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Novay_Z [31]
4 years ago
10

Simplify the expression 7 squareroot of 7 -2 squareroot of 28 A. 3 squaroot 28

Mathematics
1 answer:
Genrish500 [490]4 years ago
5 0
In mathematical form, It would be: 7√7 - 2√28
= 7√7 - 2√4√7
= 7√7 - 4√7
= 3√7

In short, Your Answer would be Option B

Hope this helps!
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F(x) = (x + 5)2(x - 9)(x + 1)
irga5000 [103]

Answer:

x = 5  and  ( 5 , − 9 )

Explanation:

The equation of a parabola in  vertex form  is.

2 2 y = a ( x − h ) 2 + k 2 2

Where  ( h , k )  are the coordinates of the vertex and a  is a multiplier

f ( x ) = ( x − 5 ) 2 − 9  is in this form

with  ( h , k ) = ( 5 , − 9 )

⇒ vertex  = ( 5 , − 9 )

the axis of symmetry passes through the vertex is vertical  with equation

x = 5 ←  axis of symmetry  graph{(y-x^2+10x-16)(y-1000x+5000)=0 [-20, 20, -10, 10]}

8 0
2 years ago
These three. Please. I don’t know how to do them
IRINA_888 [86]

Answer:

23)

P = 20x^2 + 14xy^2

A = 70x^3y^2

24)

P = 8a^3b + 4ab

A = 4a^4b^2

25)

A = 18x^3y


Step-by-step explanation:

23)

P = 2(10x^2 + 7xy^2)

P = 20x^2 + 14xy^2

A = 10x^2 x 7xy^2

A = 70x^3y^2


24)

P = 2a^3b + 6a^3b  + 4ab

P = 8a^3b + 4ab

A = 1/2(2a^3b)(4ab)

A = 4 a^4 b^2


25)

A = 1/2(4xy+8xy)(3x^2)

A = 1/2(12xy)(3x^2)

A = 18x^3y

3 0
3 years ago
Maxaad dooneysa inaad ogaato
Gnom [1K]
Ma jiraan wax hadda, laakiin mahad.
8 0
3 years ago
Read 2 more answers
A= 1/1 bh solve for h
8_murik_8 [283]

\text{Hello there!}\\\\\text{Solve for h:}\\\\A=\frac{1}{1}bh\\\\\text{Multiply both sides by 1}\\\\A=1bh\\\\\text{Divide both sides by 1b}\\\\\frac{A}{1b}=h\,\,or\,\,\frac{A}{b}=h\\\\\boxed{h=\frac{A}{b}}

4 0
4 years ago
Find the general solution of the nonhomogeneous differential equation x^2y''-2y=3(x^2) -1, (x>0).
Ira Lisetskai [31]

Answer:

G.S=C_1\frac{1}{x}+C_2x^2+x^2logx+\frac{1}{2}

Step-by-step explanation:

We are given that non-homogeneous differential equation

x^2y''-2y=3(x^2)-1

It is Cauchy Euler equation

Substitute x=e^t  x>0

Auxillary equation

D'(D'-1)-2=0

D'^2-D'-2=0

(D'-2)(D'+1)=0

D'-2=0 \implies D'=2

D'+1=0\implies D'=-1

Complementary solution

y=C_1e^{-t}+C_2e^{2t}

y=C_1\frac{1}{x}+C_2x^2

Particular solution

y_p=\frac{3e^{2t}}{D'^2-D'-2}-\frac{e^{0t}}{D'^2-D'-2}

y_p=te^{2t}+\frac{1}{2}=x^2logx+\frac{1}{2}

G.S=C_1\frac{1}{x}+C_2x^2+x^2logx+\frac{1}{2}

Hence, general solution G.S=C_1\frac{1}{x}+C_2x^2+x^2logx+\frac{1}{2}

6 0
4 years ago
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