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2 years ago

Combing like terms -14+12+9z-16z =

Mathematics

2 answers:

Temka [501]2 years ago

8 0

Combining like terms is combining numbers with the same variables.

= -14 + 12 + 9z - 16z

= (-14 + 12) + (9z - 16z)

= -2 - 7z

~Put into standard form

-7z - 2

Best of Luck!

tigry1 [53]2 years ago

5 0

Answer:

-6/7

Step-by-step explanation:

-14+12+9z-16z=0

-6-7z=0

Add 7z to both sides

-6=7z

Divide both sides by 7

z=-6/7

Therefore the value for z is -6/7

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Someone know this question??

Bogdan [553]

Hhhhmmmmm...

13.50 for food
2.40 for drink

Since there is a voucher for 1/3 of food,

13.50 multiplied by 1/3 is 4.5 which is the discount

13.50-4.5=9

9 is the final cost for the food.

Add 9 to 2.40 for the drink which results in a total of 11.40

To find the amount charged for service you multiply 11.40 by .15

11.40 × .15 = 1.71

You add 1.71 to 11.40

The total is 13.11

7 0

3 years ago

What are the factors of 2a³ + a² + 2a + 1?

morpeh [17]

Answer:

The answer is (a²+1)(2a+1)

4 0

2 years ago

Would appreciate help on this please and thanks

bonufazy [111]

Since it is an isosceles trapezoid;
AB=CD which means;
5y-3=6y-17
get the variables to one side and you get:
-3=y-17
Then add 17 to both sides and
14=y
Therefore y=14 is your answer

8 0

3 years ago

Plss answer!! <br>MERRY Christmas!!! <br>Critical Question. AAAAA

liberstina [14]

Answer:

Step-by-step explanation:

abc = 1

We have to prove that,

$\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

We take left hand side of the given equation and solve it,

$\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}$

Since, abc = 1,

$\frac{1}{c}=ab$ and c = $\frac{1}{ab}$

By substituting these values in the expression,

$\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+\frac{1}{c}}+\frac{1}{1+c+\frac{1}{a}}=\frac{1}{1+a+\frac{1}{b}}+\frac{1}{1+b+ab}+\frac{1}{1+\frac{1}{ab}+\frac{1}{a}}$

$=\frac{b}{b+ab+1}+\frac{1}{1+b+ab}+\frac{ab}{ab+1+b}$

$=\frac{1+b+ab}{1+b+ab}$

$=1$

Which equal to the right hand side of the equation.

Hence, $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

5 0

2 years ago

Find the value of the six trig functions given the triangle below.

Y_Kistochka [10]

$\text{Here,}\\\\\text{Opposite side,}~ O = 2\\\\\text{Hypotenuse,}~ H = \sqrt 5 \\\\\text{Adjacent side,}~ A = \sqrt{5-4} = \sqrt 1 =1 ~~ ;[\text{By using Pythagorean theorem}]\\\\\\\sin \theta = \dfrac{O}H = \dfrac 2{\sqrt 5}\\\\\\\cos \theta = \dfrac{A}{H} = \dfrac{1}{\sqrt 5} \\ \\\\ \tan \theta = \dfrac{O}{A} = \dfrac 21 = 2$

$csc \theta = \dfrac{1}{\sin \theta} = \dfrac{\sqrt 5}2\\\\\\\sec \theta = \dfrac{1}{\cos \theta} = \sqrt 5 \\\\\\\cot \theta = \dfrac 1{\tan \theta} = \dfrac 12$

5 0

2 years ago