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tensa zangetsu [6.8K]
3 years ago
15

Expand and simplify (x-5)(x+5)

Mathematics
1 answer:
BigorU [14]3 years ago
4 0
The answer is X^2 - 25 click on the box for more details or explain.

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Now write 12% as a fraction in simplest
Rama09 [41]
Answer: 3/25


Explanation:
12% as a fraction is 12/100. 12/100 divided by 4 is 3/25.

Hope this helped <3
5 0
3 years ago
Read 2 more answers
Peta makes a painting on a parallelogram-shape canvas. What is the area of this parallelogram? 1.53 ft² 3.74 ft² 5.27 ft² 22.09
Nikitich [7]
The answer is 5.27ft²
6 0
3 years ago
I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be
Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: g'(-1)\,(1) since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: g'(-1)= 7 as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

6 0
4 years ago
Please help me with 8 &amp; 9. I don't quite understand how to fill them in. If someone could help me I would gladly appreciate
attashe74 [19]

Complete Truth Table:

8).

p                       q                                 ~ p                        ~ p  and  q

T                        T                                   F                                 T

T                         F                                   F                                 F

F                          T                                 T                                F

F                           F                                  T                                 T







9.)      

p                      q                  ~ q                          p    or     ~ q

T                       T                     F                                 F

T                        F                   T                                  F

F                        T                     F                                  F

F                         F                    T                                    T









Hope that helps!!!!!!!!!!!!                                       : )



8 0
3 years ago
Read 2 more answers
50 POINTS 10 QUESTIONS! <br><br> Solve each quadratic equation using factoring<br><br> PLEASE HELP!!
Tanzania [10]

Answer:

Quadratic Equation | Factoring

Solve each quadratic equation using factoring.

1) v² + 5v + 6 = 0

doing middle term factorisation

v²+(3+2)v+6=0

v²+3v+2v+6=0

v(v+3)+2(v+3)=0

(v+3)(v+2)=0

either

<u>v=-3</u>

<u>or</u>

<u>v=-2</u>

2) g² - 3g = 4

keeping all terms in one side

g²-3g-4=0

doing middle term factorisation

g²-(4-1)g-4=0

g²-4g+g-4=0

g(g-4)+1(g-4)=0

(g-4)(g+1)=0

either

<u>g=4</u>

<u>or</u>

<u>g=-1</u>

3)w² + 4w = 0

w(w+4)=0

either

<u>w=0</u>

<u>or</u>

<u>w=-4</u>

4) s² - 8s + 12 = 0

doing middle term factorisation

s²-(6+2)+12=0

s²-6s-2s+12=0

s(s-6)-2(s-6)=0

(s-6)(s-2)=0

either

<u>s=6</u>

<u>or</u>

<u>s=2</u>

5) x ²+ 2x - 35 = 0

doing middle term factorisation

x²+(7-5)x-35=0

x²+7x-5x-35=0

x(x+7)-5(x+7)=0

(x+7)(x-5)=0

either

<u>x=-7</u>

<u>or</u>

<u>x=5</u>

6) r(r + 2) = 99

opening bracket

r²+2r=99

keeping all terms in one side

r²+2r-99=0

r²+(11-9)r-99=0

r²+11r-9r-99=0

r(r+11)-9(r+11)=0

(r+11)(r-9)=0

either

<u>r=-11</u>

<u>or</u>

<u>r=9</u>

7)k(k-4)=-3

opening bracket

k²-4k=-3

keeping all terms in one side

k²-4k+3=0

k²-(3+1)k+3=0

k²-3k-k+3=0

k(k-3)-1(k-3)=0

(k-3)(k-1)=0

either

k=3

or

k=1

8)t²+ 3t + 2 = 0

doing middle term factorisation

t²+(2+1)t+2=0

t²+2t+t+2=0

t(t+2)+1(t+2)=0

(t+2)(t+1)=0

either

<u>t</u><u>=</u><u>-</u><u>2</u>

<u>or</u>

<u>t</u><u>=</u><u>-</u><u>1</u>

9)m ^ 2 - 81 = 0

m²=81

doing square root in both side

\sqrt{m²}=\sqrt{9²}

<u>m=±9</u>

<u>either</u>

<u>m</u><u>=</u><u>9</u>

<u>or</u>

<u>m</u><u>=</u><u>-</u><u>9</u>

10) h²- 17h + 70 = 0

doing middle term factorisation

h²-(10+7)h+70=0

h²-10h-7h+70=0

h(h-10)-7(h-10)=0

(h-10)(h-7)=0

either

<u>h</u><u>=</u><u>1</u><u>0</u>

<u>or</u>

<u>h</u><u>=</u><u>7</u>

6 0
3 years ago
Read 2 more answers
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