All categories

3 years ago

7y-9=3y+19. Need help

Mathematics

2 answers:

CaHeK987 [17]3 years ago

8 0

Answer: y = 7.

Step-by-step explanation:

First you need to combine like terms. The like terms in this is 7y & 3y and -9 & 19. So to combine those like terms, you first need to subtract 3y on both sides because to cancel out the 3y in the right side, you need to perform the opposite operation on the positive 3. And 3y - 3y is 0 so thats why you subtract 3y on both sides. When you subtract 3y from 7y on the left side (because 3y is the like term of 7y and what operation you do on one side you have to do on the other side of the equation) you get 4y - 9 = 19. Now you still need to combine the like terms -9 and 19 so you need to add 9 on both sides to get 4y = 28. Since 4 is being multiplied by "y" to get "y" by itself in the left side of the equation, you need to divide 4 on both sides. 28 ÷ 4 = 7, so your final answer would be y = 7!

I hope this is helpful :)

alexgriva [62]3 years ago

3 0

Answer:

y=7

Step-by-step explanation:

7y-3y=19+9

4y=28

y=7

You might be interested in

PLZ HELP!!!!!!! Evaluate. 43−4÷2⋅5 20 40 54 150

vitfil [10]

Isnt it 33 ////??? thats what i got.

6 0

3 years ago

Read 2 more answers

Find the rate of change of the function.

Mila [183]

The answer is B

Step by step explanation:

8.0

6 0

3 years ago

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o

Keith_Richards [23]

Answer:

32.1 g

Step-by-step explanation:

In each 3 grams of C, there are 2 grams of A and 1 gram of B. So, for some amount C, the amount remaining of A is 40 -(2C/3), and the amount remaining of B is (50 -C/3). Since the reaction rate is proportional to the product of these amounts, we have ...

C' = k(40 -2C/3)(50 -C/3) = (2k/9)(60 -C)(150 -C) . . . for some constant k

This is separable differential equation with a solution of the form ...

ln((150 -C)/(60 -C)) = at + b

We know that C(0) = 0, so b=ln(150/60) = ln(2.5). And we know that C(10) = 20, so ln(130/40) = 10a +ln(2.5) ⇒ a = ln(1.3)/10

Then our equation for C is ...

ln((150 -C)/(60 -C)) = t·ln(1.3)/10 +ln(2.5)

For t=20, this is ...

ln((150 -C)/(60 -C)) = 2ln(1.3) +ln(2.5) = ln(2.5·1.3²) = ln(4.225)

Taking antilogs, we have ...

(150 -C)/(60 -C) = 4.225

1 +90/(60 -C) = 4.225

C = 60 -90/3.225 ≈ 32.093 . . . . . grams of product in 20 minutes

In 20 minutes, about 32.1 grams of C are formed.

7 0

3 years ago

1+1/9×100÷2+89what will the answer be need help

kirza4 [7]

95.5555555556

Hope this helps :D

4 0

3 years ago

Read 2 more answers

Pls help trig question!!

s2008m [1.1K]

$4\sin^2 x - 3 = 3 \sin x\\\\\implies 4\sin^2 x -3 \sin x -3 =0\\\\\implies 4u^2 -3u -3 =0~~~~~~~~~~~~~~;[\text{Set} ~ \sin x = u]\\\\\implies u = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4\cdot 4 \cdot (-3)}}{2(4)}~~~~~~;[\text{Apply quadratic formula}]\\\\\\\implies u = \dfrac{3\pm\sqrt{57}}8\\\\\implies \sin x = \dfrac{3\pm\sqrt{57}}8~~~~~~;[\text{Substitute back}~ u = \sin x]\\\\\text{Now,}\\\\\sin x = \dfrac{3+\sqrt{57}}8\\\\\text{No solution for}~ x\in \mathbb{R} ~\text{Since}~ -1\leq \sin x \leq 1\\$

$\sin x = \dfrac{3 - \sqrt{57}}8\\\\\implies x = n\pi + (-1)^n \sin^{-1} \left(\dfrac{3-\sqrt{57}}8\right)\\\\\\\text{ When n = 1,2 and}~ [0,2\pi]},\\\\\\x= \pi - \sin^{-1} \left(\dfrac{3-\sqrt{57}}8 \right)~ \text{and} ~x = 2\pi + \sin^{-1} \left(\dfrac{3-\sqrt{57}}8 \right)\\\\\text{In decimal,}\\\\x= 3.75~~~~ \text{and}~~~ x =5.68$

5 0

2 years ago