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photoshop1234 [79]

2 years ago

8

You deposit money in an account that pays 5% interest compounded yearly. Find the balance after 5 years for the given initial am

ount (This is not multiple choice, you need to do a, b, and c)
a. $250
b. $300
c. $350
** Round to the NEAREST whole number**

2 answers:

mezya [45]2 years ago

5 0

B I really believe you just got to read it carefully

Paraphin [41]2 years ago

5 0

B your answer would be that you have 300 left

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Hey yall so im taking a test and imm very confused on hwo to find the letter n for the equation

kotegsom [21]

Answer:

n = -7,6

Step-by-step explanation:

n(n+1)+3=45

distributive property: a(b+c) = ab+ac

n²+n+3 = 45

n²+n -42 = 0

factor now

(n+7)(n-6) = 0

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3 years ago

Help guys pls 22x²+13x+1

myrzilka [38]

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2 years ago

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Adrian made 3 carrot cakes .He cut each cake into fourths.How many 1/4size cake pieces does he have?

Igoryamba

3 divided by 1/4 =
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2 years ago

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The phone company NextFell has a monthly cellular plan where a customer pays a flat monthly fee and then a certain amount of mon

zaharov [31]

The equation in the form y = mx + b is y = 0.35x + 36.

<h3>How to represent a linear equation?</h3>

The equation formed from the expression is a linear equation and should be expressed in the slope intercept form.

y = mx + b

where

m = slope
b = y-intercept

Therefore,

y = mx + b

where

x = the number of monthly minutes used
y = the total monthly of the Next fell plan

using the coordinates, let's find m

(470, 200.5)(590, 242.5)

Hence,

m = 242.5 - 200.5 / 590 - 470

m = 42 / 120

m = 0.35

Therefore,

y = 0.35x + b

200.5 = 0.35(470) + b

200.5 - 164.5 = b

b = 36

Finally, the equation in the form y = mx + b is y = 0.35x + 36.

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7 0

1 year ago

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

<u>F</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>2</u>

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u> $\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$ </u>

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

<u>S</u><u>e</u><u>c</u><u>o</u><u>n</u><u>d</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u> $\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$ </u>

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>4</u>

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>3</u>

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>4</u>

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

3 0

2 years ago