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3 years ago

find two consecutive even numbers such that the sum of 3 times the smaller and 5 times the larger is 106

1 answer:

4 0

To find that you would do (such that x is the unknown) 3x+5(x+2)=106
and if you do this you will find that x=12

so the two numbers you are looking for are 12 and 14

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I need help with three and four

mart [117]

Answer:

3: $a^{2}+2ab+b^{2}$

4: $8x^{3} -12x^{2}+6x-1$

Step-by-step explanation:

Question 3 can be re-written as:

$(a+b)(a+b)$

which equals:

$a^{2}+2ab+b^{2}$

Question 4 can be re-written as:

$(2x-1)(2x-1)(2x-1)$

Doing the first 2 brackets gives us:

$(4x^{2} -4x+1)(2x-1)$

Expanding these two:

$8x^{3} -4x^{2} -8x^{2} +4x+2x-1$

Simplifying it:

$8x^{3} -12x^{2}+6x-1$

4 0

2 years ago

Read 2 more answers

Choose the true statement. -2 < -3 -2 > -3 -2 = -3Choose the true statement.

Marrrta [24]

-2 < -3 (I think so anyway)

7 0

3 years ago

PLESE HELP ill GIVE BRANILY<br><br> Solve for x. 3.5(7-x)+32=106-1.5(4x+18)

murzikaleks [220]

Answer:

X=9

Step-by-step explanation:

3.5(7−x)+32=106−1.5(4x+18)

Step 1: Simplify both sides of the equation.

3.5(7−x)+32=106−1.5(4x+18)

(3.5)(7)+(3.5)(−x)+32=106+(−1.5)(4x)+(−1.5)(18)(Distribute)

24.5+−3.5x+32=106+−6x+−27

(−3.5x)+(24.5+32)=(−6x)+(106+−27)(Combine Like Terms)

−3.5x+56.5=−6x+79

Step 2: Add 6x to both sides.

−3.5x+56.5+6x=−6x+79+6x

2.5x+56.5=79

Step 3: Subtract 56.5 from both sides.

2.5x+56.5−56.5=79−56.5

2.5x=22.5

Step 4: Divide both sides by 2.5.

2.5x

2.5

22.5

2.5

x=9

4 0

3 years ago

Read 2 more answers

Karen wants to join Buff Gym. There is a $100 initiation fee, and then a $40 monthly fee. A) Determine the equation that express

REY [17]

Sorry for the bad handwriting.

5 0

2 years ago

Please I need help with this!!!!!!!

inysia [295]

Answer:

Step-by-step explanation:

1). Step 4:

$x=5^{\frac{4}{3}}=(5^4)^{\frac{1}{3}}$

$x=\sqrt[3]{5^4}$ [Since, $a^{\frac{1}{3}}=\sqrt[3]{a}$ ]

$x=\sqrt[3]{5\times 5\times 5\times 5}$

Step 5:

$x=\sqrt[3]{(5)^3\times 5}$

$x=\sqrt[3]{5^3}\times \sqrt[3]{5}$

2). He simplified the expression by removing exponents from the given expression.

3). Let the radical equation is,

$(3x-1)^{\frac{1}{5}}=2$

Step 1:

$(3x-1)^{\frac{1}{5}\times \frac{5}{1} }=2^{\frac{5}{1}}$

Step 2:

$(3x-1)=2^5$

Step 3:

$3x=32+1$

Step 4:

$x=11$

4). By substituting $x=11$ in the original equation.

$(3\times 11-1)^{\frac{1}{5}}=(32)^\frac{1}{5}$

$=(2^5)^\frac{1}{5}$

$=2$

There is no extraneous solution.

3 0

2 years ago

find two consecutive even numbers such that the sum of 3 times the smaller and 5 times the larger is 106​

find two consecutive even numbers such that the sum of 3 times the smaller and 5 times the larger is 106