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OLEGan [10]
2 years ago
6

Evaluate 3x-4y when x=5 and y=-1

Mathematics
1 answer:
balandron [24]2 years ago
8 0

Answer:

19 is the answer

Step-by-step explanation:

3x - 4y

= <em>3</em><em> </em><em>×</em><em> </em><em>5</em><em> </em><em>-</em><em> </em><em>4</em><em> </em><em>×</em><em> </em><em>(</em><em>-1</em><em>)</em>

= <em>1</em><em>5</em><em> </em><em>+</em><em> </em><em>4</em><em> </em>

= <em>1</em><em>9</em>

<em>hope</em><em> </em><em>this</em><em> </em><em>answer</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em><em> </em>

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(-3x)²-(3x-2)²-6*(2x-1.5)
Ilya [14]
Without the instructions, I can only assume that you are to "expand the given expression."

Following order of operations rules requires that (3x-2)^2 and -6(2x-1.5) be evaluated first.

(3x-2)^2 = 9x^2 - 12x + 4   and    -6(2x-1.5) = -12x + 9

Then we have 9x^2 - (9x^2 - 12x + 4) -12x + 9

the 9x^2 terms cancel, leaving  us with 12x - 4 - 12x + 9 = 5 (answer)

4 0
3 years ago
a friend is having trouble with exponents and asks you to review some worked probelems look over the problems and correct any mi
Stella [2.4K]

We have the following expression:

2x^2y\cdot3x^5y^2

By combining similar terms, we get

2\cdot3\cdot x^2\cdot x^5\cdot y\cdot y^2

now, 2 times 3 is 6 and to add exponents, both the exponents and variables should be alike:

x^2\cdot x^5=x^{2+5}=x^7

Similarly,

y\cdot y^2=y^{1+2}=y^3

Therefore, the answer is

6x^7y^3

4 0
1 year ago
What is the answer to 2h/15=20?
GarryVolchara [31]
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7 0
3 years ago
Find lim ?x approaches 0 f(x+?x)-f(x)/?x where f(x) = 4x-3
Whitepunk [10]

If f(x)=4x-3:

\displaystyle\lim_{\Delta x\to0}\frac{(4(x+\Delta x)-3)-(4x-3)}{\Delta x}=\lim_{\Delta x\to0}\frac{4\Delta x}{\Delta x}=4

If f(x)=4x^{-3}:

\displaystyle\lim_{\Delta x\to0}\frac{\frac4{(x+\Delta x)^3}-\frac4{x^3}}{\Delta x}=\lim_{\Delta x\to0}\frac{\frac{4x^3-4(x+\Delta x)^3}{x^3(x+\Delta x)^3}}{\Delta x}

\displaystyle=\lim_{\Delta x\to0}\frac{4x^3-4(x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3)}{x^3\Delta x(x+\Delta x)^3}

\displaystyle=\lim_{\Delta x\to0}\frac{-12x^2\Delta x-12x(\Delta x)^2-4(\Delta x)^3}{x^3\Delta x(x+\Delta x)^3}=-\frac{12}{x^4}

7 0
3 years ago
Plz answer these questions without putting links and files
kotykmax [81]

Answer:

11/8 and 21/16

Step-by-step explanation:

3 x 1 = 3

4 x 2= 8

3/8+3/4(convert to eighth's)

3/8+6/8=11/8

1 1/8 = 9/8

9 x 1 = 9

8 x 2 = 16

9/16+3/4(convert to sixteenth's)

9/16+12/16

21/16

---

hope it helps

8 0
3 years ago
Read 2 more answers
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