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ivann1987 [24]
3 years ago
6

Help please...........................plz

Mathematics
1 answer:
katen-ka-za [31]3 years ago
7 0
To find the answers, input the values in the x row in the equation to get your answers. Here are your answers in order from -3 to 3.

-8
-6
-4
-2
0
2
4

Hope this helps!
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Hep meh plz!!!!!!!!!!!!! 25 points
Makovka662 [10]
D. X=10 all you have to do is pick an answer and plug it in
8 0
3 years ago
Izzy gets a loan of $2,800 with an APR of 3.5% . She will repay the loan in monthly payments for 9 months. To find the total amo
djyliett [7]
First, change percent into decimal value
Percent means per hundred
APR = 3.5%
APR = 3.5/100
APR = 0.035
The APR is 0.035 of the total loan

Second, we need to find interest rate for 9 months
interest rate = 9/12 × 0.035
interest rate = 3/4 × 0.035
interest rate = 0.75 × 0.035

Third, determine what to put into calculator
She can put it like this
interest = interest rate × total loan
interest = 0.75 × 0.035 × 2,800

Or she can put it like this
interest = 0.02625 × 2,800
4 0
4 years ago
Complete the table and determine whether the data is proportional and explain why
Valentin [98]

Answer:

in distance it goes 3,6,9,12,15

Step-by-step explanation:

thats all i know i hope my answer anwsers nyour questionand have a great time and get a lot of A+ ✍✍

7 0
4 years ago
Find the range of the function:<br><br> f(x) = x + 3, for x ≠ 1
Tju [1.3M]

Answer:

All Real Numbers except for 4

Step-by-step explanation:

Objective: Specify domain and range of a function.

The equation is a linear equation. A linear equation range is all real numbers. However x cannot equal 1 so let plug in x to find where the y value CANT be.

f(1) = 1 + 3 = 4

So the range is

All Real Numbers except for 4.

3 0
3 years ago
A=(2 3)<br> (-1 4)<br><br> Calculate A^2-6A+11I
disa [49]

Answer:

A² - 6A + 11 I = \left[\begin{array}{ccc}0&0\\0&0\end{array}\right]

Step-by-step explanation:

Given the matrix

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

Calculate A² - 6A + 11 I

A^2 = A*A= \left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] *\left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] = \left[\begin{array}{ccc}2*2-3*1&2*3+3*4\\-1*2-4*1&-1*3+4*4\end{array}\right] =\left[\begin{array}{ccc}1&18\\-6&13\end{array}\right]

6A=6*\left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] =\left[\begin{array}{ccc}12&18\\-6&24\end{array}\right]

11 I = 11 * \left[\begin{array}{ccc}1&0\\0&1\end{array}\right] =\left[\begin{array}{ccc}11&0\\0&11\end{array}\right]

∴ A² - 6A + 11 I = \left[\begin{array}{ccc}1&18\\-6&13\end{array}\right] -\left[\begin{array}{ccc}12&18\\-6&24\end{array}\right] +\left[\begin{array}{ccc}11&0\\0&11\end{array}\right] =\left[\begin{array}{ccc}0&0\\0&0\end{array}\right]

8 0
3 years ago
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