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Alexandra [31]
3 years ago
13

If the divisor is 40 what is the least three-digit dividend tag would give a remainder of 4

Mathematics
1 answer:
fgiga [73]3 years ago
4 0
Let the three digit dividend be abc , 

then abc = 40*Q+4

So, abc is "40 times Q (the quotient, a number) plus 4"

we try a few values of Q for which 40Q+4 is a 3 digit number:

for Q=2

40*2+4=80+4=84

for Q=3

40*3+4=120+4=124

Answer: 124
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What are the solutionso of 8x^2 = 6 + 22x
liraira [26]
You can get all the terms on one side, and then solve for x by any means:
0 = -8x^2+22x+6

Factor out a -2:
0 = -2(4x^2-11x-3)

Factor this equation:
0 = 4x^2-11x-3

I will use the AC method. To use it, first multiply a and c (in ax^2 + bx +c):
4*-3 = -12

Now, look for two numbers that multiply to -12 and add to -11. Obviously these numbers are -12 and 1. (-12*1 = -12 and -12+1  = -11). Now, because of rules, you set it up in a Punnett square:

4x^2        -12x
x              -3

Now, we find common factors of the terms in rows:
     _x_______-3__
4x| 4x^2        -12x
1   | x              -3

So, you can use this to write an equivalent expression to the quadratic given:
(x-3)(4x+1)

Now, we known the factors are (solve for x): 3 and -1/4.
3 0
3 years ago
Read 2 more answers
Sam has $2.25 in quarters and dimes, and the total number of coins is 12. How many quarters and how many dimes?
a_sh-v [17]
N₁*$0.10 + n₂*$0.25 = $2.25

n₁ + n₂ = 12

-----------

This means that:

n₁ = 12 - n₂

And:

(12 - n₂)*$0.10 + n₂*$0.25 = $2.25

12*$0.10 - n₂*$0.10 + n₂*$0.25 = $2.25

$1.20 + n₂*$0.05*(5-2) = $2.25

$1.20 + n₂*$0.05*3 = $2.25

$1.20 + n₂*$0.15 = $2.25

n₂*$0.15 = $2.25 - $1.20

n₂*$0.15 = $1.05

n₂ = ($1.05)/($0.15)

n₂ = 7

If n₂ = 7:

n₁ + 7 = 12

Therefore, n₁ = 5.

---------

Answer:

5 dimes and 7 quarters.
8 0
3 years ago
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7(x+3)=-2(x-5) <br> Help pls?
olganol [36]

Answer:

x = -11/9

Step-by-step explanation:

7(x+3)=-2(x-5)

Distribute

7x+21 = -2x+10

Add 2x to each side

7x+21 +2x = -2x+10+2x

9x +21 = 10

Subtract 21 from each side

9x +21-21 = 10-21

9x = -11

Divide each side by 9

9x/9 = -11/9

x = -11/9

3 0
3 years ago
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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
Find all zeros<br>f(x)=x^4+6x^3+173x^2+332x+164​
zheka24 [161]
The zeros are (-1,0)
6 0
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