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Aleonysh [2.5K]
2 years ago
15

Please answer correctly

Mathematics
1 answer:
katrin [286]2 years ago
5 0

Answer:

-5/2

Step-by-step explanation:

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If two zeroes of the polynomial x4-6x3-26x2+138x-35 are x+3 and x-3. Find the other zeroes
ASHA 777 [7]

Answ

Step-by-step explanation:

5 0
2 years ago
What is 2/9 divide 9/10=
Ipatiy [6.2K]

To divide fractions:

Flip the second fraction

Multiply the numerators

Multiply the denominators

Simplify


2/9*10/9

2*10=20

9*9=81

20/81


Hope this helps!!

6 0
2 years ago
Read 2 more answers
Write the equation for the following relation. R = {(x, y): (4, 5), (8, 7), (12, 9), (16, 11), . . .} and show work
timama [110]

Let

A(4,5) B(8,7) C(12,9) D(16,11)

1) Find the slope AB

m=(y2-y1)/(x2-x1)

m=(7-5)/(8-4)=0.5

2) Find the slope BC

m=(9-7)/(12-8)=0.5

3) Find the slope CD

m=(11-9)/(16-12)=0.5

The points represent a linear function

so

<u>Find the equation of the line with m=0.5 and the point A(4,5)</u>

we know that

y-y1=m*(x-x1)

y-5=0.5*(x-4)

y=0.5*x-2+5

y=0.5*x+3

therefore

<u>the answer is</u>

The equation is equal to y=0.5*x+3

8 0
2 years ago
Find the value of x<br>9.10.12,x, 20.25 ; The median is 14.​
Elodia [21]

Answer: 16  

Step-by-step explanation:

9,10,12,x ,20, 25  so we know the median is 14 which is the middle number.

Since the numbers are 6 it means there will be 2 middle numbers and if you divide them it will be 14.

we know one of the middle numbers is 12 and we trying to find x .

So we could use the equation x  + 12 = 2(14)

                                                  x + 12 = 28

                                                     -12   -12  

                                                x= 16  so X is 16.       check:  16+12=28/2 is 14.

6 0
3 years ago
Find two matrices A and B such that 2A-3B= [5 0
amm1812

The easy answer: If, say, B is the zero matrix, i.e.

B=\begin{bmatrix}0&0\\0&0\end{bmatrix}

then

2A=\begin{bmatrix}5&0\\-1&2\end{bmatrix}\implies A=\begin{bmatrix}\dfrac52&0\\\\-\dfrac12&1\end{bmatrix}

3 0
3 years ago
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