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xxTIMURxx [149]
3 years ago
10

Find the sum Adding and Subtracting Polynomials (2n2−5n−6)+(−n2−3n+11)=

Mathematics
2 answers:
AnnyKZ [126]3 years ago
7 0
<h2><em><u>EXPLANATION :-</u></em></h2>

= (2 {n}^{2} - 5n - 6) + (  - {n}^{2} - 3n  + 11)

= 2 {n}^{2}  - 5n - 6 -  {n}^{2} - 3n  + 11

= 2 {n}^{2} -  {n}^{2}  - 5n - 3n - 6 + 11

= (n²-8n+5)→Answer

I hope that it answer your question...

<u>Mark me Brainliest</u>

Anika [276]3 years ago
6 0
Collect like terms
2n2 i’m gonna assume is 2n^2
and n2 is 2n
2n^2-10n+5
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( please help this is the last question and i have 4 min left, thank you for the help!)
Naddik [55]

Answer:

Step-by-step explanation:

-4x∧2  -1

-4x squared minus one

4 0
2 years ago
Mariana jarred 42 liters of jam after 6 days. How much jam did Mariana jar if she spent 7
Alexeev081 [22]
She would’ve made 49 jars of jam
4 0
3 years ago
Read 2 more answers
Solve for x enter the solutions from the least to greatest x2+7=43 lesser x= Greater x=
Snowcat [4.5K]

Answer:

x = ±6

Step-by-step explanation:

Step 1: Write out equation

x² + 7 = 43

Step 2: Subtract 7 on both sides

x² = 36

Step 3: Square root both sides

√x² = √36

x = ±6

7 0
3 years ago
a. Determine whether the Mean Value Theorem applies to the function f (x )equals ln 15 xf(x)=ln15x on the given interval [1 comm
ivolga24 [154]

Answer:

(a)

The function f is continuous at [1,e] and differentiable at (1,e), therefore

the mean value theorem applies to the function.

(b)

c = e-1 \\ = 1.71828

Step-by-step explanation:

(a)

The function f is continuous at [1,e] and differentiable at (1,e), therefore

the mean value theorem applies to the function.

(b)

You are looking for a point c   such that

\frac{1}{c} = \frac{\ln(15e)-\ln(15*1)}{e-1} = \frac{\ln(15e/15)}{e-1} = \frac{\ln(e)}{e-1} = \frac{1}{e-1}

You have to solve for c  and get that

c = e-1 \\ = 1.71828

5 0
3 years ago
Can someone help me please?? in Algblra2 - Variations, Progression, and Theorems
VashaNatasha [74]
If (y-1) is a factor of f(y), f(y)=0 when y=1.  So if you find that f(1)=0, then (y-1) is a factor of f(y).

f(y)=y^3-9y^2+10y+5

f(1)=1-9+10+5=7

Since f(1)=7, (y-1) is not a factor.
5 0
3 years ago
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