We need to write the given statement in math terms:
y increased by 4
Recall that increased means that we are adding 4 to the previous number.
Then the mathematical expression is:
y + 4
The quantity y has been increased by 4 units.
Use properties of exponents to simplify the expression. Express answers in exponential form with positive exponents only. Assume that any variables in denominators are not equal to zero.(-2x^4 y^-4)(-5x^4 y^2)
Given:
The given expression is
[tex](-2x^4y^{-4})(-5x^4y^2)[/tex]Required:
We have to simplify the given expression.
Explanation:
We know that when the same variables are multiplied then their power will add. We use this and proceed as:
[tex]\begin{gathered} (-2x^4y^{-4})(-5x^4y^2) \\ \\ =(-2)(-5)x^{4+4}y^{-4+2} \end{gathered}[/tex][tex]=10x^8y^{-2}[/tex]Final answer:
Hence the final answer is
[tex]10x^8y^{-2}[/tex]The height of a ball above the ground as a function of time is given by the functionℎ()=−32^2+8+3where h is the height of the ball in feet and t is the time in seconds. When does the ball hit the ground? Round to the nearest thousandth (3 places past the decimal).
The height of the ball is modeled by
[tex]h(t)=-32t^2+8t+3[/tex]When the ball hit the ground, the value of h(t) becomes 0.
Now, let us solve this equation.
[tex]\begin{gathered} h(t)=0 \\ -32t^2+8t+3=0 \\ 32t^2-8t-3=0 \\ t=\frac{8\pm\sqrt[]{64+384}}{64} \\ =0.456,-0.206 \end{gathered}[/tex]So, the ball hit the ground after 0.456 seconds
May I need help with plotting. I have tried may ways but I still could not get the correct answers or plot the graph right.
Solution:
Given the figure as shown below:
The large dot is labeled as A as shown above.
The Point in the original figure is:
[tex](5,6)[/tex]When the figure is translated 6 units left and 8 units down, we have the point in final figure to be
[tex](5,6)\to(5-4,6-8)\implies(-1,-2)[/tex]Hence,
the point in original figure is (5, 6),
the point in final figure is (-1, -2).
If f(x)=(4x-6)/x , what is the average rate of change of f(x) over the interval [-3, 6]?А.1/3В.1/9 C. -1/3D. -3
Answer:
C. -1/3
Explanation:
Given the function:
[tex]f(x)=\frac{4x-6}{x}[/tex]To determine the average rate of change of f(x) over the interval [-3, 6]:
First, we evaluate f(-3) and f(6):
[tex]\begin{gathered} f(-3)=\frac{4(-3)-6}{-3}=\frac{-12-6}{-3}=\frac{-18}{-3}=6 \\ f(6)=\frac{4(6)-6}{6}=\frac{24-6}{6}=\frac{18}{6}=3 \end{gathered}[/tex]The average rate of change over [-3,6] is:
[tex]\begin{gathered} \text{Rate}=\frac{f(-3)-f(6)}{-3-6} \\ =\frac{6-3}{-3-6} \\ =\frac{3}{-9} \\ =-\frac{1}{3} \end{gathered}[/tex]The correct choice is C.
At the grocery store, Garrett bought 1.75 pounds ground beef at $6.99 a pound, 1.35 pounds ofbroccoli at $2.49 a pound, and one Vidalia onion at $0.49. Which could be the total amount of hispurchase?A. $9.97B. $15.97C. $16.08D. $16.96Violet by the dresser at a yard sale for $40. She decides to sell it online for a profit. If island wants to sell the dresser for at least $65, what is the smallest mark up she can make it to the original price of the dresser?A. 60%B. 62.5%C. 65%D. 66.7%
At the grocery store, Garrett bought 1.75 pounds ground beef at $6.99 a pound, 1.35 pounds of broccoli at $2.49 a pound, and one Vidalia onion at $0.49. Which could be the total amount of his purchase?
Total Cost of Beef = 6.99 * 1.75 = 12.2325
Total cost of Broccoli = 2.49 * 1.35 = 3.3615
Total Cost of Onion = 0.49
The sum total = 12.2325 + 3.3615 + 0.49 = 16.084
Rounded to the nearest cent, that is,
$16.08AnswerCOn 1 16. are ) Lola collects blood donations at a clinic. 7 of the donations are Type O, 3 16 8 1 Type A, and are Type AB. The 16 remaining are Type B. What part of the blood donations are Type B? On 8 5 O n=5 1 whole 16 On= 11 16 7 16 ထမ
Answer:
n = 1/8
Explanation:
We're told that 7/16 of the blood donations Lola connected are Type O, 3/8 are type A, 1/16 are type AB and the remaining are type B.
Let n represent the type B blood donations.
So we can go ahead and find n by adding all the fractions and equating it to 1;
[tex]\frac{7}{16}+\frac{3}{8}+\frac{1}{16}+n=1[/tex]To clear the fractions, let's multiply both sides of the equation by the LCM which is 16, we'll have;
[tex]\begin{gathered} 7+6+1+16n=16 \\ 14+16n=16 \\ 16n=16-14 \\ n=\frac{2}{16} \\ \therefore n=\frac{1}{8} \end{gathered}[/tex]Therefore, type B blood donations are 1/8
In one month, Logan and Matt, sold two types of couches: economy and fancy. The following lists show theamount each salesman sold and the price of each item type:Economy FancyPriceLogan 69 59Economy $407Matt 62 80$878How much revenue did each salesman earn for his company?Logan: $FancyMatt: $
We are given that there are two salesmen, Logan and Matt.
To determine the revenue, that is total sales made by each of these sales men, we can draw up a table showing the quantity of couches sold multiplied by the amount each was sold for. The details are;
[tex]\begin{gathered} \text{Prices;} \\ \text{Economy}-\text{ \$407} \\ \text{Fancy}-\text{ \$878} \end{gathered}[/tex][tex]\begin{gathered} \text{Salesman; Logan} \\ \text{Economy}-69 \\ \text{Fancy}-59 \end{gathered}[/tex][tex]\begin{gathered} \text{Salesman; Matt} \\ \text{Economy}-62 \\ \text{Fancy}-80 \end{gathered}[/tex]The total sales made by Logan can now be calculated as shown;
[tex]\begin{gathered} \text{Logan} \\ \text{Economy}=69\times407=28083 \\ \text{Fancy}=59\times878=51802 \\ \text{Total}=28083+51802 \\ \text{Total}=79,885 \end{gathered}[/tex]The total sales made by Matt can also be calculated as shown;
[tex]\begin{gathered} \text{Matt} \\ \text{Economy}=62\times407=25234 \\ \text{Fancy}=80\times878=70240 \\ \text{Total}=25234+70240 \\ \text{Total}=95,474 \end{gathered}[/tex]ANSWER:
Therefore, the revenue made by each salesman is shown below;
Logan: $79,885
Matt: $95,474
Write the equation of the line that has a slope of-2/3 and passes through the point (3, 10)
Given:
The slope of the line = m = -2/3
The line passes through the point (3, 10)
We will write the equation of the line with the point-slope form:
[tex]\begin{gathered} y-k=m\cdot(x-h) \\ y-10=-\frac{2}{3}(x-3) \end{gathered}[/tex]Convert the point-slope form into the slope-intercept form
Solve the equation for y
[tex]\begin{gathered} y-10=-\frac{2}{3}x-\frac{2}{3}\cdot(-3) \\ y-10=-\frac{2}{3}x+2 \\ y=-\frac{2}{3}x+2+10 \\ \\ y=-\frac{2}{3}x+12 \end{gathered}[/tex]so, the answer will be:
[tex]y=-\frac{2}{3}x+12[/tex]1. Which properties would you use to find the value of x in the equation 8(3x – 6.5) = 56?
Distributive property.
What is the value of 1/2 x2 -10 when x = 6?
The numerical value of expressions
We are given the expression:
[tex]P=\frac{1}{2}x^2-10[/tex]It's required to find the value of the expression when x=6
Substituting:
[tex]P=\frac{1}{2}6^2-10[/tex]Since 6 squared is 36:
[tex]P=\frac{1}{2}36-10=18-10=8[/tex]The value is 8
PLEASE RESPOND ASAP!!!! Four tangents are drawn from E to two concentric circles. A, B, C, and D are the points of tangency. Can you name as many pairs of congruent triangles as possible and tell how you can show each pair is congruent?
There are 2 pairs of congruent angles (equal)
Triangles that have their sides inside the same circle are congruent triangles:
The ones with sides CD and AB (outside ones, inside the big circle)
And the ones with sides BO and OC (interior ones, inside the small circle)
Evaluate the expression when b=-3 and y=7. 4b-y
-19
ExplanationStep 1
Let
b=-3
y=7
Step 2
replace
[tex]\begin{gathered} 4b-y \\ 4(-3)-7 \\ -12-7 \\ -19 \end{gathered}[/tex]so , the answer is -19
I hope this helps you
Solve the following problem for the specified variable C=1/3h(s+b)for s
We are given the following equation:
[tex]C=\frac{1}{3}h(s+b)[/tex]To solve for "s" we will first multiply both sides by 3:
[tex]3C=h(s+b)[/tex]Now we divide both sides by h:
[tex]\frac{3C}{h}=s+b[/tex]Now we subtract b to both sides:
[tex]\frac{3C}{h}-b=s[/tex]And thus we solve for "s".
1. Ling's home is located at (2, 3), her grandma's house is located at (2, -3), and the store is located at (-2, 0). Plot the locations on a coordinate grid.
Explanation:
Ordered pairs are always (x, y). So the first number is the x-coordinate and the second number is the y-coordinate.
Answer:
determine if the relation is a function. if not, identify two ordered pairs as proof
We have a vertical line x = 4.
If we define a function as y = f(x), then for f(x) to be a function, there is one and only value of f(x) for each value of x.
Taking that into account, in this relation, we have infinite values of f(x) for the same value of x.
For example, (4,0) and (4,10) are part of the relation. Both ordered pairs have the same value of x and different values of y, so they are enough proof to convlude that x = 4 is not a function.
Information is given about a polynomial f(x) whose coefficients are real numbers. Find the remaining zeros of f. Degree 3; zeros 9, -5-i
Given:
The degree of the polynomial f is 3.
the zeros are 9 and -5-i.
We know that the degree of the polynomial = the number of zeros of the polynomial.
The number of zeros of the polynomial =3.
We have two zeros, we need to find the one zero of the given polynomial.
Recall that the complex zeros are always in pairs.
The conjugate of -5-i is
[tex]-5+i[/tex]Hence the remaining zero of the polynomial is
[tex]-5+i[/tex]Antonio threw a ball with an upward velocity of 6 meters per second from a height of 8 meters. The formula below describes this situation. Which is closest to the time it will take the ball to hit the ground? h(x) = – 4.9t^2 + 60t + 8
Antonio threw a ball with an upward velocity of 6 meters per second from a height of 8 meters. The formula below describes this situation. Which is closest to the time it will take the ball to hit the ground? h(x) = – 4.9t^2 + 60t + 8
we know that
when the ball hit the ground, the function h(x) is equal to zero
so
– 4.9t^2 + 6t + 8=0
using the quadratic formula
[tex]t=\frac{-b\pm\sqrt[\square]{b^2-4ac}_{}}{2a}[/tex]we have
a=-4.9
b=6
c=8
substitute the values in the formula
[tex]\begin{gathered} t=\frac{4.9\pm\sqrt[\square]{(-4.9)^2-4(-4.9)(8)}_{}}{2(-4.9)} \\ \\ t=\frac{4.9\pm\sqrt[\square]{180.81}_{}}{-9.8} \\ \\ t=\frac{4.9\pm13.45_{}}{-9.8} \\ \\ t1==\frac{4.9+13.45_{}}{-9.8}=-1.87\text{ -}\longrightarrow\text{ the time can not be negative} \\ t2=\frac{4.9-13.45_{}}{-9.8}=0.87\text{ sec} \end{gathered}[/tex]therefore
the answer is
0.87 seconds
Both (×-3)(×+3) and (3-×) (3+×) contain a sum and a difference with each factor only containing a 3 and an x.If each expression is rewritten in standard form, will the two expressions be the same? Explain or show your reasoning.
The given expression is called a difference of perfect squares because those factors are equivalent to such difference, let's solve them
[tex]\begin{gathered} (x-3)(x+3)=x^2-9 \\ (3-x)(3+x)=9-x^2 \end{gathered}[/tex]According to the expressions above, the expressions are not equivalent because the terms are changed.
Hence, the two expressions are not the same because they are equivalent to similar but different expressions.a.) Determine the perimeter of triangle ACE.b.) Let point G represent the intersection point of the height, CG, and the base. Determine the coordinates of point G.
The perimeter of a figure is the sum of the length of all its sides.
In the triangle ACE, we don't know the lengths of the sides, but we do know the coordinates of the points A, C, and E in the cartesian plane.
Given two points P and Q, we can find the distance between them using:
[tex]\begin{gathered} \begin{cases}P={(x_P},y_P) \\ Q={(x_Q},y_Q)\end{cases} \\ . \\ Distance_{PQ}=\sqrt{(x_P-x_q)^2+(y_P-y_Q)^2} \end{gathered}[/tex]In this case, we have the points:
A = (3, 6)
C = (7, 2)
E = (1, 0)
Then we need to find the distances: AE, EC, and AC. Using the formula for distance:
AE:
[tex]AE=\sqrt{(3-1)^2+(6-0)^2}=\sqrt{2^2+6^2}=\sqrt{4+36}=\sqrt{40}=2\sqrt{10}[/tex]EC:
[tex]EC=\sqrt{(1-7)^2+(0-2)^2}=\sqrt{(-6)^2+2^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}[/tex]AC:
[tex]AC=\sqrt{(3-7)^2+(6-2)^2}=\sqrt{(-4)^2+4^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}[/tex]Now, we know the lengths of all the sides of the triangle. To find the perimeter, we need to add them:
[tex]AE+EC+AC=2\sqrt{10}+2\sqrt{10}+4\sqrt{2}=4\sqrt{10}+4\sqrt{2}=4(\sqrt{10}+\sqrt{2})[/tex]a) The perimeter is: 4(√10 + √2) units.
Now, to find the coordinates of point G, let's make a diagram:
I rotated the triangle so it lays over the base AE, indicated in a.
We have divided the triangle into two separate right triangles. We'll use the pythagorean theorem to solve this.
For triangle AGC:
[tex]AC^2=AG^2+h^2[/tex]For triangle CGE:
[tex]EC^2=GE^2+h^2[/tex]Now, we can note:
[tex]GE=AE-AG[/tex]Using this in the equation for the triangle CGE:
[tex]EC^2=(AE-AG)^2+h^2[/tex]We have the two equations:
[tex]\begin{cases}AC^2=AG^2+h^2{} \\ EC^2=(AE-AG)^2+h^2\end{cases}[/tex][tex]\begin{cases}AC^2=AG^2+h^2{} \\ EC^2=(AE-AG)^2+h^2\end{cases}[/tex][tex]\begin{cases}AC^2=AG^2+h^2{} \\ EC^2=(AE-AG)^2+h^2\end{cases}[/tex]We know the values of AC, EC and AE. Then:
[tex]\begin{cases}(4\sqrt{2)}^2=AG^2+h^2{} \\ (2\sqrt{10})^2=(2\sqrt{10}-AG)^2+h^2\end{cases}[/tex]Now simplify both equations and solve for h².
Triangle AGC:
[tex](4\sqrt{2)}^2=AG^2+h^2{}\Rightarrow32-AG^2=h^2[/tex]Triangle CGE:
[tex]\begin{gathered} (2\sqrt{10})^2=(2\sqrt{10}-AG)^2+h^2 \\ . \\ 4\cdot10=(2\sqrt{10})^2-2\cdot2\sqrt{10}AG+AG^2+h^2 \\ . \\ 40=40-4\sqrt{10}AG+AG^2+h^2 \\ . \\ 4\sqrt{10}AG-AG^2=h^2 \end{gathered}[/tex]We have two equations:
[tex]\begin{cases}32-AG^2=h^2{} \\ 4\sqrt{10}AG-AG^2=h^2{}\end{cases}[/tex]Now, we can equate both equations:
[tex]32-AG^2=4\sqrt{10}AG-AG^2[/tex]And solve:
[tex]\begin{gathered} 32=4\sqrt{10}AG \\ . \\ AG=\frac{32}{4\sqrt{10}}=\frac{4\sqrt{10}}{5} \end{gathered}[/tex]We know the distance between point A and point G.
We can use to find the distance between point E and point G:
[tex]\begin{gathered} GE=AE-AG \\ Thus \\ GE=2\sqrt{10}-\frac{4\sqrt{10}}{5}=\frac{6\sqrt{10}}{5} \end{gathered}[/tex]Now, we can use the formula for distance between two points, to create a system of equations.
Let be G:
[tex]G=(x_G,y_G)[/tex]Distance between points A and G:
[tex]\frac{4\sqrt{10}}{5}=\sqrt{(3-x_G)^2+(6-y_G)^2}[/tex]Distance between points E and G:
[tex]\frac{6\sqrt{10}}{5}=\sqrt{(1-x_G)^2+(0-y_G)^2}[/tex]We have the system of euqations:
[tex]\begin{gathered} \frac{4\sqrt{10}}{5}=\sqrt{(3-x_G)^2+(6-y_G)^2{}} \\ . \\ \frac{6\sqrt{10}}{5}=\sqrt{(1-x_G)^2+(0-y_G)^2} \end{gathered}[/tex]Let's take the first equation and simplify it:
[tex]\begin{gathered} (\frac{4\sqrt{10}}{5})^2=(\sqrt{(3-x_G)^2+(6-y_G)^2})^2 \\ . \\ \frac{16\cdot10}{25}=(3-x_G)^2+(6-y_G)^2 \\ . \\ \frac{32}{5}=(3-x_G)^2+(6-y_G)^2 \end{gathered}[/tex]Now the second equation:
[tex]\begin{gathered} (\frac{6\sqrt{10}}{5})^2=(\sqrt{(1-x_G)^2+(-y_G)^2})^2 \\ . \\ \frac{36\cdot10}{25}=(1-x_G)^2+y_G^2^ \\ . \\ y_G^2=\frac{72}{5}-(1-x_G)^2 \\ . \\ y_G^2=\frac{67}{5}-2x_G+x_G^2 \end{gathered}[/tex]Taco Palace wanted to determine what proportion of its customers prefer soft-shell tacos. Out of 350 customers, 133 of them (that is, 38% of them) chose soft-shell tacos. What is the 95% confidence interval of the true proportion of customers who prefer soft-shell tacos? Round your answer to the nearest hundredth (or nearest whole percent).
We have to find the 95% confidence interval for the proportion.
We first need the sample proportion, the standard error and the critical value of z.
The sample proportion is p=0.38.
[tex]p=X/n=133/350=0.38[/tex]The standard error of the proportion is:
[tex]\begin{gathered} \sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.38*0.62}{350}} \\ \sigma_p=\sqrt{0.000673}=0.0259 \end{gathered}[/tex]The critical z-value for a 95% confidence interval is z=1.96.
Now we can can calculate the margin of error (MOE) as:
[tex]MOE=z\cdot\sigma_p=1.96\cdot0.0259=0.0509[/tex]Then, the lower and upper bounds of the confidence interval are:
[tex]\begin{gathered} LL=p-z\cdot\sigma_p=0.38-0.0509=0.3291 \\ UL=p+z\cdot\sigma_p=0.38+0.0509=0.4309 \end{gathered}[/tex]The 95% confidence interval for the population proportion is (0.3291, 0.4309).
When expressed as percentage, we can conclude that the 95% confidence interval is (33%, 43%).
Answer: the 95% confidence interval is (33%, 43%).
A tram moved downward 45 meters in 5 seconds at a constant rate what is the change in the trams elevation each second
Explanation
to find the rate of change per second we can use:
[tex]\text{rate}=\frac{change\text{ in distance}}{\text{change in time}}[/tex]then, it moved 45 meters in 5 seconds, replace
[tex]undefined[/tex]If I get paid $19.90 per hour and I get a 3% raise, how much am I now making per hour?
Answer:
[tex]\text{ New hourly pay = \$20.50}[/tex]Explanation:
Here, we want to know the new hourly pay
Mathematically, we have it that:
[tex]\text{ New hourly pay = old hourly pay + (percentage increase in hourly pay}\times old\text{ hourly pay)}[/tex]We can write that mathematically as:
[tex]\begin{gathered} \text{New hourly Pay = 19.90 + (}\frac{3}{100}\times19.90) \\ \\ \text{New hourly pay = \$20.497 }\approx\text{ \$20.50} \end{gathered}[/tex]A hotel manager recorded the percentage of rooms that were occupied each day over a period of 25 days. The data she collected is shown on the graph.The manager recognizes that the data approximates a transformation of the parent sine function, Y = sin(e).Which value is closest to the amplitude of the transformed function?
Amplitude = (maximum - minimum)/2
From the graph,
maximum = 70
minimum = 40
Amplitude = (70 - 40)/2 = 30/2
Amplitude = 15
A regular octagon (8-sided polygon) has an area of 690 sq. ft. and a side length of 20feet. Find the length of the apothem to the nearest foot.
Given:
[tex]\begin{gathered} s=\text{side length} \\ s=20\text{ fe}et \\ a=\text{ apothem length} \\ n=\text{ number of sides of a polygon} \\ n=8 \end{gathered}[/tex]Apothem length.
[tex]a=\frac{s}{2\tan (\frac{180}{n})}[/tex][tex]\begin{gathered} a=\frac{20}{2\tan (\frac{180}{8})} \\ a=\frac{20}{2\tan (22.5}) \\ a=\frac{20}{2\times0.4142} \\ a=\frac{20}{0.828} \\ a=24.142 \end{gathered}[/tex]Length of the apothem is 24.142
math help!
Average rate of change of q from 4 to 5= ?
The estimate of the function g average rate of change is 6
How to calculate the average rate of change of g?From the question, the interval is given as
Rate of q from 4 to 5
This can be rewritten as
a = 4 to b = 5
This can also be represented as
(a, b) = (4, 5)
From the attached table of values, we have
g(4) = 2
g(5) = 8
The estimate of the average rate of change g is calculated using the following rate formula
Rate = [g(b) - g(a)]/[b - a]
This gives
Rate = [g(5) - g(4)]/[5 - 4]
So, we have
Rate = [8 - 2]/[5 - 4]
Evaluate
Rate = 6
Hence, the average rate of change of g is 6
Read more about average rate of change at
brainly.com/question/8728504
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If the point (-1, 3) is reflected across the line x = 4,what will be the coordinates of the image?
To answer this question we have to draw a graph:
the initial point is (-1;3), the reflection is on x =4.
Since it is reflected across an x line, the y component stays the same.
The distance from the line of reflection (x=4) to the image it's equal to the distance from the line of reflection to the initial point.
In this case, that distance is 5 units on the x-axis.
So, the new coordinates are:
(-1+5+5,3)= (9,3)
What is the sequence and the common difference/ ratio for number 12
The given sequence is 600, -300, 100, -25.
It is a recurrence sequence and is given by
[tex]undefined[/tex]The function F(x)=6/x is one to oneFind parts A and B
Answer
(a)
[tex]f^{-1}(x)=\frac{6}{x}[/tex]Explanation
Given function:
[tex]f(x)=\frac{6}{x}[/tex](a) To find f⁻¹(x)
[tex]\begin{gathered} \text{Let y }=f(x) \\ \text{This implies, y }=\frac{6}{x} \\ x=\frac{6}{y} \\ \text{Note that x }=f^{-1}(y) \\ f^{-1}(y)=\frac{6}{y} \\ \therefore f^{-1}(x)=\frac{6}{x} \end{gathered}[/tex](b) To show that
[tex]f(f^{-1}(x))=x\text{ and }f^{-1}(f(x))=x[/tex][tex]\begin{gathered} To\text{ find }f(f^{-1}(x)) \\ \text{Subtitute }x=f^{-1}(x)\text{ into }f(f^{-1}(x)) \\ \text{Since }f(x)=\frac{6}{x},\text{ it follows that} \\ f(f^{-1}(x))=\frac{6}{f^{-1}(x)}=6\div\frac{6}{x}=6\times\frac{x}{6}=x \end{gathered}[/tex]Also,
[tex]\begin{gathered} to\text{ find }f^{-1}(f(x)) \\ \text{Substitute }x=^{}f(x)\text{ into }f^{-1}(f(x)) \\ \text{Since }f^{-1}(x)=\frac{6}{x},\text{ it follows that} \\ f^{-1}(f(x))=\frac{6}{f(x)}=6\div\frac{6}{x}=6\times\frac{x}{6}=x \end{gathered}[/tex]Therefore, it is verified that
[tex]f(f^{-1}(x))=f^{-1}(f(x))=x[/tex]Gordon had $200,000 in a CD at Lotsa Loot Bank, which just failed. If theFDIC insurance limit per depositor,per bank, is $250,000, how muchwill Gordon get back?
SOLUTION:
The amount Gordon will get back is $200,000
The amount Gage had in the Credit Deposit CD at the Lots - a--Loot Bank that just failed =$200,000
The Federal Deposit Insurance Corporation (FDIC) insurance limit per depositor per bank = $250,000
Therefore, the maximum amount that can be refunded by FDIC per depositor per bank in the event of the bank failure = $250,000
Therefore, the amount Gage get can get back is ≤ $250,000 whereby Gordon's deposit is $200,000, the amount Gage will get back= $200,000.
Final answer:
$200000
pls just put the answers.
Answer:
Step-by-step explanation:
y intercept -1
slope is -x
equation is y=-x-1
negative slope
