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trapecia [35]
3 years ago
12

Pamela deposited $30 in a savings account earning 5% interest, compounded annually.

Mathematics
1 answer:
makvit [3.9K]3 years ago
4 0

\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$30\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ t=years\dotfill &3 \end{cases} \\\\\\ I=(30)(0.05)(3)\implies I=4.5

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Write the equation of the line perpendicular to 2x - 6y = 12 that passes through the point (-3,0).
Trava [24]

Answer:

y = -3x -9

Step-by-step explanation:

slope = 1/3

perpendicular slope = -3

y = mx + b

0 = -3(-3) + b

-9 = b

y = -3x -9

7 0
3 years ago
Read 2 more answers
PLEASE HELP THIS QUESTION IS SO ANNOYING!!
nikklg [1K]
ANSWER- 23 , 76


Explanation- just did it :)
6 0
3 years ago
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Can someone help me with my algebra?
SIZIF [17.4K]

Answer:

The top graph

Solutions:

-2

0

Step-by-step explanation:

The given quadratic function in factored form is

f(x) =  - (x + 2)(x - 2)

This is a parabola that has x-intercepts at (-2,0) and (2,0)

This parabola opens downward because the leading coefficient is less than 1.

The second function is

g(x) = 2 |x  + 2|

This is an absolute value function with vertex at (-2,0).

Therefore the graph that shows the solution to f(x)=g(x) is the top graph.

Hence the solution is x=-2,x=0

4 0
3 years ago
Determine as a linear relation in x, y, z the plane given by the vector function F(u, v) = a + u b + v c when a = 2 i − 2 j + k,
Ostrovityanka [42]

Answer:

2x - y - 3z = 0

Step-by-step explanation:

Since the set

{i, j}  = {(1,0), (0,1)}

is a base in \mathbb{R}^2

and F is linear, then

<em>{F(1,0), F(0,1)}  </em>

would be a base of the plane generated by F.

F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k

F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k

Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k

We need a normal vector which is the cross product of 3i+2k and 4i-j+3k

(3i+2k)X(4i-j+3k) = 2i-j-3k

The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by

2(x-3) -1(y-0) -3(z-2) = 0

or what is the same

2x - y - 3z = 0

3 0
3 years ago
If 1/2x+1/3=4 What is the value of 3x+2y
ivolga24 [154]

Answer:

x=3/22 ; y=-9/44

Step-by-step explanation:

1/2x+1/3=4

=>1/2x=4-1/3

=>1/2x=(12-1)/3

=>1/2x=11/3

=>22x=3

=>x=3/22....(1)

then,

3x+2y

2y=-3 . (3/22)

y=-9/44

6 0
3 years ago
Read 2 more answers
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