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3 years ago

What's the answer to -5-r=1+2r

2 answers:

7 0

-5-r=1+2r
Add your r to both sides of the equal sign
-5=1+3r
Subtract your 1 on both sides of the equal sign
-6=3r
Divide your 3 on both sides of the equal sign
-2=r
HOPE THIS HELPS YOU! ^_^

Stella [2.4K]3 years ago

6 0

-5-r=1+2r
-r-2r=1+5
-3r=6
r=-2

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Find the value to the nearest tenth

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Answer:

twenty five point four ....25.4

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3 years ago

Which equation is equivalent to x + 8 = 21? (x + 8) × 4 = 21 (x + 8) × 4 = 21 ÷ 4 (x + 8) × 4 = 21 × 2 (x + 8) × 4 = 21 × 4

Drupady [299]

Answer:

(x + 8) × 4 = 21 × 4

Explanation:

x + 8 = 21

21 - 8 = 13

(13 + 8) × 4 = 21 × 4

21 × 4 = 21 × 4

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84 = 84

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2 years ago

Pamela's age is two times Jiri's age. The sum of their ages is <br> 33<br> . What is Jiri's age?

notka56 [123]

Let x is Jiri's age
2x is Pamela's age

x + 2x = 33
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x =11
Answer: Jiri's age is 11

4 0

3 years ago

Read 2 more answers

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 $ax_1+bx_2$ is 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 $R^3$

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3 years ago

Given data below, what is the mean absolute deviation? <br> 58,88,40,60,72,66,80,48

JulijaS [17]

Mean is 64 and the mean absolute deviation is 12.5

3 0

3 years ago