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1 year ago

Please Solve 8=1/3(a+5)

Mathematics

2 answers:

olga2289 [7]1 year ago

7 0

Answer:

a=19

Step-by-step explanation:

8=1/3(a+5)

To solve for a, multiply each side by 3

3*8=3*1/3(a+5)

24 = a+5

Subtract 5 from each side

24-5 = a+5-5

19 =a

e-lub [12.9K]1 year ago

6 0

Answer:

a = 19

Step-by-step explanation:

Given the equation of:

$\displaystyle{8 = \dfrac{1}{3}(a+5)}$

First, multiply both sides by 3. That will clear the denominator which makes it easier to solve for a variable:

$\displaystyle{8 \cdot 3= \dfrac{1}{3}(a+5) \cdot 3}\\\\\displaystyle{24= a+5}$

Subtract both sides by 5 to isolate the variable:

$\displaystyle{24-5=a+5-5}\\\\\displaystyle{19=a}$

Therefore, the value of a is 19.

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Solve 4(t+1)=6t-1 A) 1 1/2 B) 2 1/2 C) 1 D) 0

-BARSIC- [3]

Answer:

answer is B...2 1/2

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

4.9t+0.3=5.6t+0.72 I need help on my work lol

Whitepunk [10]

The first step that we must take before attempting to solve the problem is to understand what the problem is asking us to do and what is given to us to help accomplish that goal. Although it does not explicitly state that we must solve for t, this is usually what the problem statement would be asking if we just receive and expression like this. What is given to us to accomplish that goal is the expression $4.9t+0.3=5.6t+0.72$ .

Now that we have completed that step, we can move onto the next part which is actually solving the problem. The next step that we should take when solving for the unknown, in this case t, is to subtract 4.9t from both sides.

Subtract 4.9t from both sides

$4.9t+0.3=5.6t+0.72$

$(4.9t-4.9t)+0.3=(5.6t-4.9t)+0.72$

$0.3=(5.6t-4.9t)+0.72$

$0.3=(0.7t)+0.72$

Now that we got all of the t's to one side, let us isolate t completely and the next step that we should take is to subtract 0.72 from both sides.

Subtract 0.72 from both sides

$0.3=0.7t+0.72$

$0.3 - 0.72=0.7t+0.72 - 0.72$

$0.3 - 0.72=0.7t$

$-0.42=0.7t$

The final step that we need to take to isolate t would be to divide both sides by 0.7 which would remove the coefficient from the unknown variable t and divide 0.7 from -0.42

Divide both sides by 0.7

$-0.42=0.7t$

$\frac{-0.42}{0.7}=\frac{0.7t}{0.7}$

$\frac{-0.42}{0.7}=t$

$-0.6=t$

Therefore, after fully narrowing down the solution we were able to determine that the solution of the unknown variable or t is equal to -0.6

5 0

1 year 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 $ax_1+bx_2$ 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:

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