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3 years ago

How do you solve this equation shown? 5x - 3(x + 9) + 10 = 4

Mathematics

2 answers:

PolarNik [594]3 years ago

7 0

Answer:

$x = \frac{21}{2}$

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

5x − 3(x+9) + 10 = 4

5x + (−3)(x) + (−3)(9) + 10 = 4 (Distribute)

5x + −3x + −27 + 10 = 4

(5x+−3x) + (−27+10) = 4 (Combine Like Terms)

2x + −17 = 4

2x − 17 = 4

Step 2: Add 17 to both sides.

2x − 17 + 17 = 4 + 17

2x = 21

Step 3: Divide both sides by 2.

$\frac{2x}{2} = \frac{21}{2}$

$x =\frac{21}{2}$

Hope it helps!!

timofeeve [1]3 years ago

6 0

X=21/2
X=10.5
X=10 1/2
There all right
Pls mark me brainliest

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X + y + w = b

White raven [17]

Answer:

(a) a=6 and b≠ $\frac{11}{4}$

(b)a≠6

Step-by-step explanation:

writing equation in agumented matrix form

$\begin{bmatrix}1 &1 & 0 &1 &b\\ 2 &3 & 1 &5 &6\\ 0& 0 & 1 &1 &4\\ 0& 2 & 2&a &1\end{bmatrix}$

now $R_{2} =R_{2}-2\times R_{1}$

$\begin{bmatrix}1 &1& 0 &1 &b\\ 0 &1& 1 &3 &6-2b\\ 0& 0 & 1 &1 &4\\ 0& 2 & 2&a &1\end{bmatrix}$

now $R_{4} =R_{4}-2\times R_{2}$

$\begin{bmatrix}1 &1& 0 &1 &b\\ 0 &1& 1 &3 &6-2b\\ 0& 0 & 1 &1 &4\\ 0& 0 & 0 &a-6 &4b-11\end{bmatrix}$

a) now for inconsistent

rank of augamented matrix ≠ rank of matrix

for that a=6 and b≠ $\frac{11}{4}$

b) for consistent w/ a unique solution

rank of augamented matrix = rank of matrix

a≠6

c) consistent w/ infinitely-many sol'ns

rank of augamented matrix = rank of matrix < no. of variable

for that condition

a=6 and b=[tex]\frac{11}{4}

then rank become 3 which is less than variable which is 4.

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4 years ago

1 over 4 minus 1 over 8

fgiga [73]

I hope this helps you

1^4-1^8

1-1

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3 years ago

Laurissa rolls two number cubes, each with the numbers 1 through 6. Laurissa adds the numbers that appear in the roll. Which two

defon

Answer:

Let's analyze the possible sums of both dices, i will use the notation:

Dice1 + Dice 2 = sum.

also remember that we have 2 dices, with 6 options each.

So the total number of combinations is 6*6 = 36

we have 36 possible outcomes.

I will start at the extremes, the minimum that we can sum is 2, and the maximum is 12, then:

We can have 2 if:

1 + 1 = 2

only one permutation.

and 12 if:

6 + 6 = 12

Again, only one permutation.

so 2 and 12 have the same chance (1 out of 36)

now, to have 3 we can have:

2 + 1 = 3

1 + 2 = 3.

and to have 11

5 + 6 = 11

6 + 5 = 11

Again, 3 and 11 have the same probability (2 out of 36 options)

And now we can see a pattern.

4 and 10 will have the same chance.

5 and 9 will have the same chance

6 and 8 will have the same chance

7 is the only number that has an unique chance (and it has the largest one)

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4 years ago

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dedylja [7]

216 degrees.
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s344n2d4d5 [400]

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