f(n) = f(n-1) + 3
substitute n= 2 in the above
f(2) = f (2-1) + 3
= f(1) + 3
= 5 + 3
= 8
substitute n = 3 in the formula
f(3) = f(3-1) + 3
= f(2) + 3
= 8 + 3
= 11
substituite n = 4
f(4) = f(4-1) + 3
= f(3) + 3
= 11 + 3
= 14
The first four terms are 5, 8, 11 and 14
i need help with this problem..Yolanda took out a 30-year mortgage for $80,000 at 10% How much wills he pay over money year? i assume 2666.66
Step 1:
Most mortgages are also simple interest loans, although they can certainly feel like compound interest. In fact, all mortgages are simple interest except those that allow negative amortization. An important thing to pay attention to is how the interest accrues on the mortgage.
Step 2:
[tex]Interest\text{ = }\frac{Prt}{100}\text{ }[/tex]Step 3:
Write the given data
P = $80000
t = 30 years
r = 10%
Step 4
[tex]\begin{gathered} \text{Interest = }\frac{80000\text{ }\times\text{ 30 }\times\text{ 10}}{100} \\ \text{Interest = \$240000} \end{gathered}[/tex]Final answer
Interest = $240000
the price of a calculator with discount of 45% which fraction is equal to 45%
Explanation
The price of the calculator is discounted 45%al
Since f is parallel to line g, use the diagram to the right right to answer the following question (I need help with problem D and the graph right next to it )
Given,
The line f and g are parallel lines.
a)The measure of angle 2 is 117 degree.
By exterior atlernate angle property,
[tex]\begin{gathered} \angle2=\angle7 \\ \angle7=117^{\circ} \end{gathered}[/tex]The measure of angle 7 is 117 degree.
b)The measure of angle 4 is 68 degree.
By sum of adjacent angle between two parallel lines property,
[tex]\begin{gathered} \angle4+\angle6=180^{\circ} \\ \angle6=180^{\circ}-68^{\circ} \\ \angle6=112^{\circ} \end{gathered}[/tex]The measure of angle 6 is 112 degree.
c)The measure of angle 5 is 32 degree.
By alternate interior angle property,
[tex]\begin{gathered} \angle4=\angle5^{} \\ \angle4=32^{\circ} \end{gathered}[/tex]The measure of angle 4 is 32 degree.
d)The measure of angle 7 is 121 degree.
By corresponding angle property,
[tex]\begin{gathered} \angle7=\angle3^{} \\ \angle3=121^{\circ} \end{gathered}[/tex]The measure of angle 3 is 121 degree.
Rick shoots a basketball at an angle of 60' from the horizontal. It leaves his hands 6 feet from the ground with a velocity of 25 ft/s.Step 1 of 2: Construct a set of parametric equations describing the shot. Answer
Solution:
Given:
[tex]\begin{gathered} Initial\text{ velocity,}u=25ft\text{ /s} \\ \theta=60^0 \end{gathered}[/tex]
The parametric equations are gotten by first resolving the velocity into horizontal and vertical components.
Recall;
[tex]\begin{gathered} speed=\frac{distance}{time} \\ distance=speed\times time \end{gathered}[/tex]Hence, the parametric equations are:
[tex]\begin{gathered} x=(25cos60)t \\ y=(25sin60)t+6 \end{gathered}[/tex]write the following degrees in radian measure (a) 80 degrees(b) 120 degrees (c) 90 degrees (d) 270 degrees(e) 135 degrees
Answer
(a) 80 degrees = (4π/9) radians = 0.444π radians = 1.397 radians.
(b) 120 degrees = (2π/3) radians = 0.667π radians = 2.095 radians.
(c) 90 degrees = (π/2) radians = 0.50π radians = 1.571 radians.
(d) 270 degrees = (3π/2) radians = 1.50π radians = 4.714 radians.
(e) 135 degrees = (3π/4) radians = 0.75π radians = 2.357 radians.
Explanation
To do degree to radians conversion, we need to first note that
360° = 2π radians
So, for each of these cases, if we let the value of the angle given in degree be x in radians, then we can easily solve for x for each ot them
(a) 80 degrees
80° = x radians
360° = 2π radians
We can write a mathematical relationship by cross multiplying
(360) (x) = (80) (2π)
360x = 160π
Divide both sides by 360
(360x/360) = (160π/360)
x = (4π/9) = 0.444π
80° = (4π/9) radians = 0.444π radians = 1.397 radians
(b) 120 degrees
120° = x radians
360° = 2π radians
We can write a mathematical relationship by cross multiplying
(360) (x) = (120) (2π)
360x = 240π
Divide both sides by 360
(360x/360) = (240π/360)
x = (2π/3) = 0.667π
120° = (2π/3) radians = 0.667π radians = 2.095 radians
(c) 90 degrees
90° = x radians
360° = 2π radians
We can write a mathematical relationship by cross multiplying
(360) (x) = (90) (2π)
360x = 180π
Divide both sides by 360
(360x/360) = (180π/360)
x = (π/2) = 0.50π
90° = (π/2) radians = 0.50π radians = 1.571 radians
(d) 270 degrees
270° = x radians
360° = 2π radians
We can write a mathematical relationship by cross multiplying
(360) (x) = (270) (2π)
360x = 540π
Divide both sides by 360
(360x/360) = (540π/360)
x = (3π/2) = 1.50π
270° = (3π/2) radians = 1.50π radians = 4.714 radians
(e) 135 degrees
135° = x radians
360° = 2π radians
We can write a mathematical relationship by cross multiplying
(360) (x) = (135) (2π)
360x = 540π
Divide both sides by 360
(360x/360) = (270π/360)
x = (3π/4) = 0.75π
135° = (3π/4) radians = 0.75π radians = 2.357 radians
Hope this Helps!!!
Jill mixes two types of concentrations of HCI (hydrochloric acid): 0.375 liters of 25% hydrochloric acid, and 0.625 liters of 65% hydrochloric acid. What is the HCI concentration of the mixed solution? O 56% O 40% O 50% O 446
Given:
Jill mixes two types of concentrations of HCI (hydrochloric acid):
a.) 0.375 liters of 25% hydrochloric acid and 0.625 liters of 65% hydrochloric acid.
To be able to find the final HCL concentration, we will be generating the following formula:
[tex]\text{ Original + Added = Result}[/tex][tex](0.375)(\frac{25}{100})\text{ + (0.625)(}\frac{65}{100})\text{ = (x)}(0.375\text{ + 0.625)}[/tex]Where,
x = the final concentration of HCL
Let's find x,
[tex](0.375)(\frac{25}{100})\text{ + (0.625)(}\frac{65}{100})\text{ = (x)}(0.375\text{ + 0.625)}[/tex][tex](0.375)(0.25)\text{ + (0.625)(0.65) = (x)(}1)[/tex][tex]0.09375\text{ + 0.40625 = x}[/tex][tex]0.5\text{ = x }\rightarrow\text{ x = 0.5}[/tex][tex]\text{ x = 0.5 x 100 = 50\%}[/tex]Therefore, the final concentration of the mixed solution is 50%.
What is the probability of rolling a die one time and having it land on a number greater than 4?
Let:
n = Number of outcomes = 6
A = roll a number greater than 4 = 2
Therefore:
[tex]\begin{gathered} P(A)=\frac{A}{n} \\ P(A)=\frac{2}{6}=\frac{1}{3}=0.33 \end{gathered}[/tex]Answer:
33%
If A={g,y,m,n,a,s,t,i,c} and U={a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z}, find A′.
Step 1:
What is the complement of set A' ? are elements or members of set A that are not in the universal set U.
Step 2
Set A = {g,y,m,n,a,s,t,i,c}
Universal set U = {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z}
Step 3:
A' = {b, d, e, f, h, j, k, l, o, p, q, r, u, v, w, x, z}
Final answer
A' = {b, d, e, f, h, j, k, l, o, p, q, r, u, v, w, x, z}
Two angles are supplementary angles if the sum of their measures is 180. Find the measures of two supplementary angles if the measure of one angle is 4 degrees less than three times the other.What are the measures of the two angles?
Answer:
The measures of the two angles are 46° and 134°.
Explanation:
Let x and y be the two angles:
x+y=180
It is mentioned that the measure of one angle is 4 degrees less than three times the other.
Let:
y=3x-4
Then, we substitute y=3x-4 into x+y=180.
So,
[tex]\begin{gathered} x+y=180 \\ x+(3x-4)=180 \\ \text{Simplify and rearrange} \\ x+3x-4=180 \\ 4x=180+4 \\ 4x=184 \\ x=\frac{184}{4} \\ \text{Calculate} \\ x=46 \end{gathered}[/tex]We substitute x=46 into x+y=180. So,
[tex]\begin{gathered} x+y=180 \\ 46+y=180 \\ \text{Simplify and rearrange} \\ y=180-46 \\ \text{Calculate} \\ y=134 \end{gathered}[/tex]Therefore, the measures of the two angles are 46° and 134°.
Hint: You should have three pairs of congruent corresponding angles AND three pairs of congruent corresponding sides.
The corresponding pairs of the two congruent triangles △AEC ≅ △BFD are:
∠A ≅ ∠B∠E ≅ ∠F∠C ≅ ∠DAE ≅ BFEC ≅ FDAC ≅ BDWhat is the congruency of triangles?Two triangles are said to be congruent if all three corresponding sides and all three corresponding angles have the same size. You can move, flip, twist, and turn these triangles to produce the same effect.So, the corresponding pairs of the given congruent triangles will be:
We know that all angles and sides are equal.Then, we have:
∠A ≅ ∠B∠E ≅ ∠F∠C ≅ ∠DAE ≅ BFEC ≅ FDAC ≅ BDTherefore, the corresponding pairs of the two congruent triangles △AEC ≅ △BFD are:
∠A ≅ ∠B∠E ≅ ∠F∠C ≅ ∠DAE ≅ BFEC ≅ FDAC ≅ BDKnow more about the congruency of triangles here:
https://brainly.com/question/29003974
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Use the information given to enter an equation in standard form.Slope is 4, and (3,6) is on the line.
Answer
The equation in the point slope form
y - 6 = 4 (x - 3)
Simplifying further,
y - 6 = 4x - 12
y = 4x - 12 + 6
y = 4x - 6
OR
y - 4x = -6
OR
-4x + y = -6
OR
-4x + y + 6 = 0
Explanation
The general form of the equation in point-slope form is
y - y₁ = m (x - x₁)
where
y = y-coordinate of a point on the line.
y₁ = This refers to the y-coordinate of a given point on the line
m = slope of the line.
x = x-coordinate of the point on the line whose y-coordinate is y.
x₁ = x-coordinate of the given point on the line
For this question,
Slope = m = 4
(x₁, y₁) = (3, 6)
x₁ = 3
y₁ = 6
y - y₁ = m (x - x₁)
y - 6 = 4 (x - 3)
Simplifying further,
y - 6 = 4x - 12
y = 4x - 12 + 6
y = 4x - 6
Hope this Helps!!!
How do you use the following formulas for an equation like this?
Let's begin by listing out the information given to us:
|CD| = 9.5, |CE| = 13.75, |AC| = 13.75 + 5.5 = 19.25, |BC| = 9.5 + x
Using Triangle proportionality theorem, we have:
[tex]\begin{gathered} |CE|\colon|EA|=|CD|\colon|DB| \\ 13.75\colon5.5=9.5\colon x\Rightarrow\frac{13.75}{5.5}=\frac{9.5}{x} \\ 13.75x=5.5\cdot9.5 \\ \frac{13.75x}{13.75}=\frac{5.5\cdot9.5}{13.75}\Rightarrow x=\frac{5.5\cdot9.5}{13.75} \\ x=3.8 \\ \\ \therefore|BD|=3.8 \end{gathered}[/tex]if the 4 in 47,502 was changed to a 7 how much would the value changed
You have the following number given in the exercise:
[tex]47,502[/tex]According to the information given in the exercise, the digit 4 (located in the ten thousands place) was changed to a 7. Then now it is:
[tex]77,502[/tex]In order to find how much the value would change, you must find the difference (the difference, by definition, is the result of a subtraction).
In this case, knowing the value of the digits, you can set up the following subtraction:
[tex]70,000-40,0000[/tex]Solving the subtraction, you get the following difference:
[tex]=30,000[/tex]Therefore, the answer is:
[tex]30,000[/tex]Answer:30,000
(Please message me if I got this wrong)
Step-by-step explanation: Take 47,502 and turn it into 77,502. If you minus 47,502, you can see the difference/how much the value changed, which is 30,000
A musician plans to perform 5 selections. In how many ways can the musician arrange the musical selections?
Given:
A musician plans to perform 5 selections.
To find the total number of possible ways he can arrange the musical selections:
At the first time, there are 5 possibilities to make the musical selections.
At the second time, there will be 4 possibilities to make the musical selections.
At the third time, there will be 3 possibilities to make the musical selections.
At the fourth time, there will be 2 possibilities to make the musical selections.
At the fifth time, there will be 1 possibility to make the musical selections.
So, we have,
[tex]\begin{gathered} ^5C_1\times^4C_1\times^3C_1\times^2C_1\times^1C_1=5\times4\times3\times2\times1 \\ =120\text{ ways} \end{gathered}[/tex]Hence, the answer is 120 ways.
Rewrite cot 24° in terms of its cofunction.cot 24° =(Type an exact answer. Simplify your answer. Type any angle measures in degrees.
Given the angle = 24
the complementary angles have a sum of 90
The complemantry angle of 24 will be = 90 - 24 = 66
so, cot 24 = tan 66
So, the answer will be tan 66
using the points that are given, what is the slop of this line?
what would be the cost of 707 square feet of topsoil at the price of $1.25 per square foot
Answer:
$883.75
Explanation:
The cost of 1 square foot of topsoil = $1.25.
Thus, the cost of 707 square feet of topsoil:
[tex]\begin{gathered} =707\times\$1.25 \\ =\$883.75 \end{gathered}[/tex]Find the coordinates of the vertices of the figure after the given transformation: T<2,4> translation 3A. J′(1,−1),E′(1,1),V′(4,3)B. J′(0,−2),E′(0,0),V′(3,2)C. J′(−2,−1),E′(−2,1),V′(1,3)D. J′(−3,−2),E′(−3,0),V′(0,2)
According to the given rule of transformation, any point (x,y) is transformed as follows:
[tex](x,y)\rightarrow(x+2,y+4).[/tex]Now, the coordinates of the vertices of the triangle are:
[tex]V(-1,-1),J(-4,-5),E(-4,-3).[/tex]Therefore:
[tex]\begin{gathered} V^{\prime}(-1+2,-1+4), \\ J^{\prime}(-4+2,-5+4), \\ E^{\prime}(-4+2,-3+4). \end{gathered}[/tex]Simplifying the above result, you get:
[tex]J^{\prime}(-2,-1),E^{\prime}\left(−2,1\right),V^{\prime}\left(1,3\right).[/tex]Answer: [tex]J^{\prime}(-2,-1),E^{\prime}(-2,1),V^{\prime}(1,3).[/tex]which number is in the tenths place : 123.456?Round the following number to the tenths place: 123.456?
Answer:
The number in the tenths place is
[tex]4[/tex]Rounding the number to the tenths place, we have;
[tex]123.5[/tex]Explanation:
Given the number;
[tex]123.456[/tex]The number in the tenths place is the first number after the decimal point. Which is;
[tex]4[/tex]Rounding the number to the tenths place, we have;
[tex]123.456\approx123.5\text{ (to the nearest tenth)}[/tex]6) 1,4,9,_,25,_,_,_,81Explain and fill the sequence, write the explicit and recursive formula for the sequence
Answer:
Explanation:
Here, we want to fill the sequence, write the recursive and explicit formulae
From the sequence, we can see that each of the numbers are perfect squares
Depending on the term, the number is squared
Take for example, 1^2 is 1, 2^2 is 4
The correct way of filling is thus to raise the term number to 2
So, we have to fill for the 4th term, the 6th term, the 7th term and the 8th term
We have that as follows:
[tex]\begin{gathered} 4thterm=4^2\text{ = 16} \\ 6thterm=6^2\text{ = 36} \\ 7thterm=7^2\text{ = 49} \\ 8thterm=8^2\text{ = 64} \end{gathered}[/tex]The sequence can then be written as:
[tex]1,4,9,16,25,36,49,64,81[/tex]Now, we want to write the explicit and recursive formula
The explicit formula is written in a way that it does not consider the term before the present term
We can easily write that as:
[tex]T_n=n^2[/tex]For the recursive formula, we write it as a mathematical function that takes into account the term before or after the current term
A point to note that there are odd number differences that increase by 3 as we move from term to term
We can see that:
Term 2 minus Term 1 is 3
Term 3 minus Term 2 is 5
Term 4 minus Term 3 is 7
Term 5 minuus Term 4 is 9
Thus, we have the recursive formula as:
[tex]\begin{gathered} T_n=T_{(n-1)}\text{ + n + n-1} \\ T_n=T_{(n-1)\text{ }}+\text{ 2n-1} \end{gathered}[/tex]Find the distance from the point to the line Y=-1x-3 and Q (2,3)
Solution
We can do the following:
Ax + By + C= 0
Rewriting the line we got:
1x +1y +3=0
And the point is: (x= 2, y= 3)
and we can use the following formula:
[tex]d=\frac{|Ax_1+By_1+C|}{\sqrt[]{A^2+B^2}}=\frac{|1\cdot2+1\cdot3+3|}{\sqrt[]{1^2+1^2}}=\frac{8}{\sqrt[]{2}}\cdot\frac{\sqrt[]{2}}{\sqrt[]{2}}=4\sqrt[]{2}[/tex]
Find the value of x in the circle below:(8x – 2)°(12.2 - 8)°1001.588.29
The total sum of the angles must give 360 then:
[tex]\begin{gathered} 360=(8x-2)+90+(12x-8)+100 \\ 360=(8x+12x)+(100-2+90-8) \\ 360=20x+180 \\ 20x=360-180 \\ 20x=180 \\ x=\frac{180}{20} \\ x=9º \end{gathered}[/tex]A = bh; solve for h
Given the equation:
[tex]A=b\cdot h[/tex]It's required to solve it for h, that is, isolate h as the only letter on the left side of the equation.
First, swap sides.
[tex]b\cdot h=A[/tex]Then, divide both sides of the equation by b:
[tex]\frac{b\cdot h}{b}=\frac{A}{b}[/tex]Simplify the left side:
[tex]h=\frac{A}{b}[/tex]An accountant executive had car expenses of $1025.58 for insurance, $1805.82 for gas, $37.92 for oil, and $288.27 for maintenance during the year in which 11,320 miles were driven. Find the cost per mile for these four items taken as a group. Round to the nearest tenth of a cent.
Answer:
The cost per mile for each and all the expenses is;
[tex]\begin{gathered} \text{For insurance: = 9.1 cents/mile} \\ \text{For gas: = 16.0 cents/mile} \\ \text{ For oil: = 0.3 cents/mile} \\ \text{ For maintenance: = 2.5 cents/mile} \\ \text{ Total cost per mile: = 27.9 cents/mile} \end{gathered}[/tex]Explanation:
Given that;
An accountant executive had car expenses of $1025.58 for insurance, $1805.82 for gas, $37.92 for oil, and $288.27 for maintenance during the year.
The sum of the four expenses is;
[tex]\begin{gathered} T=\text{ \$1025.58 + \$1805.82 + \$37.92 + \$288.27} \\ T=\text{ \$3157.59} \end{gathered}[/tex]During the year it travels 11,320 miles.
The cost per mile for each of the items are;
For Insurance;
[tex]\begin{gathered} \frac{\text{ \$}1025.58}{11320} \\ =\text{ \$0.09 per mile} \\ =9.1\text{ cents/mile} \end{gathered}[/tex]For gas;
[tex]\begin{gathered} \frac{\text{ \$1805.82}}{11320} \\ =\text{ \$0.1595 per mile} \\ =16.0\text{ cent/mile} \end{gathered}[/tex]For oil;
[tex]\begin{gathered} \frac{\text{ \$37.92}}{11320} \\ =0.0033 \\ =0.3\text{ cents/mile} \end{gathered}[/tex]For maintenance;
[tex]\begin{gathered} \frac{\text{ \$288.27}}{11320} \\ =0.0254 \\ =2.5\text{ cents/mile} \end{gathered}[/tex]The total cost per mile will be;
[tex]\begin{gathered} \frac{\text{ \$3157.59}}{11320} \\ =0.2789 \\ =27.9\text{ cents/mile} \end{gathered}[/tex]The cost per mile for each and all the expenses is;
[tex]\begin{gathered} \text{For insurance: = 9.1 cents/mile} \\ \text{For gas: = 16.0 cents/mile} \\ \text{ For oil: = 0.3 cents/mile} \\ \text{ For maintenance: = 2.5 cents/mile} \\ \text{ Total cost per mile: = 27.9 cents/mile} \end{gathered}[/tex]Write –9 43/100 as a decimal number
Let's rewrite the mixed number as a fraction, using the following formula:
[tex]a\frac{b}{c}=\frac{a\cdot c+b}{c}_{}[/tex]So:
[tex]-(9\frac{43}{100})=-(\frac{9\cdot100+43}{100})=-(\frac{900+43}{100})=-\frac{943}{100}[/tex]To write -943/100 we can use long division, or since we are dividing by 100 we can simply move the decimal point two units to the left, so:
[tex]-\frac{943}{100}=-9.43[/tex]Answer:
-9.43
Mead II) 5x + 5 = 45 5x 45-5 12) 31-8x + 8 = 80 31-8x1+5290 278-30 wwxt8-80-8 24x272 15x1=40 5x -40 X> 512 5* = -40 X = -8 x=8 and 2-8 X-3 513+4x) = -115
Please provide an image of the question. If you still want an explanation for problem 13, please don't close the session before I am done with the session.
In our previous session we were analysing the following absolute value equation when the session was abruptly closed:
3 | - 8 x | + 8 = 80
we subtracted 8 to both sides to isolate the term in "x" on the left
3 | - 8 x | = 80 - 8
3 | - 8 x| = 72
divide both sides by 3:
|- 8 x| = 72 / 3
|- 8 x| = 24
Now remove the absolute value bars considering:
1) That the expression inside it is POSITIVE so we DON'T change anything when removing the bars:
- 8 x = 24
now we divide boths sides by "- 8"
x = 24 / (- 8)
x = - 3
Then, when x = -3 notice that the expression - 8 x becomes -8 * (-3) = 24
corroborating that the expression inside the bars was a posotive number.
2) Considering the case that "- 8 x" is NEGATIVE , then when removing the absolute value bars we CHANGE the sign of the expression into:
8 x
Then we have:
8 x = 24
divide by 8 both sides
x = 24 / 8
x = 3
Then we have TWO answers to this equation:
x = -3 and x = 3
both satisfy the equation.
f(x) = 3x^2 + 6x - 59(x) = 4x^3 - 5x^2+ 6Find ( f + g)(x).
3x² + 6x - 5 = f(x)
+
4x³ - 5x² + 6 = g(x)
----------------------------
4x³ - 2x² + 6x + 1 = (f+g)(x)
Consider the two triangles, which are not drawn to scaleFor the two triangles to be similar by angle-angle similarity, which values could be x be?A. 27 or 115B. 38 or 77C. 52 or 77D. 115 or 153
Recall that the Angle-Angle criterion states that two triangles are similar if two of their angles are congruent.
Now, recall that the interior angles of a triangle add up to 180 degrees, therefore, the other side of the largest triangle has measure
[tex]180^{\circ}-115^{\circ}-38^{\circ}=27^{\circ}.[/tex]Therefore, the values that x could be are:
[tex]27\text{ or 115.}[/tex]Answer: First option.
[tex]27\text{ or 115.}[/tex]If m = 2 and n = 3 then, evaluate 1*m*3 + 2*n*2 + 4
Given:
[tex]1\times m\times3+2\times n\times2+4[/tex]Substitute the value of m=2 and n=3 into the
Find the length of a diagonal ofa square with sides of 10inches long.[?]V ] inches
the Given:
The side of the square is a = 10 inches.
Explanation:
The length of the diagonal of the square is determined by Pythagoras theorem.
[tex]\begin{gathered} d=\sqrt[]{a^2+a^2} \\ =\sqrt[]{2a^2} \\ =\sqrt[]{2}a \end{gathered}[/tex]Substitute 10 for a in the equation to determine the length of the diagonal of the square.
[tex]\begin{gathered} d=\sqrt[]{2}\cdot10 \\ =10\sqrt[]{2} \end{gathered}[/tex]So the answer is,
[tex]10\sqrt[]{2}[/tex]