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3 years ago

11

Ana deposits $1,500 in a savings account that earns 4% interest annually. Which of the following equations models the amount of

money,A, she’ll have in 5 years?
A. A=1,500(0.96)^5

B. A=1,500(1.04)^5

C. A=1,500(1.4)^5

D. A=1,500(0.04)^5

2 answers:

Naddika [18.5K]3 years ago

5 0

Answer:

B is the correct answer A=1500(1.04)^5

Oduvanchick [21]3 years ago

3 0

The answer is going to be B: A=1,500(1.04)^5

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3x-1/2y +2 2/3y -5/6x

telo118 [61]

$Solution, 3x-\frac{1}{2}y+\frac{22}{3}y-\frac{5}{6}x=\frac{13x+41y}{6}$

$Steps:$

$\mathrm{Group\:like\:terms}, =3x-\frac{5}{6}x-\frac{1}{2}y+\frac{22}{3}y$

$\mathrm{Add\:similar\:elements:}\:3x-\frac{5}{6}x=\frac{13}{6}x, =\frac{13}{6}x-\frac{1}{2}y+\frac{22}{3}y$

$\mathrm{Add\:similar\:elements:}\:-\frac{1}{2}y+\frac{22}{3}y=\frac{41}{6}y, =\frac{13}{6}x+\frac{41}{6}y$

$\mathrm{Multiply\:}\frac{13}{6}x\::\quad \frac{13x}{6}, =\frac{13x}{6}+\frac{41}{6}y$

$\mathrm{Multiply\:}\frac{41}{6}y\::\quad \frac{41y}{6}, =\frac{13x}{6}+\frac{41y}{6}$

$\mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}, =\frac{13x+41y}{6}$

The correct answer is $=\frac{13x+41y}{6}$

Hope this helps!!!

4 0

3 years ago

Jan mixed 2 pints of blue paint with 5 pints of red paint to make purple paint. He wrote the equation b= ( )r to represent the p

max2010maxim [7]

so here

ratio of mixture is actually ratio of number of pints

so

blue pints/ red pints = b/r =2/5

so

b = (2/5) r

b = (0.4) r

8 0

3 years ago

Solve (u+6)^2+4=52 , where u is a real number.

Ugo [173]

Answer:

The value of u are,

0.93, -12.93

Step-by-step explanation:

It is given an expression in variable u

(u + 6² + 4 = 52

<u>To find the solution of given expression</u>

(u + 6)·² + 4 = 52

(u + 6)² = 52 - 4

(u + 6)² = 48

u + 6 = √48

u + 6 = ±6.93

u = ±6.93 - 6

u = 6.93 - 6 or u = -6.93 - 6

u = 0.93 or u = -12.93

3 0

3 years ago

I’m really confused on how to do this

Julli [10]

Answer:

Substitute x with six in the equation and solve as normal.

Step-by-step explanation:

The format here is a bit weird, but it basically means x is six.

7 0

3 years ago

How many real roots and how many complex roots exist for the polynomial

Volgvan

Answer:

D. 4 real roots and 0 complex roots

Step-by-step explanation:

If I assume that the function you are saying is

$F(x)=x^4+x^3-5x^2+x-6$

There should be up to "4 roots," there can't be more or less than 4 total solutions. First, we need to check how many sign changes are there in this function. There are 3 positive real roots. Now lets check for negative roots.

$F(-x)=x^4-x^3-5x^2-x-6$

There are is only 1 negative real root. Since we basically have 4 real roots, and the max is 4. There should be 4 real roots and 0 complex roots.

8 0

3 years ago