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motikmotik
3 years ago
13

On April 1, 1986, Casey deposited $1150 into a savings account paying 9.6% interest, compounded quarterly. If he hasn't made any

additional deposits or withdrawals since then, and if the interest rate has stayed the same, in what year did his balance hit $2300, according to the rule of 72? A. 1993 B. 1992 C. 1994 D. 1991​
Mathematics
1 answer:
jeka57 [31]3 years ago
8 0

Answer: A. 1993

Step-by-step explanation: A P E X

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In the graph 5x2 + 3, the y-intercept is....
ira [324]

Answer:

0 zero

Step-by-step explanation:

The slope-intercept form is y=mx+b y = m x + b , where m m is the slope and b b is the y-intercept. Combine x x and 13 1 3 . Rewrite in slope-intercept form. Using the slope-intercept form, the y-intercept is 0 .

5 0
3 years ago
Factor the expression below<br>-r -8<br>​
miv72 [106K]

Answer:

-(r+8)

Step-by-step explanation:

4 0
3 years ago
What’s the x-intercept of the graph of <br> 5x-4y= 20
viva [34]

The x-intercept of the graph is (4 , 0)

Step-by-step explanation:

Let us revise the meaning of x-intercept and y-intercept

  • x-intercept means value x when y = 0
  • That means the graph intersects x-axis at point (a , 0), where a the value of x when y = 0
  • y-intercept means value of y when x = 0
  • That means the graph intersects y-axis at point (0 , b), where b is the value of y when x = 0

∵ 5x - 4y = 20 is the equation of the graph

- To find x-intercept substitute y by 0

∵ y = 0

∴ 5x - 4(0) = 20

∴ 5x - 0 = 20

∴ 5x = 20

- Divide both sides by 5

∴ x = 4

The x-intercept of the graph is (4 , 0)

Learn more:

You can learn more about x and y intercepts in brainly.com/question/4932386

#LearnwithBrainly

7 0
3 years ago
4 times the difference of 4 and a number is 24
Len [333]

Answer:

10

Step-by-step explanation:

Let "a number" = x

4(x - 4) = 24

Isolate the x. First, divide 4 from both sides

(4(x -4))/4 = (24)/4

x - 4 = 6

Add 4 to both sides

x - 4 (+4) = 6 (+4)

x = 6 + 4

x = 10

10 is your answer for "a number"

6 0
4 years 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
3 years ago
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