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3 years ago

Please an someone help me 5(−3.4k−7)+(3k+21)

Mathematics

2 answers:

Shalnov [3]3 years ago

6 0

Answer:

Step-by-step explanation:

saul85 [17]3 years ago

3 0

Answer:

4.5k

Step-by-step explanation:

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The table shows ticket prices for the local minor

Zanzabum

Answer:

2x club member ticket = 1x non-club member ticket

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3 years ago

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3x +2y=10 4x-y=6 using substitutional method

Anna11 [10]

Answer:

( 2 , 2 )

Step-by-step explanation:

Solve the eqaution :

3x +2y=10

4x-y=6

Solve the eqaution for y :

3x + 2y = 10

y = -6 + 4x

Substitute for y : 3x + 2 ( -6 + 4x ) = 10

x = 10

Substitute for the value of x : x = 2

Substitute the given value of x into the eqaution

y = -6 + 4x : y = -6 + 4 × 2

Solve the eqaution: y = -6 + 4 × 2

solve the eqaution for y : y = 2

y = 2 A possible solution : the ordered solution pair is

( 2 , 2 )

8 0

3 years ago

Please answer!!! Will award brainliest!!!! TY

-Dominant- [34]

Answer:

It said there was 5 green marbles, and it said to draw 5, and you have 5 Green marbles, so green marbles is good to pick.

Step-by-step explanation:

8 0

3 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago

Solve for x. 5(x – 10) = 30 – 15x

SIZIF [17.4K]

Hey!

The first step to solving this problem would be to distribute the parenthesis. We do this so that we can get rid of the parenthesis.

Original Equation :
5 ( x - 10 ) = 30 - 15x

New Equation {Changed by Distribution} :
5x - 50 = 30 - 15x

The next step would be to 50 to both sides. The reason we do that is to get rid of the 50 that is on the left side of the equation.

Old Equation :
5x - 50 = 30 - 15x

New Equation {Changed by Adding 50 to Both Sides} :
5x = 80 - 15x

And now we would add 15x to both sides to get rid of the -15x on the right side of the equation.

Old Equation :
5x = 80 - 15x

New Equation {Changed by Adding 15x to Both Sides} :
20x = 80

Now we divide both sides by 20 to get x on it's own.

Old Equation :
20x = 80

New Equation {Changed by Dividing Both Sides by 20} :
$\frac{20x}{20}$ = $\frac{80}{20}$

Finally, all we have to do is solve to find x.

Old Equation :
$\frac{20x}{20}$ = $\frac{80}{20}$

Solution {Old Equation Solved} :
x = 4

So, in the equation 5 ( x – 10 ) = 30 – 15x, x = 4.

Hope this helps!

- Lindsey Frazier ♥

3 0

3 years ago

Read 2 more answers