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2 years ago

Given the equation y = x – 2, you substitute 5 for x. Check all true statements below.

Mathematics

1 answer:

Annette [7]2 years ago

3 0

The correct value of the given function at x = 5 is y =3.

<h3>What is Equation?</h3>

An equation is a mathematical statement with an 'equal to =' symbol between two expressions that have equal values.

Here, the given equation :

y = x - 2

put x = 5 in the given equation, we get

y = 5 - 2

y = 3

Thus, the correct value of the given function at x = 5 is y =3.

Learn more about Equations from:

brainly.com/question/10413253

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An amortized loan of RM60,000 has annual payments for fifteen years, the first occurring exactly one year after the loan is made

Nimfa-mama [501]

9514 1404 393

Answer:

a) RM2256.09 . . . principal paid by 8th payment

b) RM1791.10 . . . . interest paid by 12th payment

Step-by-step explanation:

First of all, we need to find the payments.

The payment amount is the amount that makes the future value of the series of payments equal to the future value of the loan at the given interest rate.

The future value of a single amount is ...

FV = P(1 +r)^n . . . . . where r is the annual rate, and n is the number of years in the future

The future value of a series of payments is ...

FV = P((1 +r)^n -1)/r . . . . . where n is the number of payments of P earning annual rate r

For payments in a series that does not end at the end of the loan, the future value is the product of that of the series and the effect of the accumulation of interest for the remaining time.

The first 4 payments will have a future value at the end of the loan period of ...

s1 = X((1 +0.05)^4 -1)/0.05×(1 +0.05)^11 = X(1.05^15 -1.05^11)/0.05

s1 = 7.3717764259X

The next 5 payments will have a future value at the end of the loan period of ...

s2 = 2X((1 +0.05)^5 -1)/0.05×(1 +0.05)^6 = 2X(1.05^11 -1.05^6)/0.05

s2 = 14.8097486997X

The last 6 payments will have a future value at the end of the loan period of ...

s3 = 4X((1 +0.05)^6 -1)/0.05 = 27.20765125X

So, the total future value of the series of payments is ...

payment value = 7.3717764259X +14.8097486997X +27.20765125X

= 49.3891763756X

The future value of the loan amount after 15 years is ...

loan value = 60,000(1 +0.05)^15 = 124,735.69

In order for these amounts to be the same, we must have ...

49.3891763756X = 124,735.69

X = 124,735.69/49.3891763756 = 2,525.57

At this point, it is convenient to use a spreadsheet to find the interest and principal portions of each of the loan payments. (We find the interest charge to be greater than the payment amount for the first 4 payments. So, the loan balance is increasing during those years.)

In the attached, we have shown the interest on the beginning balance, and the principal that changes the beginning balance to the ending balance after each payment. (That is, the interest portion of the payment is on the row above the payment number.)

The spreadsheet tells us ...

A) the principal repaid in the 8th payment is RM2,256.09

B) the interest paid in the 12th payment is RM1,791.10

_____

Additional comment

The spreadsheet "goal seek" function could be used to find the payment amount that makes the loan balance zero at the end of the term.

We have used rounding to sen (RM0.01) in the calculation of interest payments. The effect of that is that the "goal seek" solution is a payment value of 2525.56707 instead of the 2525.56734 that we calculated above. The value rounded to RM0.01 is the same in each case: 2525.57.

3 0

2 years ago

The ratio of 6th grade students to 7th grade students in a soccer league is 15:10. If there are 100 students in all, how many 7t

UNO [17]

Answer:

40 7th Graders

Step-by-step explanation:

First you add all of the students to get 25. Then you divide 100 by 25 to get 4 (this is gonna be your multiplier) now multiply each part of the ration by 4, and you get 60 6th graders and 40 7th graders :D.

5 0

2 years ago

Find dx/dt when y=2 and dy/dt=1, given that x^4=8y^5-240 dx/dt=

katrin2010 [14]

Answer:

The value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

Step-by-step explanation:

The given equation is

$x^4=8y^5-240$

We need to find the value of $\frac{dx}{dt}$ .

Differentiate with respect to t.

$4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0$ $[\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0]$

$4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}$

It is given that y=2 and dy/dt=1, substitute these values in the above equation.

$4x^3\frac{dx}{dt}=40(2)^4(1)$

$4x^3\frac{dx}{dt}=40(16)(1)$

$4x^3\frac{dx}{dt}=640$

Divide both sides by 4x³.

$\frac{dx}{dt}=\frac{640}{4x^3}$

$\frac{dx}{dt}=\frac{160}{x^3}$

Therefore the value of $\frac{dx}{dt}$ is $\frac{160}{x^3}$ .

4 0

3 years ago

Simplify (6x^-2)^2(0.5x)^4... plz show work!

kakasveta [241]

Simplify (6x^-2)^2(0.5x)^4... plz show work!

 $(6x^{-2})^2(0.5x)^4= \\ \\ (6^2x^{-2*2})(0.5^4x^4)= \\ \\ (36x^{-4})(0.0625x^4)= \\ \\ (36*0.0625)(x^{-4}x^4)= \\ \\ (36*0.0625)(x^{-4+4})= \\ \\ (36*0.0625)(x^{0})= \\ \\ (36*0.0625)= \boxed{2.25} \\ \\$

8 0

3 years ago

A pond had an initial population of 120 fish. The number of fish is exponentially decreasing by one-fourth each year. If y repre

Kruka [31]

1) Number of fish in the pond: y
Number of years: x

A pond had an initial population of 120 fish. Then when x=0, y=120. The graph must begin at point (x,y)=(0,120). Only the graph above at right and below at left begin in this point (0,120).

The number of fish is exponentially decreasing by one-fourth each year. Then the first year the number of fish must decrease 120*(1/4)=120/4=30, and the number of fish after the first year must be:
120-30=90=120(1-1/4)=120(4-1)/4=120(3/4). Then when x=1, y=90. The point is (x,y)=(1,90)
In the graph above at right when x=1, y is between 24 and 36. y=90 is not in this interval. then this graph is not the correct.
In the graph below at left when x=1, y is between 84 and 96. y=90 is in this inverval. Then this is the correct graph.

Answer: The graph of the solution set for this situation is the graph below at left.

2) The equation has the form:
y=y0(r)^x
Where y0 is the initial population and r is the rate of reduction. In this case:
y0=120 and
r=1-1/4=(4-1)/4→r=3/4
Then the equation modeled by the graph is:
y=120(3/4)^x

Answer: The equation modeled by the graph is that above at right:
y=120(3/4)^x

7 0

3 years ago