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Mekhanik [1.2K]

2 years ago

6

Write the equation of a line with a slope of 5 and a y-intercept of (0, 3). Write the equation in slope-intercept form.

1 answer:

sladkih [1.3K]2 years ago

7 0

The answer is y=-5x+3

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Solve equation by completing the square 5n^2-3n-10=15

VARVARA [1.3K]

These are the two answers you could get

5 0

2 years ago

What is the written form of the number 62?

mr Goodwill [35]

A. Sixty-two is the written form of the number 62. It is hyphenated.

5 0

2 years ago

if content settles for 1 1/2 hours before the liquid is remove .You fill the tank at 1:45 a.m when do you remove the liquid

artcher [175]

U fill the tank at 1:45 a.m......then u wait 1 1/2 hrs....

1:45 + 1 hr = 2:45.....2:45 + 30 minutes (because 1/2 hr = 30 minutes) =
3:15.....so u remove the liquid at 3:15 a.m.

3 0

2 years ago

VWX and NOP are similar. If mV=44 and mP=66, what is mW

Elena L [17]

It is given that the two figures are similar. This means that mV is equal to mN, mW is equal to mO, and mX is equal to mP. The sum of the angles of a triangle is equal to 180 degrees.

<span> mV + mW + mX = 180</span>

Substituting the known variables and the conditions given in the description of the similar triangles,

<span> 44 + mW + 66 = 180</span>

<span>The value of mW in the equation is 70 degrees. </span>

3 0

3 years ago

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years f

Anni [7]

Answer:

Part A) Annual $\$66,480.95$

Part B) Semiannual $\$66,862.38$

Part C) Monthly $\$67,195.44$

Part D) Daily $\$67,261.54$

Step-by-step explanation:

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

so

Part A) Annual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=1$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{1})^{1*5}$

$A=\$47,400(1.07)^{5}$

$A=\$66,480.95$

Part B) Semiannual

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=2$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{2})^{2*5}$

$A=\$47,400(1.035)^{10}$

$A=\$66,862.38$

Part C) Monthly

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=12$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{12})^{12*5}$

$A=\$47,400(1.0058)^{60}$

$A=\$67,195.44$

Part D) Daily

in this problem we have

$t=5\ years\\ P=\$47,400\\ r=0.07\\n=365$

substitute in the formula above

$A=\$47,400(1+\frac{0.07}{365})^{365*5}$

$A=\$47,400(1.0002)^{1,825}$

$A=\$67,261.54$

8 0

3 years ago