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Nadya [2.5K]
3 years ago
8

How can i get the answer to 5 X 70 = 5 X ____ tens = ____ tens = _____

Mathematics
1 answer:
Maru [420]3 years ago
3 0
Multiple 5x70 to get 350. Which makes the equation 350=5x and then you have to divide both sides by 5 to get x by itself.  so 5x/5 and 350/5 makes x=70
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Let p and q be different prime numbers. How many positive factors will (p^2 * q^4)^3 have?
Naily [24]

Answer:

  91

Step-by-step explanation:

p^6×q^12 will have (6+1)(12+1) = 7×13 = 91 positive integer divisors.

7 0
3 years ago
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Divide £50 into the ratio 1:4
Helga [31]
Don't know how to type the symbole infront of the 50 so I will subsitute
£50=$50

so 1:4
add it up
1+4=5

so 5 total units
$50=5 units
divide by 5
$10=1 unit

so 1:4=10:10 times 4=10:40

the answer is 10:40

8 0
3 years ago
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I will give you brainliest, 5 stars, and a thanks! Plus ten points please answer quickly!!
svlad2 [7]

Pre-Image:

(-4, 2), (4, 2), (-3, -1), (3, -1)

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(-8, 4), (8, 4), (-6, -2), (6, -2)

6 0
2 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
What is - 3/5 x = 15
Keith_Richards [23]

Answer:

X = 25 because 25 · 3/5, you will get 15

5 0
3 years ago
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