To be able to determine the bags of fertilizer that the gardener will need, let's first determine the area of the garden.
Since the shape of the garden is a trapezoid, we will be using the following formula:
[tex]\text{ Area = }\frac{1}{2}H(B_1+B_2)[/tex]We get,
[tex]\text{ Area = }\frac{1}{2}H(B_1+B_2)[/tex][tex]\text{ = }\frac{1}{2}(50)(70\text{ + 40)}[/tex][tex]\text{ = }\frac{1}{2}(50)(110\text{) = }\frac{50\text{ x 110}}{2}[/tex][tex]\text{ = }\frac{5,500}{2}[/tex][tex]\text{ Area = }2,75m^2[/tex]Let's determine how many bags of fertilizer will be used.
[tex]\text{ No. of Bags of Fertilizer = }\frac{\text{ Area of Garden}}{\text{ Area that a Bag of Fertilizer can cover}}[/tex]We get,
[tex]\text{ = }\frac{2,750(m^2)}{125\frac{(m^2)}{\text{bag}}}[/tex][tex]\text{ No. of Bags of Fertilizer = }22\text{ Bags}[/tex]Therefore, the gardener will be needing 22 Bags of Fertilizer.
Convert y = 9x2 + 108x - 72 to vertex form by completing the square.
Answer:
Expressing the equation in vertex form we have;
[tex]y=9(x+6)^2-396[/tex]Vertex at (-6,-396)
Explanation:
We want to convert the quadratic equation given to vertex form by completing the square.
[tex]y=9x^2+108x-72[/tex]The vertex form of quadratic equation is of the form;
[tex]f(x)=a(x-h)^2+k[/tex]To do this by completing the square;
Firstly, let's add 72 to both sides of the qeuation;
[tex]\begin{gathered} y+72=9x^2+108x-72+72 \\ y+72=9x^2+108x \end{gathered}[/tex]Them we will add a number that can make the right side of the equation a complete square to both sides;
Adding 324 to both sides;
[tex]\begin{gathered} y+72+324=9x^2+108x+324 \\ y+396=9x^2+108x+324 \end{gathered}[/tex]factorizing the right side of the equation;
[tex]\begin{gathered} y+396=9(x^2+12x+36) \\ y+396=9(x+6)(x+6) \\ y+396=9(x+6)^2 \end{gathered}[/tex]Then, let us subtract 396 from both sides;
[tex]\begin{gathered} y+396-396=9(x+6)^2-396 \\ y=9(x+6)^2-396 \end{gathered}[/tex]Therefore, expressing the equation in vertex form we have;
[tex]y=9(x+6)^2-396[/tex]Vertex at (-6,-396)
Find the Midpoint of the two given endpoints of (-5, 6) and (9,7)
1) Given those endpoints, we can
Graph the system and find the vertices (corners of the darkest shaded area, where the lines intersect) of the region.f(x) = 2x-3f(x) 3XS-2(0, -3). (2.0), (0,0)(-2, 3), (-2,-6), (4,3)(0, -3), (-2,3), (4,3)(3,-2).(-2,-6), (3, 4)
We are given the following system of inequalities:
[tex]\begin{gathered} f(x)\ge\frac{3}{2}x-3 \\ f(x)\le3 \\ x\le-2 \end{gathered}[/tex]We are told to plot the graphs and find the coordinate of the vertices.
In order to find the vertices we need to plot each inequality.
Plot 1:
[tex]f(x)\ge\frac{3}{2}x-3[/tex]In order to plot this inequality, we simply choose two points because a line can be created with only two points.
The way to choose these two points, is to set x = 0 and find f(x) and set f(x) = 0 and find x. These would help us find the y-intercept and x-intercept respectively.
Let us perform this operation:
[tex]\begin{gathered} f(x)\ge\frac{3}{2}x-3 \\ \text{set x = 0} \\ f(x)\ge\frac{3}{2}(0)-3 \\ f(x)\ge-3 \\ \\ \text{set f(x)=0} \\ 0\ge\frac{3}{2}x-3 \\ \text{add 3 to both sides} \\ 3\ge\frac{3}{2}x \\ \therefore3\times\frac{2}{3}\ge x \\ \\ x\le2 \end{gathered}[/tex]From the above, we just need to plot (0, -3) and (2, 0) to find the inequality plot.
We can see the forbidden region. This is the region that does not conform to the inequalities.
Next, we move to the next system of inequality.
Plot 2:
[tex]f(x)\le3[/tex]Here, we just draw the line f(x) = 3 and then shade the forbidden region as well.
The forbidden region here is above the line because that is the region where f(x) is greater than 3, hence we shade it off.
Finally, the last inequality:
[tex]x\le-2[/tex]Plot 3:
We simply plot the line x = -2 and then shade the forbidden region
After plotting all three, we shall have the following:
j
The points V1, V2, and V3 where the lines meet are the coordinates of the vertices.
A picture of the vertices is attached below:
Thus, the vertices are (-2, 3), (4, 3) and (-2, -6)
The final answer is Option 2
Find the length of the arc, S, on the circle of radius are intercepted by central angle zero. Express the arc length in terms of X. Then round your answer to two decimal places. Radius, our equals 8 inches; central angle, zero equals 135°. First convert the degree measure into radians. Then use the formula S equals 0R, where S is the arc length zero is the measure of the central angle in radians and are is the radius of the circle
The length of an arc subtended by a central angle and 2 radii is
[tex]S=r\theta[/tex]Where:
r is the radius
Cita is the central angle in radian
Since the radius of the circle is 8 inches, then
[tex]r=8[/tex]Since the arc is subtended by a central angle of 135 degrees, then
[tex]\begin{gathered} \theta=135\times\frac{\pi}{180} \\ \theta=\frac{3}{4}\pi \end{gathered}[/tex]Substitute them in the rule above
[tex]\begin{gathered} S=8\times\frac{3}{4}\pi \\ S=6\pi \end{gathered}[/tex]The length of the arc is 6pi
We will find it in 2 decimal places
[tex]\begin{gathered} S=6\pi \\ S=18.85 \end{gathered}[/tex]The length of the arc is 18.85 inches
Find the exact length of the floor clearance, using metres.
the length of the floor clearance is 2.8 m
Explanationas we have 2 similar triangles ( ABC and AED) we can set a proportion
Step 1
a)let
[tex]ratio=\frac{vertical\text{ side}}{horizontal\text{ side}}[/tex]so
for triangle ABC ( divide the given measure by 100 to obtain meters)
[tex]ratio_1=\frac{0.30\text{ m}}{0.40\text{ m}}=\frac{3}{4}[/tex]and
for triangle AED
let
[tex]ratio_2=\frac{2.1}{floor\text{ clearance}}[/tex]Step 2
as the ratio is the same, set the proportion
[tex]\begin{gathered} ratio_1=ratio_2 \\ \frac{3}{4}=\frac{2.1}{floor\text{ clerance}} \\ solve\text{ for floor clearance} \\ floor\text{ }cleareance=\frac{2.1*4}{3} \\ floor\text{ }cleareance=2.8\text{ m} \end{gathered}[/tex]therefore, the length of the floor clearance is 2.8 m
I hope this helps you
The equation that models Earth's elliptical orbit around the sun is (x+2.5)^2/22,350.25+y^2/22,344=1 in millions of kilometers. If the sun is located at one focus and it’s coordinates are (0,0), find Earth's farthest distance from the sun in millions of kilometers.
Given the equation of the elliptical orbit is (x+2.5)^2/22,350.25+y^2/22,344=1.
This equation can be written as
[tex]\begin{gathered} \frac{(x+2.5)^2}{22350.25}+\frac{y^2}{22344}=1 \\ \frac{(x+2.5)}{(149.5)^2}+\frac{y^2}{(149.479)^2}=1 \end{gathered}[/tex]Now, if we shift this path by 2.5 units to left then we get
[tex]\frac{x^2}{(149.5)^2}+\frac{y^2}{(149.479)^2}=1[/tex]The farthest distance of the earth from the sun will be 149.5 - 2.5 = 147 million of kilometers
Thus, option C is correct.
Elizabeth wraps a gift box in the shape of a square pyramid. The figure below shows a net for the gift box. 6 in 6.8 in
The wrapping paper used by Elizabeth is equal to the area of the square pyramid which is 127.84 in.².
Dimension of the square base:
Side = 6.8 in.
Area of the base = 6.8 in. × 6.8 in.
A = 46.24 in.²
Dimension of the triangle:
Base = 6.8 in.
Height = 6 in.
Area of 1 triangle = 1/2 × 6.8 in. × 6 in.
A (triangle) = 20.4 in.²
Area of 4 triangles = 4 × 20.4 in.²
A' = 81.6 in.²
Total area of the square pyramid = A + A'
T = 46.24 in.² + 81.6 in.²
T = 127.84 in.²
Therefore, the wrapping paper used by Elizabeth is equal to the area of the square pyramid which is 127.84 in.².
Learn more about area here:
https://brainly.com/question/25292087
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Your question is incomplete. Please refer the content below:
Elizabeth wraps a gift box in the shape of a square pyramid.
The figure below shows a net for the gift box.
How much wrapping paper did she use?
how to solve y=2x+3 y=2x+1
how to solve y=2x+3
y=2x+1
In this problem, we have two parallel lines with different y-intercept, ( two different parallel lines) so the lines don't intersect
That means------> the system has no solution
Using a graphing tool
see the attached figure
please wait a minute
Remember that, when solving a system by graphing, the solution is the intersection point. in this problem, the system has no solution , because the lines don't intersect
How many factors are there for 36? What do you notice about the number of factors of 36 and the number of arrays Courtney can make with the photos
ANSWER:
9 factors
9 arrays
STEP-BY-STEP EXPLANATION:
The factors of the number 36 are:
[tex]1,2,3,4,6,9,12,18,36[/tex]Which means that there are a total of 9 factors.
The arrays would be:
1 by 36
36 by 1
2 by 18
18 by 2
3 by 12
12 by 3
4 by 9
9 by 4
6 by 6
There are a total of 9 arrays , we can see that the number of arrays is equal to the number of factors
Gene bought a living room suite for P85,000. He agreed to pay in 5 months at 12% simple interest rate. How much will he pay for the furniture?
Given the cost of the room 85000.
Rate of interest 12%
time =5 months=5/12 year
IN 5 months he will pay as interest
[tex]85000\times\frac{12}{100}\times\frac{5}{12}=4250[/tex]In total, he will pay
[tex]85000+4250=89250[/tex]Usain Bolt ran the 2012 Olympic 100m race and 9.63 seconds if he runs at this rate on a road with a speed limit of 25 miles per hour how will his speed compared to the speed limit justify your answer 25 meter equals 40234 m a h r
Let's calculate his velocity:
v = d/t
v = 100m/9.63s = 10.38m/s
The speed limit of the road is 25mi/h
Let's make a conversion:
[tex]\frac{25mi}{h}\times\frac{1609.34m}{1mi}\times\frac{1h}{3600s}=\text{ 11.18m/s}[/tex]Therefore, we can conclude, that the speed limit is greater than the velocity of Usain Bolt, since:
11.18m/s > 10.38m/s
given f(x)=e^-x^3 find the vertical and horizontal asymptotes
Given:
[tex]f\mleft(x\mright)=e^{-x^3}[/tex]To find the vertical and horizontal asymptotes:
The line x=L is a vertical asymptote of the function f(x) if the limit of the function at this point is infinite.
But, here there is no such point.
Thus, the function f(x) doesn't have a vertical asymptote.
The line y=L is a vertical asymptote of the function f(x) if the limit of the function (either left or right side) at this point is finite.
[tex]\begin{gathered} y=\lim _{x\rightarrow\infty}e^{-x^3} \\ =e^{-\infty} \\ y=0 \\ y=\lim _{x\rightarrow-\infty}e^{-x^3} \\ y=e^{\infty} \\ =\infty \end{gathered}[/tex]Thus, y = 0 is the horizontal asymptote for the given function.
2. Damian is buying movie tickets to a movie. The tickets cost $4.35 per ticket. Damian has $40.00. What is the greatest amount of tickets he can buy?
Given:
Cost of one ticket is, c = $4.35.
Total amount with Damian is, T = $40.00.
The objective is to find the number of tickets Damian can buy with this total amount.
Consider the number of tickets as x.
The equation for this situation can be represented as,
[tex]\begin{gathered} \text{Total amount=cost per ticket}\times\text{number of tickets } \\ T=c\times x \end{gathered}[/tex]Now, substitute the given values in the above equation.
[tex]\begin{gathered} 40=4.35x \\ x=\frac{40}{4.35} \\ x=9.19 \\ x\approx9 \end{gathered}[/tex]Hence, Damian can buy maximum 9 tickets with total cost of $40.00.
Round to the nearest whole number(I) 18.32 (li) 224.9 (ili) 3.511
I
Answer:
18
Explanation:
18.32
To round to the nearest whole number, we would consider the term immediately after the decimal point. If it is greater than or equal to 5, the last term before the decimal point increases by 1. If it is less than 5, the last term remains the same. In this case, 3 is less than 5. Thus, 8 remains the same. Thus, to the nearest whole number, the answer is
18
hi I don’t understand this question,can u do it step by step? Thanks!
The rule of the division of differentiation is
[tex]\frac{d}{dx}(\frac{u}{v})=\frac{u^{\prime}v-uv^{\prime}}{v^2}[/tex]The given function is
[tex]y=f(x)=\frac{x^2+3x+3}{x+2}[/tex]a)
Let u the numerator and v the denominator
[tex]\begin{gathered} u=x^2+3x+3 \\ u^{\prime}=2x+3 \end{gathered}[/tex][tex]\begin{gathered} v=x+2 \\ v^{\prime}=1 \end{gathered}[/tex]Substitute them in the rule above
[tex]\begin{gathered} \frac{dy}{dx}=\frac{(2x+3)(x+2)-(x^2+3x+3)(1)}{(x+2)^2} \\ \frac{dy}{dx}=\frac{2x^2+7x+6-x^2-3x-3}{(x+2)^2} \\ \frac{dy}{dx}=\frac{x^2+4x+3}{(x+2)^2} \\ \frac{dy}{dx}=\frac{(x+3)(x+1)}{(x+2)^2} \end{gathered}[/tex]We will differentiate dy/dx again to find d^2y/dx^2
[tex]\begin{gathered} u=x^2+4x+3 \\ u^{\prime}=2x+4 \end{gathered}[/tex][tex]\begin{gathered} v=(x+2)^2=x^2+4x+4 \\ v^{\prime}=2x+4 \end{gathered}[/tex]Then substitute them in the rule above
[tex]\begin{gathered} \frac{d^2y}{dx^2}=\frac{(2x+4)(x^2+4x+4)-(x^2+4x+3)(2x+4)}{(x^2+4x+4)^2} \\ \frac{d^2y}{dx^2}=\frac{(2x+4)\lbrack x^2+4x+4-x^2-4x-3\rbrack}{(x^2+4x+4)^2} \\ \frac{d^2y}{dx^2}=\frac{(2x+4)\lbrack1\rbrack}{(x^2+4x+4)^2} \\ \frac{d^2y}{dx^2}=\frac{(2x+4)}{(x+2)^4} \\ \frac{d^2y}{dx^2}=\frac{2(x+2)}{(x+2)^4} \\ \frac{d^2y}{dx^2}=\frac{2}{(x+2)^3} \end{gathered}[/tex]b)
The turning point is the point that has dy/dx = 0
Equate dy/dx by 0 to find the values of x
[tex]\begin{gathered} \frac{dy}{dx}=\frac{(x+3)(x+1)}{(x+2)^2} \\ \frac{dy}{dx}=0 \\ \frac{(x+3)(x+1)}{(x+2)^2}=0 \end{gathered}[/tex]By using the cross multiplication
[tex]\begin{gathered} (x+3)(x+1)=0 \\ x+3=0,x+1=0 \\ x+3-3=0-3,x+1-1=0-1 \\ x=-3,x=-1 \end{gathered}[/tex]Substitute x by -3 and -1 in f(x) to find y
[tex]\begin{gathered} f(-3)=\frac{(-3)^2+3(-3)+3}{-3+2} \\ f(-3)=\frac{3}{-1} \\ y=-3 \end{gathered}[/tex][tex]\begin{gathered} f(-1)=\frac{(-1)^2+3(-1)+3}{-1+2} \\ f(-1)=\frac{1}{1} \\ y=1 \end{gathered}[/tex]The turning points are (-3, -3) and (-1, 1)
c)
To find the inflection point equate d^2y/dx^2 by 0 to find x
[tex]\begin{gathered} \frac{d^2y}{dx^2}=\frac{2}{(x+2)^3} \\ \frac{d^2y}{dx^2}=0 \\ \frac{2}{(x+2)^3}=0 \end{gathered}[/tex]By using the cross multiplication
[tex]2=0[/tex]Which is wrong 2 can not be equal to zero, then
NO inflection point for the curve
d)
Since the denominator of the curve is x + 2, then
Equate it by 0 to find the vertical asymptote
[tex]\begin{gathered} x+2=0 \\ x+2-2=0-2 \\ x=-2 \end{gathered}[/tex]There is a vertical asymptote at x = -1
Since the greatest power of x up is 2 and the greatest power of down is 1, then there is an Oblique asymptote by dividing up and down
[tex]\begin{gathered} \frac{x^2+3x+3}{x+2}=x+1 \\ y=x+1 \end{gathered}[/tex]The Oblique asymptote is y = x + 1
No horizontal asymptote
e)
This is the graph of y = f(x)
This is the graph of y = f(IxI)
f)
For the curve
[tex]y=\frac{x^2-3x+3}{2-x}[/tex]Take (-) sign as a common factor down, then
[tex]\begin{gathered} y=\frac{(x^2+3x+3)}{-(-2+x)} \\ y=-\frac{(x^2-3x+3)}{(x-2)} \end{gathered}[/tex]Since the sign of y is changed, then
[tex]y=-f(x)[/tex]Then it is the reflection of f(x) about the y-axis we can see it from the attached graph
The red graph is f(x)
The purple graph is -f(x) which is the equation of the last part
the graph shows the relationship between the length of time Ted spends knitting and the number of scars he Knits. what does 16 meak in this situation ( ima send a picture of the graph )
we have the point (1,16)
that means
1 scar
16 hours
is the option C
What about takes four hours to travel 128 km going upstream and return it takes two hours going down stream what is the rate of the boat in Stillwater and what is the rate of a Current
Since the rate = distance/time
Since the distance is 128 km
Since the time of upstream is 4 hours
Then the rate of the boat in the still water is
[tex]\begin{gathered} R_s=\frac{128}{4} \\ R_s=32km\text{ per hour} \end{gathered}[/tex]Since the boat took 2 hours downstream, then
The rate of the current is
[tex]\begin{gathered} R_c=\frac{128}{2} \\ R_c=64km\text{ per hour} \end{gathered}[/tex]Logarithmic help is needed. Be sure to note the differences between logarithmic and exponential forms in each equation.
The pattern in converting logarithmic form to exponential form and vice versa is this:
[tex]y=b^x\leftrightarrow\log _by=x[/tex]For the first exponential equation that is 16 = 8^4/3, our y = 16, b = 8, and x = 4/3. Let's plug this in the logarithmic pattern.
[tex]\begin{gathered} \log _by=x \\ \log _816=\frac{4}{3} \end{gathered}[/tex]The logarithmic form of the first equation is log₈ 16 = 4/3.
Now, let's move to the second one.
[tex]\log _5(15,625)=6[/tex]b = 5, y = 15, 625, and x = 6. Let's plug these in to the exponential pattern.
[tex]\begin{gathered} y=b^x \\ 15,625=5^6 \end{gathered}[/tex]Hence, the exponential form of the second equation is 15, 625 = 5⁶.
Ted has run 12 miles this month. each day he wants to run 3 miles until he reaches his goal of 48 miles. Write an equation and solve
If x is the number of days Ted is going to run, the given situation can be written in an algebraic way as follow:
12 + 3x = 48
12 because this is the number of miles Ted has already ran, 3x is the number of miles Ted has run after x days, and 48 because he wants to reach his goal of 48 miles.
In order to determine how many days he need to reach his goal, you solve the previous equation for x, just as follow:
12 + 3x = 48 subtract 12 both sides
3x = 48 - 12 simplify
3x = 36 divide by 3 both sides
x = 36/3
x = 12
Then, Ted needs 12 days to reach his goal of 48 miles
Are the triangles below congruent?If so, write a congruence statement and say why
Answer:
123(8)67/#;#--#442627!* aopqppwiue
Find each angle measure in the figure. (x + 30) The angle measures are and : Use the equation to justify your answer. x + (x +30) + 2x =(can you write the answers on the picture please)
x = 50º
x +30º = 80º
2x = 100º
1) Since in any triangle the Sum of the interior angles is equal to 180º, we can write for this triangle:
x + 2x + x +30 = 180º Combine like terms
3x +30 = 180 Subtract 30 from both sides
3x = 150
x = 150/3
x= 50
2) So the angles are:
x = 50º
x +30º = 80º
2x = 100º
3) And the answer is 50º, 80º, and 100º
Reduce the rational expression to lowest terms. If it is already in lowest terms, enter the expression in the answer box. Also, specify any restrictions on the variable.y³ - 2y² - 9y + 18/y² + y - 6Rational expression in lowest terms:Variable restrictions for the original expression: y
ANSWER
[tex]\begin{gathered} \text{ Rational expression in lowest terms: }y-3 \\ \\ \text{ Variable restrictions for the original expression: }y\ne2,-3 \end{gathered}[/tex]EXPLANATION
We want to reduce the rational expression to the lowest terms:
[tex]\frac{y^3-2y^2-9y+18}{y^2+y-6}[/tex]First, let us factor the denominator of the expression:
[tex]\begin{gathered} y^2+y-6 \\ \\ y^2+3y-2y-6 \\ \\ y(y+3)-2(y+3) \\ \\ (y-2)(y+3) \end{gathered}[/tex]Now, we can test if the factors in the denominator are also the factors in the numerator.
To do this for (y - 2), substitute y = 2 in the numerator. If it is equal to 0, then, it is a factor:
[tex]\begin{gathered} (2)^3-2(2)^2-9(2)+18 \\ \\ 8-8-18+18 \\ \\ 0 \end{gathered}[/tex]Since it is equal to 0, (y - 2) is a factor. Now, let us divide the numerator by (y -2):
We have simplified the numerator and now, we can factorize by the difference of two squares:
[tex]\begin{gathered} y^2-9 \\ \\ y^2-3^2 \\ \\ (y-3)(y+3) \end{gathered}[/tex]Therefore, the simplified expression is:
[tex]\frac{(y-2)(y-3)(y+3)}{(y-2)(y+3)}[/tex]Simplify further by dividing common terms. The expression becomes:
[tex]y-3[/tex]That is the rational expression in the lowest terms.
To find the variable restrictions, set the denominator of the original expression to 0 and solve for y:
[tex]\begin{gathered} y^2+y-6=0 \\ \\ y^2+3y-2y-6=0 \\ \\ y(y+3)-2(y+3)=0 \\ \\ (y-2)(y+3)=0 \\ \\ y=2,\text{ }y=-3 \end{gathered}[/tex]Those are the variable restrictions for the original expression.
write as a product:y raised to the 3 - y raised to the 5
We will have the following:
[tex]y^3-y^5=y^3(1-y^2)[/tex]***Explanation***
In order to solve we factor the common values.
We can see that the smallest exponent is 3 and if we subtract y^3 from both values we will have that
[tex]y^3-y^5=y^3\cdot1-y^3\cdot y^2[/tex]So, we can see that they share the same common value, thus:
[tex]y^3(1-y^2)[/tex]We also must remember that:
[tex]y^a\cdot y^b=y^{a+b}[/tex]what can you prove congruent from your given
Answer:
a. line OL bisects angle MLN
b. triangle MLO is congruent to triangle OLN
c. line OL proves that they are congruent through reflexive property.
Step-by-step explanation:
hope this helps!
Complete the table for the given rule. Rule:y is 2 more than 4 times x
We have been given the relationship between x and y to be
y = 4x + 2
To complete the table, we will substitute the value of x = 0, 2, and 4 into the equation
when x = 0
y = 4 x 0 + 2 = 0+ 2 = 2
when x = 2
y = 4 x 2 + 2 = 10
when x = 4
y = 4 x 4 + 2 = 18
The answer is given below
(n-1)9. Expand11-11 + 2 + 3 + 4 + 5 + 60 + 1 + 2 + 3 + 4 + 5 + 60 + 1 + 2 + 3 +4 + 5AB(-1) + (-2) + (-3) + (-4) + (-5) + (-6)
To expand the given summation, we proceed as follows:
[tex]\begin{gathered} \text{Given:} \\ \sum ^6_{n\mathop=1}(n-1) \\ \Rightarrow\text{ }\sum ^6_{n\mathop{=}1}(n)-\text{ }\sum ^6_{n\mathop{=}1}(1) \\ \text{Now:} \\ \sum ^6_{n\mathop{=}1}(n)\text{ is the sum of the first six natural numbers (1,2,3,4,5,6)} \\ \text{And:} \\ \sum ^6_{n\mathop{=}1}(1)\text{ is simply (6}\times1)--That\text{ is, the number 1 added to itself six times } \\ \text{Therefore, we have:} \\ \Rightarrow\text{ }\sum ^6_{n\mathop{=}1}(n)-\text{ }\sum ^6_{n\mathop{=}1}(1) \\ \Rightarrow(1+2+3+4+5+6)-(1+1+1+1+1+1) \\ \Rightarrow(1+2+3+4+5+6)-(6) \\ \Rightarrow(1+2+3+4+5) \\ \end{gathered}[/tex]Therefore:
[tex]\sum ^6_{n\mathop{=}1}(n-1)\text{ = 1+2+3+4+5}[/tex]So, the correct option is option C
This is because the sum: 0+1+2+3+4+5 gives the same value as the sum: 1+2+3+4+5
Give the slope and the y intercept of the line 92 2y - 3 = 0. Slope = y intercept = 0, Enter your answers as integers or as reduced fractions in the form A/B
Answer
Slope = (-9/2)
y-intercept = (-3/2)
Explanation
The slope and y-intercept form of the equation of a straight line is given as
y = mx + b
where
y = y-coordinate of a point on the line.
m = slope of the line.
x = x-coordinate of the point on the line whose y-coordinate is y.
b = y-intercept of the line.
So, to answer this question, we will express the equation of the line given in this form.
-9x - 2y - 3 = 0
-2y = 9x + 3
Divide through by -2
(-2y/-2) = (9x/-2) + (3/-2)
y = (-9/2)x + (-3/2)
Hope this Helps!!!
There is 1 teacher for every 18 students on a school trip. How many teachers are there if 72 students go ve values to create a proportion that can be used to solve the problem.
Proportion 1 : 18
x : 72
Ratio
1 teacher / 18 students = number of teacher / 72 students
1/18 = x/ 72
x= 72/ 18
x = 4
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Answer
if 72 students go, 4 teaches are required.
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AB||CD, BE=DE, AE=CE, andAB=CD. Can we conclude thatthe two triangles are congruent?YesNo
The answer is YES
Because their three sides are of equal lenght
Deb's Diner offers its clients a choice of regular and diet soda. Last night, the diner served 34 sodas in all, 50% of which were regular. How many regular sodas did the diner serve?