All categories

2 years ago

Which graph best shows y=1/4x-2 y=7/4x+4 what graph is correct

Mathematics

1 answer:

lesya692 [45]2 years ago

4 0

Solution:

Given:

$\begin{gathered} y=\frac{1}{4}x-2 \\ y=\frac{7}{4}x+4 \end{gathered}$

The graph of the system of equations is shown below;

The solution exists at (-4,-3).

From the options, the graph with a similar solution set is;

Therefore, OPTION D is correct.

You might be interested in

What is the value of x A. 20 B. 34 C. 60 D. 69

Grace [21]

Answer:

Step-by-step explanation:

The given angles are vertical and thus congruent, so

3x = x + 40 ( subtract x from both sides )

2x = 40 ( divide both sides by 2 )

x = 20 → A

4 0

4 years ago

Rearrange the formula to make x the subject: <img src="https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7Bx%20%2B%202%7D%7Bx%20-

Lana71 [14]

Answer:

Rearranging Formula. This unit covers rearranging formulae. The manipulation of algebraic expressions is an important ... Changing the subject of a formula .

Step-by-step explanation:

8 0

3 years ago

Paul lands a home mortgage of $58,00 at a 7.40% interest rate. He plans to pay this after 3 years. What will the total principle

Elena-2011 [213]

Answer:

The answer to the question is

The total principles + interest payment is $64,854.00

Step-by-step explanation:

To solve the question, we not that

Home mortgage value = $58000

Interest rate, r = 7.4 % annually = 7.4/12 % monthly or 0.62 % monthly

Duration of payment = 3 years = 3 × 12 =36 months

Monthly payment is given by

$M = P\frac{r(1+r)^n}{(1+r)^n-1}$ = $58000\frac{0.00062(1.00062)^{36}}{(1.00062)^{36}-1}$ = $1,801.5

Cumulative interest payed is given by

M×n - P = 1801.5 × 36 - P =$6, 854

Total principles + interest payment = M × n = $64,854.00

6 0

3 years ago

Find the function y1 of t which is the solution of 16y′′−81y=0 with initial conditions y1(0)=1,y′1(0)=0.

TEA [102]

Consider this option.
Make a design according the local requirements.

7 0

3 years ago

1. You are saving to buy a new house in 7 years. If you invest $4,500 now at 5.5% interest compounded

motikmotik

Answer:

Part 1) $\$6,595.94$

Part 2) $\$3,449.23$

Part 3) $\$17,040.06$

Part 4) $\$20,773.90$

Part 5) The Option A is the best way to invest the money by $4,223.94 than Option B

Step-by-step explanation:

Part 1)

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=7\ years\\ P=\$4,500\\ r=5.5\%=5.5/100=0.055\\n=4$

substitute in the formula above

$A=4,500(1+\frac{0.055}{4})^{4*7}$

$A=4,500(1.01375)^{28}$

$A=\$6,595.94$

Part 2)

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=2\ years\\ P=\$3,200\\ r=3.75\%=3.75/100=0.0375$

substitute in the formula above

$A=3,200(e)^{0.0375*2}$

$A=\$3,449.23$

Part 3)

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=18\ years\\ A=\$40,000\\ r=4.75\%=4.75/100=0.0475\\n=12$

substitute in the formula above

$40,000=P(1+\frac{0.0475}{12})^{12*18}$

$40,000=P(\frac{12.0475}{12})^{216}$

$P=40,000/[(\frac{12.0475}{12})^{216}]$

$P=\$17,040.06$

Part 4)

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=7\ years\\ A=\$30,000\\ r=5.25\%=5.25/100=0.0525$

substitute in the formula above

$30,000=P(e)^{0.0525*7}$

$30,000=P(e)^{0.3675}$

$P=30,000/(e)^{0.3675}$

$P=\$20,773.90$

Part 5)

Option A

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=8\ years\\ P=\$11,500\\ r=5.6\%=5.6/100=0.056\\n=2$

substitute in the formula above

$A=11,500(1+\frac{0.056}{2})^{2*8}$

$A=11,500(1.028)^{16}$

$A=\$17,889.07$

Option B

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=5\ years\\ P=\$11,500\\ r=3.45\%=3.45/100=0.0345$

substitute in the formula above

$A=11,500(e)^{0.0345*5}$

$A=11,500(e)^{0.1725}$

$A=\$13,665.13$

Compare the options

Option A ------> $\$17,889.07$

Option B -----> $\$13,665.13$

Option A > Option B

Find out the difference

$\$17,889.07-$13,665.13=$4,223.94$

therefore

The Option A is the best way to invest the money by $4,223.94 than Option B

3 0

3 years ago