 or 5 - t = 0
8 + t = 0
t = -8
We cannot have negative values for t.
5 - t = 0
-t = -5 *(-1)
t = 5
The object hits the ground when t = 5, which was easy to find since the equation was already factored by it's roots.
A loan of $43,000 is made at 5.25% interest, compounded annually. After how many years will the amount due reach $64,000 or more? (Use the calculator provided if necessary.)Write the smallest possible whole number answer.
Answer:
8 years
Explanation:
For a compound interest loan compounded annually, the amount due after t years is calculated using the formula:
[tex]A(t)=P(1+r)^t\text{ where }\begin{cases}P={Loan\;Amount} \\ {r=Annual\;Interest\;Rate}\end{cases}[/tex]We want to find when the amount due will reach $64,000 or more.
[tex]43000(1+0.0525)^t\geq64,000[/tex]The equation is solved for t:
[tex]\begin{gathered} \text{ Divide both sides by }43000 \\ \frac{43,000(1+0.0525)^t}{43000}\geqslant\frac{64,000}{43000} \\ (1.0525)^t\geq\frac{64}{43} \\ \text{Take the log of both sides:} \\ \log(1.0525)^t\geqslant\log(\frac{64}{43}) \\ \text{By the power law of logarithm:} \\ \implies t\operatorname{\log}(1.0525)\geq\operatorname{\log}(\frac{64}{43}) \\ \text{ Divide both sides by }\operatorname{\log}(1.0525) \\ t\geq\frac{\operatorname{\log}(\frac{64}{43})}{\operatorname{\log}(1.0525)} \\ t\geq7.77 \end{gathered}[/tex]The number of years when the amount due will reach $64,000 or more is 8 years.
Solve the following system of equations by graphing. y = –1∕2x – 2 y = –3∕2x + 2
We can see the solution in the graph as follows:
If tan theta = 4/3 and pi
Given that tan theta = 4/3 and theta lies in the third quadrant.
[tex]\pi<\theta<\frac{3\pi}{2}[/tex]Divide the compound inequality by 2.
[tex]\frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{4}[/tex]This means theta/2 lies in the second quadrant. So, cos theta/2 and sec theta/2 are negative.
Use trigonometric identities to find sec theta.
[tex]\begin{gathered} \sec \theta=\sqrt[]{1+\tan ^2\theta} \\ =\sqrt[]{1+(\frac{4}{3})^2} \\ =\sqrt[]{1+\frac{16}{9}} \\ =\sqrt[]{\frac{25}{9}} \\ =-\frac{5}{3} \end{gathered}[/tex]we know that cosine is the inverse of secant. So, cos theta = -3/5.
now, using the half-angle formula, we have to find cos theta/2,
[tex]\begin{gathered} \cos (\frac{\theta}{2})=-\sqrt[]{\frac{1+\cos x}{2}} \\ =-\sqrt[]{\frac{1-\frac{3}{5}}{2}} \\ =-\sqrt[]{\frac{\frac{2}{3}}{2}} \\ =-\sqrt[]{\frac{1}{3}} \end{gathered}[/tex]PLEASE HELP DUE SOON ONLY Q6 I have the work for the rest.
Given:
Diameter=40 feet
so radius=20 feet(d=2r)
Height of 2nd tank= 130 feet
Required:
Volume of sphere
Volume of cylinder
Explanation:
First of all we are going to calculate volume of sphere=
[tex]\begin{gathered} \frac{4}{3}\pi r^3 \\ =\frac{4}{3}\times3.14\times20\times20\times20 \\ =33,493.33feet^3 \end{gathered}[/tex]Then we are going to calculate volume of cylinder=
[tex]\begin{gathered} =\pi r^2h \\ =3.14\times20\times20\times130 \\ 163,280feet^3 \end{gathered}[/tex]and the vertically cross- section of tank 2 is a rectangle.
Required answer:
volume of sphere is 33,493.33
volume of cylinder=163,280
and the shape is rectangle.
Use the Factor Theorem to find all real zeros for the given polynomial function and one factor. (Enter your answers as a comma-separated list.)f(x) = 4x3 − 19x2 + 29x − 14; x − 1
Given:
The polynomial and one factor
[tex]f(x)=4x^3-19x^2+29x-14[/tex]Required:
Use the Factor Theorem to find all real zeros for the given polynomial function and one factor.
Explanation:
We have one factor, we will us that
[tex]\begin{gathered} =\frac{4x^3-19x^2+29x-14}{x-1} \\ \text{ It can be written as } \\ =(x-1)(4x^2-15x+14) \\ \text{ So, roots are} \\ =1,2,\frac{7}{4} \end{gathered}[/tex]Answer:
answered the question.
I am thinking of a number. It has two digits. When I reverse the digits and then add the new number to the original number I get 33. What is the number?
Let x and y be the digits.
The original number has two digits that means that one is the tens and the other the ones, in this case let x be the tens and y the ones, then we have the number:
[tex]10x+y[/tex]if we reverse it this means that the y become the tens and x becomes the ones then we have the number:
[tex]10y+x[/tex]And if we add them the result is 33, then we have the equation:
[tex]\begin{gathered} (10x+y)+(10y+x)=33 \\ 11x+11y=33 \\ x+y=3 \\ y=3-x \end{gathered}[/tex]This means that y has to be 3-x. Now, since we both numbers to have two digits x can't be zero nor 3. Then has to be 1 or 2.
If x=1 then y=2 and the original number is 12.
If x=2 then y=1 and the original number is 21.
Notice how in both cases we get the other one when reversed, therefore the numbers we are looking for are 12 and 21.
2 1' 15 = 1 2. 2 1 1 = 1 5 . (Type a whole number, fraction, or mixed number.)
Step 1
Write out your question.
[tex]1\frac{2}{5}\frac{.}{.}\text{ 1}\frac{1}{2}[/tex]Step 2
Convert mixed fractions to improper fractions.
[tex]\frac{7}{5}\text{ }\frac{.}{.}\text{ }\frac{3}{2}[/tex]Step 3
Convert division to multiplication and invert the fraction after the division.
[tex]\begin{gathered} =\text{ }\frac{7}{5}\text{ x }\frac{2}{3} \\ =\text{ }\frac{7\text{ x 2}}{5\text{ x 3}} \\ =\text{ }\frac{14}{15} \end{gathered}[/tex]What is the missing exponent?w^3 x w^? = w^-6
Let the missing exponent be a,
[tex]w^3\times w^a=w^{-6}[/tex]From the law of indices stated below which satisfies the above equation,
[tex]\begin{gathered} x^a\times x^b=x^{a+b} \\ \text{relating the equation to the formula,} \\ w^3\times w^a=w^{-6} \\ w^{3+a}_{}=w^{-6} \\ \text{solving the exponents,} \\ 3+a=-6 \\ \text{Collect like terms} \\ a=-6-3 \\ a=-9 \end{gathered}[/tex]Hence, the missing exponent a is -9.
simplifyx^-1 X (y^-8 X z^5)^3------------------------------x^-4 X y^-3 X z^6
Given:
[tex]\frac{x^{-1}\left(y^{-8}z^5\right)^3}{x^{-4}y^{-3}z^6}[/tex]Simplify:
[tex]\frac{x^{-1}y^{-24}z^{15}}{x^{-4}y^{-3}z^6}[/tex]And:
[tex]x^{-1-(-4)}y^{-24-(-3)}z^{15-6}=x^3y^{-21}z^9[/tex]Re order:
[tex]\frac{x^3z^9}{y^{21}}[/tex]Answer:
[tex]\frac{x^{3}z^{9}}{y^{21}}[/tex]clarify each of the following triangles by their angles in sides
SOLUTION
From the diagram below,
Triangle A is right isosceles, since two angles are equal, two sides are equal, and there is a right-angle (90 degrees angle) present.
Triangle B is obtuse scalene, since none of the sides are equal, and one of the angles is greater than 90 degrees.
Triangle C is acute scalene, since none of the sides are equal, and all the angles are less than 90 degrees.
Triangle D is acute isosceles, since two angles and two sides are equal, and all the angles are less than 90 degrees.
Solve the systems using subsitution for 1 and 2Solve the system using elimination for question 3
3x+y=2 (a)
6x+2y= 11 (b)
Solve equation (a) for y :
3x+y = 2
y= 2-3x
Replace the y value on (b)
6x+2(2-3x) =11
6x+4-6x=11
4=11
the system has no solution.
Find the infinite sum of the geometric sequence with a=3,r=3/6 if it exists.S∞=
ANSWER
[tex]S_{\infty}=6[/tex]EXPLANATION
Given:
1. First term (a) = 3
2. Common ration (r) = 3/6
Desired Outcome:
Infinite sum of the geometric sequence.
The formula to calculate the infinite sum of the geometric sequence
[tex]S_{\infty}=\frac{a(1-r^n)}{1-r}[/tex]Now, as n approaches infinity,
[tex]1-r^n\text{ approaches 1}[/tex]So,
[tex]\frac{a(1-r^n)}{1-r}\text{ approaches }\frac{a}{1-r}[/tex]Therefore,
[tex]S_{\infty}=\frac{a}{1-r}[/tex]Substitute the values
[tex]\begin{gathered} S_{\infty}=\frac{3}{1-\frac{3}{6}} \\ S_{\infty}=\frac{3}{1-\frac{1}{2}} \\ S_{\infty}=\frac{3}{\frac{1}{2}} \\ S_{\infty}=6 \end{gathered}[/tex]Hence, the infinite sum of the geometric sequence is 6.
Determine whether triangle DEF with vertices D(6, -6), E(39, -12), and F(24, 18) isscalene (no congruent sides), isosceles (two congruent sides), or equilateral (threecongruent sides).
We have three given points. We need to graph them, and then find the distances between them.
We need to remember that we can classify the triangles according to their sides:
1. A triangle with three congruent sides is an equilateral triangle.
2. A triangle with two congruent sides is an isosceles triangle.
3. A triangle with no congruent sides is a scalene triangle.
Additionally, we know that a segment is congruent to other when it has the same size as the other.
Then we can graph the three points as follows:
Now, we need to find the distances between the sides of the triangle using the distance formula as follows:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]This is the distance formula for points (x1, y1) and (x2, y2).
Finding the distance between points D and EThe coordinates for the two points are D(6, -6) and E(39,-12), and we can label them as follows:
• (x1, y1) = (6, -6) and (x2, y2) = (39, -12)
Then we have:
[tex]\begin{gathered} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ d=\sqrt{(39-6)^2+(-12-(-6))^2} \\ \\ d=\sqrt{(33)^2+(-12+6)^2} \\ \\ d=\sqrt{33^2+(-6)^2}=\sqrt{1089+36}=\sqrt{1125} \\ \\ d_{DE}=\sqrt{1125}\approx33.5410196625 \end{gathered}[/tex]Therefore, the distance between points D and E is √1125.
And we need to repeat the same steps to find the other distances.
Finding the distance between points E and FWe can proceed similarly as before:
The coordinates of the points are E(39, -12) and F(24, 18)
• (x1, y1) = (39, -12)
,• (x2, y2) = (24, 18)
Then we have:
[tex]\begin{gathered} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ d=\sqrt{(24-39)^2+(18-(-12))^2} \\ \\ d=\sqrt{(-15)^2+(18+12)^2}=\sqrt{(-15)^2+(30)^2}=\sqrt{225+900} \\ \\ d_{EF}=\sqrt{1125}\approx33.5410196625 \end{gathered}[/tex]Then the distance between points E and F is √1125.
Finding the distance between F and D
The coordinates of the points are F(24, 18) and D(6, -6)
• (x1, y1) = (24, 18) and (x2, y2) = (6, -6)
Then we have:
[tex]\begin{gathered} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ \\ d=\sqrt{(6-24)^2+(-6-18)^2}=\sqrt{(-18)^2+(-24)^2}=\sqrt{324+576} \\ \\ d=\sqrt{900}=30 \\ \\ d_{FD}=30 \end{gathered}[/tex]Now, we have the following measures for each of the sides of the triangle:
[tex]\begin{gathered} \begin{equation*} d_{DE}=\sqrt{1125}\approx33.5410196625 \end{equation*} \\ \\ d_{EF}=\sqrt{1125}\approx33.5410196625 \\ \\ d_{FD}=30 \end{gathered}[/tex]Therefore, in summary, according to the results, we have two sides that are congruent (they have the same size). Therefore, the triangle DEF is an isosceles triangle.
find the missing side. triangle trig
Answer:
see below; round to the correct number of digits (I couldn't see this part)
Step-by-step explanation:
cos(28°) = x / 14
x = 14cos(28°) ⇒ calculator
x ≈ 12.3612663
Answer:
x ≈ 12.36
Step-by-step explanation:
using the cosine ratio in the right triangle
cos28° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{x}{14}[/tex] ( multiply both sides by 14 )
14 × cos28° = x , that is
x≈ 12.36 ( to 2 dec. places )
Find the next term of the geometric sequence 3/2, 3/4, 3/8 , ...
Given:
Geometric sequence:
[tex]\frac{3}{2},\frac{3}{4},\frac{3}{8},.....[/tex]Find-: Next term of the geometric.
Sol:
Common ratio of a geometric sequence.
[tex]r=\frac{a_n}{a_{n-1}}[/tex]A common ratio is:
[tex]\begin{gathered} r=\frac{\frac{3}{4}}{\frac{3}{2}} \\ r=\frac{3}{4}\times\frac{2}{3} \\ r=\frac{1}{2} \end{gathered}[/tex]The next term is:
[tex]a_n=ra_{n-1}[/tex][tex]\begin{gathered} =\frac{1}{2}\times\frac{3}{8} \\ =\frac{3}{16} \end{gathered}[/tex]Suppose you found a CD that pays 2.1% interest compounded monthly for 6 years. If you deposit $12,000 now, how much will you have in the account in 6 years? (Rounded to the nearest cent.) What was the interest earned? Now suppose that you would like to have $20,000 in the account in 6 years. How much would you need to deposit now?
The interest earned is the difference between the initial amount that was deposited in the bank and the amount that is in the account after six years.
[tex]\begin{gathered} A\text{ = 12000\lparen1+}\frac{2.1\%}{12})^{6*12} \\ This\text{ is from the compound interest formula.} \\ A\text{ = 13609.89} \end{gathered}[/tex]Amount in the account: $13609.89
The interest earned is therefore: 13609.89 - 12000 = $1609.89
[tex]\begin{gathered} 20000\text{ = P\lparen1+}\frac{2.1\%}{12})^6*12 \\ 17634.24\text{ = P} \end{gathered}[/tex]To earn $20000 in six years he should invest $17634.24.
The map above is a road map of Learner County. Each centimeter on the map represents 30 miles. About how far is it from the town of Presley to Mt. Sametone on the road shown in red?A. 270 milesB. 180 milesC. 360 milesD. 390 miles
Scaling
Each centimeter on the map represents 30 miles in Learner County.
The town of Presley and Mt. Samerone are separated by a horizontal distance of 8 centimeters,
If we only consider the horizontal component of the distance, then both points are separated by 8 * 30 = 240 miles.
Since there is not an option for this number, we choose the closest, considering they are including a smaller vertical distance in the calculations, thus the answer is:
A. 270 miles
Answer:
A. 270 miles
Step-by-step explanation:
Find the GCF : (7x^2y ,x^2y^2,21x^4y^2)
The given expressions are
[tex]7x^2y,x^2y^2,21x^4y^2[/tex]We have to find the greatest common factor of the coefficients and variables.
The Greatest Common Factor between coefficients is 7 because that's the highest factor that's common.
The Greatest Common Factor between variables is
[tex]x^2y[/tex]Hence, the GCF is[tex]7x^2y[/tex]#32 At 10am, a green car leaves a house at a rate of 60 mph. At the same time, a blue carleaves the same house at a rate of 50 mph in the opposite direction. At what time will the carsbe 330 miles apart?#3b. Two bicyclists ride in the same direction. The first bicyclist rides at a speed of 8 mph.One hour later, the second bicyclist leaves and rides at a speed of 12 mph. How long will thesecond bicyclist have traveled when they catch up to the first bicyclist?
1. Determine the domain and range of the quadratic function shown in the graph and represent the domainand range using inequalities.
Domain: [ -5 , 4.5 )
Range: ( -7 , 6 ]
Using inequalities:
[tex]\text{Domain: }\lbrace-5\leq x<4.5\}[/tex][tex]\text{Range: }\mleft\lbrace-7