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Sauron [17]
3 years ago
15

10. State the independent variable and the dependent variable in each situation. Then find the rate of change for each situation

. a. Shelly delivered 12 newspapers after 20 minutes and 36 papers after 60 minutes. b. Two pounds of apples cost $3.98. Six pounds cost $11.94.
Mathematics
1 answer:
lina2011 [118]3 years ago
8 0

Answer:

<h2>i solve mo para ka namang walang utak</h2>

Step-by-step explanation:

<h2>shhhhhhhhh</h2>
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What are the solutions of the quadratic equation? 4x^2+34x+60=0
insens350 [35]

Answer:

Option 3) -6,-5/2

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

4x^{2} +34x+60=0  

so

a=4\\b=34\\c=60

substitute in the formula

x=\frac{-34(+/-)\sqrt{34^{2}-4(4)(60)}} {2(4)}

x=\frac{-34(+/-)\sqrt{196}} {8}

x=\frac{-34(+/-)14} {8}

x_1=\frac{-34(+)14} {8}=-\frac{20}{8}=-\frac{5}{2}

x_2=\frac{-34(-)14} {8}=-6

6 0
3 years ago
Can someone please answer. There is one problem. There's a picture. Thank you!
allochka39001 [22]
\dfrac{60^o}{360^o}=\dfrac{1}{6}\\\\A_O=\pi\cdot4^2=16\pi\\\\Area\ of\ the\ shaded\ sector:\\\\A=\dfrac{1}{6}\cdot16\pi=\dfrac{8}{3}\pi\approx\dfrac{8}{3}\cdot3.14\approx8.4\\\\Answer:\boxed{8.4}
8 0
3 years ago
Write an equation of a line that passes through the point (8,2) that is parallel and (b) perpendicular to the graph of the equat
oksian1 [2.3K]
For a parallel line the slope of the lines/equations will be the same but for perpendicular line the slope will be the negative reciprocal
-(1/4)*x+b
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4 0
3 years ago
Use the following matrices, A, B, C and D to perform each operation.
Vinvika [58]

Step-by-step explanation:

A=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]

B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]

C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]

D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]

1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]

2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]

3.\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]

4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]

5.\\2D+3C\\\text{This operation can't be performed because the matrices}\\\text{ are of different dimensions.}

6 0
3 years ago
You are examining your choices of banks to open a new checking account. Potential costs include monthly maintenance fees, statem
Nikitich [7]
Overdraft fees is your answer
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