ANSWER:
49.254%
STEP-BY-STEP EXPLANATION:
The first thing is to calculate the value after it has drops by 33%, like this:
[tex]\begin{gathered} 28775-28775\cdot33\% \\ \\ 28775-28775\cdot0.33 \\ \\ 28775-9495.75=19279.25 \end{gathered}[/tex]Now, we calculate what should grow by the following equation:
[tex]\begin{gathered} 19279.25+19279.25\cdot \:x=28775\: \\ \\ x=\frac{28775\:-19279.25}{19279.25} \\ \\ x=\frac{9495.75}{19279.25} \\ \\ x=0.49254\cong49.254\% \end{gathered}[/tex]The percent that should grow is 49.254%
Suppose the coordinate of p=2 and PQ=8. Whare are the possible midpoints for PQ?
The midpoint for segment PQ can be calculated as:
[tex]\frac{P+Q}{2}[/tex]Then, the midpoint of PQ is:
[tex]\frac{2\text{ + Q}}{2}=1+0.5Q[/tex]Additionally, PQ can be calculated as:
[tex]PQ=\left|Q-P\right|[/tex]So:
[tex]\begin{gathered} \left|Q-P\right|=8 \\ \left|Q-2\right|=8 \end{gathered}[/tex]It means that:
[tex]\begin{gathered} Q-2=8\text{ or } \\ 2\text{ - Q = 8} \end{gathered}[/tex]Solving for Q, we get:
Q = 8 + 2 = 10 or Q = 2 - 8 = -6
Finally, replacing these values on the initial equation for the midpoint, we get:
If Q = 10, then:
midpoint = 1 + 0.5(10) = 1 + 5 = 6
If Q = -6, then:
midpoint = 1 + 0.5(-6) = 1 - 3 = -2
The possible midpoints for PQ are 6 and -2
9. SAILING The sail on Milton's schooner is the shape of a 30°-60°-90°triangle. The length of the hypotenuse is 45 feet. Find the lengths of thelegs. Round to the nearest tenth.
The triangle is shown below:
Notice how this is an isosceles triangle.
We can find the lengths of the hypotenuse by using the trigonometric functions:
[tex]\sin \theta=\frac{\text{opp}}{\text{hyp}}[/tex]Then we have:
[tex]\begin{gathered} \sin 45=\frac{21}{hyp} \\ \text{hyp}=\frac{21}{\sin 45} \\ \text{hyp}=29.7 \end{gathered}[/tex]Therefore the hypotenuse is 29.7 ft.
Which of the following shows the expansion of sum from n equals 0 to 4 of 2 minus 5 times n ?
(−18) + (−13) + (−8) + (−3) + 0
(−3) + (−8) + (−13) + (−18) + (−23)
2 + (−3) + (−8) + (−13) + (−18)
2 + 7 + 12 + 17 + 22
The option that indicates the required sum when n equals 0 to 4 of 2 minus 5 times n, is 2 + (−3) + (−8) + (−13) + (−18) (Option C)
What is the Sum of sequences?The sum of the terms of a sequence is called a series.
From the given sum of a sequence, we are to find the sum of the given sequence from n = 0 to n = 4
When n = 0
a(0) = 2 - 5(0)
a(0) = 2 - 0
a(0) = 2
When n = 1
a(1) = 2 - 5(1)
a(1) = 2 -5
a(1) = -3
When n = 2
a(2) = 2 - 5(2)
a(2) = 2 - 10
a(2) = -8
When n = 3
a(3) = 2 - 5(3)
a(3) = 2 - 15
a(3) = -13
When n = 4
a(4) = 2 - 5(4)
a(4) = 2 - 20
a(4) = -18
Hence the required sum is 2 + (−3) + (−8) + (−13) + (−18)
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The required sum is 2 + (−3) + (−8) + (−13) + (−18) when n equals 0 to 4 of 2 minus 5 times n, which is the correct answer that would be an option (C).
The given expression is (2 - 5n)
We to determine the sum of the given sequence from n = 0 to n = 4
Let the required sum is T₀ + T₁ + T₂ + T₃ + T₄
Substitute the value of n = 0 in the expression (2 - 5n) to get T₀
⇒ T₀ = 2 - 5(0) = 2 - 0 = 2
Substitute the value of n = 1 in the expression (2 - 5n) to get T₁
⇒ T₁ = 2 - 5(1) = 2 -5 = -3
Substitute the value of n = 2 in the expression (2 - 5n) to get T₂
⇒ T₂ = 2 - 5(2) = 2 - 10 = -8
Substitute the value of n = 3 in the expression (2 - 5n) to get T₃
⇒ T₃ = 2 - 5(3) = 2 - 15 = -13
Substitute the value of n = 4 in the expression (2 - 5n) to get T₄
⇒ T₄ = 2 - 5(4) = 2 - 20 = -18
Therefore, the required sum is 2 + (−3) + (−8) + (−13) + (−18)
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The answer is 57.3 provided by my teacher, I need help with the work
Apply the angles sum property in the triangle ABC,
[tex]62+90+\angle ACB=180\Rightarrow\angle ACB=180-152=28^{}[/tex]Similarly, apply the angles sum property in triangle BCD,
[tex]20+90+\angle BCD=180\Rightarrow\angle BCD=180-110=70[/tex]From triangle ABC,
[tex]BC=AC\sin 62=30\sin 62\approx26.5[/tex]From triangle BDC,
[tex]BD=BC\cos 20=26.5\cos 20\approx24.9[/tex]Now, consider that,
[tex]\angle BDE+\angle BDC=180\Rightarrow\angle BDE+90=180\Rightarrow\angle BDE=90[/tex]So the triangle BDE is also a right triangle, and the trigonometric ratios are applicable.
Solve for 'x' as,
[tex]x=\tan ^{-1}(\frac{BD}{DE})=\tan ^{-1}(\frac{24.9}{16})=57.2764\approx57.3[/tex]Thus, the value of the angle 'x' is 57.3 degrees approximately.ang
Find the sum of the arithmetic series -1+ 2+5+8+... where n=7.A. 56B. 184C. 92D. 380Reset Selection
The arithmetic series is:
-1 + 2 + 5 + 8 + .....
The first term, a = -1
The common difference, d = 2 - (-1)
d = 3
The number of terms, n = 7
Find the sum of the arithmetic series below
[tex]\begin{gathered} S_n=\frac{n}{2}[2a+(n-1)d] \\ \\ S_7=\frac{7}{2}[2(-1)+(7-1)(3)] \\ \\ S_7=\frac{7}{2}(-2+18) \\ \\ S_7=\frac{7}{2}(16) \\ \\ S_7=56 \end{gathered}[/tex]Therefore, the sum of the arithmetic series = 56
What is 3 +4.3+45?A4늘OB.B. 7O. 8○ D. 12
solution
[tex]3+4\frac{1}{3}=7\frac{1}{3}[/tex]answer: B
Can a triangle be formed with side lengths 17, 9, and 8? Explain.
Yes, because 17 + 9 > 8
Yes, because 17 + 8 < 9
No, because 9 + 8 > 17
No, because 8 + 9 = 17
Answer:
(d) No, because 8 + 9 = 17
Step-by-step explanation:
You want to know if side lengths 8, 9, and 17 can form a triangle.
Triangle inequalityThe triangle inequality requires the sum of the two short sides exceed the length of the longest side. For sides 8, 9, 17, this would require ...
8 + 9 > 17 . . . . . . . not true; no triangle can be formed
The sum is 8+9 = 17, a value that is not greater than 17. The triangle inequality is not satisfied. So, no triangle can be formed.
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Surface are of the wood cube precision =0.00The weight of the woo cube precision =0.00 The volume was 42.87 in3
Given:
The volume of the cube is 42.87 cubic inches.
The volume of a cube is given as,
[tex]\begin{gathered} V=s^3 \\ 42.87=s^3 \\ \Rightarrow s=3.5 \end{gathered}[/tex]The surface area of a cube is,
[tex]\begin{gathered} SA=6s^2 \\ SA=6\cdot(3.5)^2 \\ SA=73.5 \end{gathered}[/tex]Answer: the surface area is 73.5 square inches ( approximately)
Solve this equation 3n+8=20
Given the equation below
[tex]3n\text{ + 8 = 20}[/tex]Step 1
Collect like terms.
[tex]\begin{gathered} 3n=20-8 \\ 3n=12 \end{gathered}[/tex]Step 2
Divide both sides of the equation obtained, by the coefficient of the unknown.
[tex]\begin{gathered} \text{The unknown is n.} \\ \text{The co}efficient\text{ of n is 3.} \\ \text{Thus,} \\ \frac{3n}{3}=\frac{12}{3} \\ \Rightarrow n=4 \end{gathered}[/tex]Hence, the value of n in the equation is 4
Use an inequality to represent the corresponding Celsius temperature that is at or below 32° F.
C ≤ 0
Explanations:The given equation is:
[tex]F\text{ = }\frac{9}{5}C\text{ + 32}[/tex]Make C the subject of the equation
[tex]\begin{gathered} F\text{ - 32 = }\frac{9}{5}C \\ 9C\text{ = 5(F - 32)} \\ C\text{ = }\frac{5}{9}(F-32) \end{gathered}[/tex]At 32°F, substitute F = 32 into the equation above to get the corresponding temperature in °C
[tex]\begin{gathered} C\text{ = }\frac{5}{9}(32-32) \\ C\text{ = }\frac{5}{9}(0) \\ C\text{ = 0} \end{gathered}[/tex]The inequality representing the corresponding temperature that is at or below 32°F is C ≤ 0
Using the distributive property, show how to decompose 8 * 78
Given any three numbers a, b, and c.
By the distributive law, we must have:
a x (b + c) = (a x b) + (a x c)
Now to find 8 x78
8 x 78 = 8 x (70 + 8) = (8 x 70) + (8 x 8) = 560 + 64 = 624
write the equation for a quadratic function in vertex form that opebs down shifts 8 units to the left and 4 units down .
STEP - BY STEP EXPLANATION
What to find?
Equation for a quadratic equation.
Given:
Shifts 8 unit to the left.
4 units down
Step 1
Note the following :
• The parent function of a quadratic equation in general form is given by;
[tex]y=x^2[/tex]• If f(x) shifts q-units left, the f(x) becomes, f(x+q)
,• If f(x) shift m-units down, then the new function is, f(x) -m
Step 2
Apply the rules to the parent function.
8 units to the left implies q=8
4 units down implies m= 4
[tex]y=(x+8)^2-4[/tex]ANSWER
y= (x+8)²- 4
I NEED HELP
5C/2 = 20
you would have to do this backwards
20 times 2 would remove the /2
5c=40
40 divided by 5
is
8
C=8
is 2÷2 4 or am I wrong
2/2 = 1
The answer would be 1
Please help me no other tutor could or understand it
We must find the equation that models the amount of medication in the bloodstream as a function of the days passed from the initial dose. The initial dose is a and we are going to use x for the number of days and M for the amount of mediaction in the bloodstream. We are going to model this using an exponential function which means that the variable x must be in the exponent of a power:
[tex]M(x)=a\cdot b^x[/tex]We are told that the half-life of the medication is 6 hours. This means that after 6 hours the amount of medication in the bloodstream is reduced to a half. If the initial dose was a then the amount after 6 hours has to be a/2. We are going to use this to find the parameter b but first we must convert 6 hours into days since our equation works with days.
Remember that a day is composed of 24 hours so 6 hours is equivalent to 6/24=1/4 day. This means that the amount of medication after 1/4 days is the half of the initial dose. In mathematical terms this means M(1/4)=M(0)/2:
[tex]\begin{gathered} \frac{M(0)}{2}=M(\frac{1}{4}) \\ \frac{a\cdot b^0}{2}=a\cdot b^{\frac{1}{4}} \\ \frac{a}{2}=a\cdot b^{\frac{1}{4}} \end{gathered}[/tex]We can divide both sides of this equation by a:
[tex]\begin{gathered} \frac{\frac{a}{2}}{a}=\frac{a\cdot b^{\frac{1}{4}}}{a} \\ \frac{1}{2}=b^{\frac{1}{4}} \end{gathered}[/tex]Now let's raised both sides of this equation to 4:
[tex]\begin{gathered} (\frac{1}{2})^4=(b^{\frac{1}{4}})^4 \\ \frac{1}{2^4}=b^{\frac{1}{4}\cdot4} \\ b=\frac{1}{16} \end{gathered}[/tex]Which can also be written as:
[tex]b=16^{-1}[/tex]Then the equation that models how much medication will be in the bloodstream after x days is:
[tex]M(x)=a\cdot16^{-x}[/tex]Using this we must find how much medication will be in the bloodstream after 4 days for an initial dose of 500mg. This basically means that a=500mg, x=4 and we have to find M(4):
[tex]M(4)=500mg\cdot16^{-4}=0.00763mg[/tex]So after 4 days there are 0.00763 mg of medication in the bloodstream.
Now we have to indicate how much more medication will be if the initial dose is 750mg instead of 500mg. So we take a=750mg and x=4:
[tex]M(4)=750mg\cdot16^{-4}=0.01144mg[/tex]If we substract the first value we found from this one we obtained the required difference:
[tex]0.01144mg-0.00763mg=0.00381mg[/tex]So the answer to the third question is 0.00381mg.
PLEASE HELP AS SOON AS POSSIBLE PLEASE!! ( one question, can whole numbers be classified as integers and rational numbers)
ANSWER
G. 10 and -2 only
EXPLANATION
-5/4 is a fraction that can't be simplified. Therefore it is not an integer.
1.25 has decimals, so it is not an integer either.
10 and -2 are are integers.
5-3/2x>1/3what is x?
Coco, this is the solution to the inequality:
5 - 3x/2 ≥ 1/3
Subtracting 5 at both sides:
5 - 3x/2 - 5 ≥ 1/3 - 5
-3x/2 ≥ 1/3 - 15/3
-3x/2 ≥ -14/3
LCD (Least Common Denominator) between 2 and 3 : 6
-9x/6 ≥ -28/6
Dividing by -9/6 at both sides:
-9x/6 / -9/6 ≥ -28/6 / -9/6
x ≥ 28/9
In consequence, the correct answer is C. x ≥ 28/9
The slope and y-intercept of the relation represented by the equation 12x-9y+12=0 are:
12x - 9y +12 =0
To find the slope and y intercept, we want to put the equation in slope intercept form
y = mx+b where m is the slope and b is the y intercept
Solve the equation for y
Add 9y to each side
12x - 9y+9y +12 =0+9y
12x+12 = 9y
Divide each side by 9
12x/9 +12/9 = 9y/9
4/3 x + 4/3 = y
Rewriting
y = 4/3x + 4/3
The slope is 4/3 and the y intercept is 4/3
Need answer if you could show work would be nice
ExpenseYearly costor rateGasInsuranceWhat is the cost permile over the course ofa year for a $20,000 carthat depreciates 20%,with costs shown in thetable, and that hasbeen driven for 10,000miles?$425.00$400.00$110,00$100.0020%OilRegistrationDepreciationB. $4.10 per mileA. $1.10 per mileC. $0.25 per mileD. $0.50 per mile
Step 1:
Find the depreciation
[tex]\begin{gathered} \text{Depreciation = 20\% of \$20,000} \\ \text{Depreciation = }\frac{20}{100}\text{ }\times\text{ \$20000} \\ \text{Depreciation = \$4000} \end{gathered}[/tex]Step 2:
Total cost = $425 + $400 + $110 + $100 + $4000
Total cost = $5035
Final answer
[tex]\begin{gathered} \text{Cost per mile = }\frac{5035}{10000} \\ \text{Cost per mile = \$0.5035} \\ \text{Cost per mile = \$0.50} \end{gathered}[/tex]Option D $0.50 per mile
I need help to find the indicated operation:g(x)= -x^2 +4xh(x)= -4x-1Find (3g-h)(-3)
We have the following functions:
[tex]\begin{gathered} g\mleft(x\mright)=-x^2+4x \\ h\mleft(x\mright)=-4x-1 \end{gathered}[/tex]And we need to find:
[tex](3g-h)(-3)[/tex]Step 1. Find 3g by multiplying g(x) by 3:
[tex]\begin{gathered} g(x)=-x^2+4x \\ 3g=3(-x^2+4x) \end{gathered}[/tex]Use the distributive property to multiply 3 by the two terms inside the parentheses:
[tex]3g=-3x^2+12x[/tex]Step 2. Once we have 3g, we subtract h(x) to it:
[tex]3g-h=-3x^2+12x-(-4x-1)[/tex]Here we have 3g and to that, we are subtracting h which in parentheses.
Simplifying the expression by again using the distributive property and multiply the - sign by the two terms inside the parentheses:
[tex]3g-h=-3x^2+12x+4x+1[/tex]Step 4. Combine like terms:
[tex]3g-h=-3x^2+16x+1[/tex]What we just found is (3g-h)(x):
[tex](3g-h)(x)=-3x^2+16x+1[/tex]Step 5. To find what we are asked for
[tex]\mleft(3g-h\mright)\mleft(-3\mright)[/tex]We need to evaluate the result from step 4, when x is equal to -3:
[tex](3g-h)(-3)=-3(-3)^2+16(-3)+1[/tex]Solving the operations:
[tex](3g-h)(-3)=-3(9)^{}-48+1[/tex][tex](3g-h)(-3)=-27^{}-48+1[/tex][tex](3g-h)(-3)=-74[/tex]Answer:
[tex](3g-h)(-3)=-74[/tex]I need help with this question... the correct answer choice
Reflection over the x-axis:
(x,y)--->(x, -y)
and the question is what is not a reflection across the x-axis.
so,
the correct option is D which is:
R'(-9, 4) ----> R'(9, -4)
Because it is a reflection over the y-axis.
In windy cold weather, the increased rate of heat loss makes the temperature feel colder than the actual temperature. To describe an equivalent temperature that more closely matches how it “feels,” weather reports often give a windchill index, WCI. The WCI is a function of both the temperature F(in degrees Fahrenheit) and the wind speed v (in miles per hour). For wind speeds v between 4 and 45 miles per hour, the WCI is given by the formula(FORMULA SHOWN IN PHOTO)A) What is the WCI for a temperature of 10 F in a wind of 20 miles per hour?B) A weather forecaster claims that a wind of 36 miles per hour has resulted in a WCI of -50 F. What is the actual temperature to the nearest degree?
Let's remember what the variables mean:
F= temperature (in Fahrenheit),
v= wind speed.
A) The formula "works" when the wind speed is between 4 and 45 miles per hour. The question asks for a wind speed of 20 miles per hour. Then, we can apply the formula. Here,
[tex]\begin{cases}F=10 \\ v=20\end{cases}[/tex]Then,
[tex]\begin{gathered} WCI(10,20)=91.4-\frac{(10.45+6.69\cdot\sqrt[]{20}-0.447\cdot20)(91.4-10)}{22}\approx\ldots \\ \ldots91.4-116.2857=-24.8857 \end{gathered}[/tex]Approximating, the answer is
[tex]-25F[/tex]B) This question is just about to find F in the provided equation after replacing the given v and WCI. Let's do that:
[tex]\begin{gathered} -50=91.4-\frac{(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F)}{22}, \\ -141.4=-\frac{(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F)}{22}, \\ -3110.8=-(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F), \\ 3110.8=(10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36)(91.4-F), \\ \frac{3110.8}{10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36}=91.4-F, \\ F=91.4-\frac{3110.8}{10.45+6.69\cdot\sqrt[]{36}-0.447\cdot36}\approx1.2 \end{gathered}[/tex]Then, the actual temperature is
[tex]1F[/tex]A square has a perimeterof 8,000 centimeters. Whatis the length of each side ofthe of the square inmeters?
Answer:
20 meters
Explanation:
The formula for calculating the perimeter of a square is expressed as
perimeter = 4s
where
s is the length of each side of the square
From the information given,
perimeter = 8,000 centimeters
Recall,
1 cm = 0.01 m
8000cm = 8000 x 0.01 = 80 m
Thus,
80 = 4s
s = 80/4
s = 20
The length of each side of the square is 20 meters
Find the exact solution to the exponential equation. (No decimal approximation)
Let's solve the equation:
[tex]\begin{gathered} 54e^{3x+3}=16 \\ e^{3x+3}=\frac{16}{54} \\ e^{3x+3}=\frac{8}{27} \\ \ln e^{3x+3}=\ln (\frac{8}{27}) \\ 3x+3=\ln (\frac{2^3}{3^3}) \\ 3x+3=\ln (\frac{2}{3})^3 \\ 3x+3=3\ln (\frac{2}{3}) \\ 3x=-3+3\ln (\frac{2}{3}) \\ x=-1+\ln (\frac{2}{3}) \\ x=-1+\ln 2-\ln 3 \end{gathered}[/tex]Therefore the solution of the equation is:
[tex]x=-1+\ln 2-\ln 3[/tex]Can you please solve the last question… number 3! Thanks!
Let us break the shape into two triangles and solve for the unknowns.
The first triangle is shown below:
We will use the Pythagorean Theorem defined to be:
[tex]\begin{gathered} c^2=a^2+b^2 \\ where\text{ c is the hypotenuse and a and b are the other two sides} \end{gathered}[/tex]Therefore, we can relate the sides of the triangles as shown below:
[tex]25^2=y^2+16^2[/tex]Solving, we have:
[tex]\begin{gathered} y^2=25^2-16^2 \\ y^2=625-256 \\ y^2=369 \\ y=\sqrt{369} \\ y=19.2 \end{gathered}[/tex]Hence, we can have the second triangle to be:
Applying the Pythagorean Theorem, we have:
[tex]22^2=x^2+19.2^2[/tex]Solving, we have:
[tex]\begin{gathered} 484=x^2+369 \\ x^2=484-369 \\ x^2=115 \\ x=\sqrt{115} \\ x=10.7 \end{gathered}[/tex]The values of the unknowns are:
[tex]\begin{gathered} x=10.7 \\ y=19.2 \end{gathered}[/tex]What would the answer be?
Nvm, I got it wrong
Applying the definition of similar triangles, the measure of ∠DEF = 85°.
What are Similar Triangles?If two triangles are similar, then their corresponding angles are all equal in measure to each other.
In the image given, since E and F are the midpoint of both sides of triangle BCD, then it follows that triangles BCD and EFD are similar triangles.
Therefore, ∠DBC ≅ ∠DEF
m∠DBC = m∠DEF
Substitute
4x + 53 = -6x + 133
4x + 6x = -53 + 133
10x = 80
10x/10 = 80/10 [division property of equality]
x = 8
Measure of ∠DEF = -6x + 133 = -6(8) + 133
Measure of ∠DEF = 85°
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-27\sqrt(3)+3\sqrt(27), reduce the expression
Explanation
[tex]-27\sqrt[]{3}+3\sqrt[]{27}[/tex]Step 1
Let's remember one propertie of the roots
[tex]\sqrt[]{a\cdot b}=\sqrt[]{a}\cdot\sqrt[]{b}[/tex]hence
[tex]\sqrt[]{27}=\sqrt[]{9\cdot3}=\sqrt[]{9}\cdot\sqrt[]{3}=3\sqrt[]{3}[/tex]replacing in the expression
[tex]\begin{gathered} -27\sqrt[]{3}+3\sqrt[]{27} \\ -27\sqrt[]{3}+3(3\sqrt[]{3}) \\ -27\sqrt[]{3}+9\sqrt[]{3} \\ (-27+9)\sqrt[]{3} \\ -18\sqrt[]{3} \end{gathered}[/tex]therefore, the answer is
[tex]-18\sqrt[]{3}[/tex]I hope this helps you
Rewrite the fraction with a rational denominator:
[tex]\frac{1}{\sqrt{5} +\sqrt{3} -1}[/tex]
Give me a clear and concise explanation (Step by step)
I will report you if you don't explain
The expression with rational denominator is [tex]\frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]
How to rewrite the fraction?From the question, the fraction is given as
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1}[/tex]
To rewrite the fraction with a rational denominator, we simply rationalize the fraction
When the fraction is rationalized, we have the following equation
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{1}{\sqrt 5 + \sqrt{3} - 1} \times \frac{\sqrt 5 - \sqrt{3} + 1}{\sqrt 5 - \sqrt{3} + 1}[/tex]
Evaluate the products in the above equation
So, we have
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{(\sqrt 5)^2 - (\sqrt{3} + 1)^2}[/tex]
This gives
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{5 - 10 - 2\sqrt 3}[/tex]
So, we have
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{- 5 - 2\sqrt 3}[/tex]
Rationalize again
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{- 5 - 2\sqrt 3} \times \frac{- 5+2\sqrt 3}{- 5 +2\sqrt 3}[/tex]
This gives
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{(-5)^2 - (2\sqrt 3)^2}[/tex]
So, we have
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{25 -12}[/tex]
Evaluate
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]
Hence, the expression is [tex]\frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]
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The cost, c(x) in dollars per hour of running a trolley at an amusement park is modelled by the function [tex]c(x) = 2.1x {}^{2} - 12.7x + 167.4[/tex]Where x is the speed in kilometres per hour. At what approximate speed should the trolley travel to achieve minimum cost? A. About 2km/h B about 3km/h C about 4km/D about 5km/hr
The equation is modelled by the function,
c(x) = 2.1x^2 - 12.7x + 167.4
The general form of a quadratic equation is expressed as
ax^2 + bx + c
The given function is quadratic and the graph would be a parabola which opens upwards because the value of a is positive
Since x represents the speed, the speed at which the he