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aev [14]
3 years ago
7

G(x) = - 4x2 + 4x + 5 Find the value of g(1).

Mathematics
1 answer:
RideAnS [48]3 years ago
6 0

Since x=1

g(1)=-4

IF YOU LIKE MY ANSWER PLS GIVE ME BRAINLIEST!!

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Almanah is baking cookies. She has a bag that contains 300 cookies. If she wants to bake a batch of 40 cookies with 6 chocolate
Mademuasel [1]

Answer:

Did you mean she has a bag that contains 300 chips? if so yes.

Step-by-step explanation:

If each cookie gets 6 chips and she makes 40 cookies, you can use multiplication to see.

6*40=240

She has plenty, 60, of chocolate chips left over

6 0
3 years ago
Which of the following sets are subspaces of R3 ?
Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

\to A={(x,y,z)|3x+8y-5z=2} \\\\\to  for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                        =3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)

The set A is not part of the subspace R^3

for point B:

\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon  B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                             =-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0

\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon  B

The set B is part of the subspace R^3

for point C: \to C={(x,y,z)|x

In this, the scalar multiplication can't behold

\to for (-2,-1,2) \varepsilon  C

\to -1(-2,-1,2)= (2,1,-1) ∉ C

this inequality is not hold

The set C is not a part of the subspace R^3

for point D:

\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)

The scalar multiplication s is not to hold

\to for (-4, 1,2)\varepsilon  D\\\\\to  -1(-4,1,2) = (4,-1,-2) ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace R^3

For point E:

\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to  a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\

The  x_1, x_2 is the arbitrary, in which ax_1+bx_2is arbitrary  

\to a(x_1,0,0)+b(x_2,0,0) \varepsilon  E

The set E is the part of the subspace R^3

For point F:

\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon  F \\\\\to  a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))

The x_1, x_2 arbitrary so, they have ax_1+bx_2 as the arbitrary \to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F

The set F is the subspace of R^3

5 0
3 years ago
BRAINLIEST PLEASE HELP np<br> If f(x)=-x^2+6x-1 and g(x)=3x^2-4x-1, find (f+g)(x)
raketka [301]

Answer:

\large\boxed{B.\ (f+g)(x)=2x^2+2x-2}

Step-by-step explanation:

f(x)=-x^2+6x-1\\\\g(x)=3x^2-4x-1\\\\(f+g)(x)=f(x)+g(x)\\\\\text{substitute:}\\\\(f+g)(x)=(-x^2+6x-1)+(3x^2-4x-1)\\\\(f+g)(x)=-x^2+6x-1+3x^2-4x-1\qquad\text{combine like terms}\\\\(f+g)(x)=(-x^2+3x^2)+(6x-4x)+(-1-1)\\\\(f+g)(x)=2x^2+2x-2

8 0
3 years ago
What is 874 base 9 multiplied by 676 base 9?​
Gekata [30.6K]

Consider the digit expansion of one of the numbers, say,

676₉ = 600₉ + 70₉ + 6₉

then distribute 874₉ over this sum.

874₉ • 6₉ = (8•6)(7•6)(4•6)₉ = (48)(42)(24)₉

• 48 = 45 + 3 = 5•9¹ + 3•9⁰ = 53₉

• 42 = 36 + 6 = 4•9¹ + 6•9⁰ = 46₉

• 24 = 18 + 6 = 2•9¹ + 6•9⁰ = 26₉

874₉ • 6₉ = 5(3 + 4)(6 + 2)6₉ = 5786₉

874₉ • 70₉ = (8•7)(7•7)(4•7)0₉ = (56)(49)(28)0₉

• 56 = 54 + 2 = 6•9¹ + 2•9⁰ = 62₉

• 49 = 45 + 4 = 5•9¹ + 4•9⁰ = 54₉

• 28 = 27 + 1 = 3•9¹ + 1•9⁰ = 31₉

874₉ • 70₉ = 6(2 + 5)(4 + 3)10₉ = 67710₉

874₉ • 600₉ = (874•6)00₉ = 578600₉

Then

874₉ • 676₉ = 578600₉ + 67710₉ + 5786₉

= 5(7 + 6)(8 + 7 + 5)(6 + 7 + 7)(0 + 1 + 8)(0 + 0 + 6)₉

= 5(13)(20)(20)(1•9)6₉

= 5(13)(20)(20 + 1)06₉

= 5(13)(20)(2•9 + 3)06₉

= 5(13)(20 + 2)306₉

= 5(13)(2•9 + 4)306₉

= 5(13 + 2)4306₉

= 5(1•9 + 6)4306₉

= (5 + 1)64306₉

= 664306₉

4 0
3 years ago
Anybody know how to do Special Right Triangles for geometry if so could you explain how to do it.
Lina20 [59]
Step 1: This is a right triangle with two equal sides so it must be a 45°-45°-90° triangle. Step 2: You are given that the both the sides are 3. If the first and second value of the ratio x:x:x√2 is 3 then the length of the third side is 3√2. Answer: The length of the hypotenuse is 3√2 inches.

3 0
3 years ago
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