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svlad2 [7]
3 years ago
7

Tuto

Mathematics
1 answer:
Damm [24]3 years ago
8 0

Answer:

8r+8pg-pq

Step-by-step explanation:

The subtractable pg cancels out one of the 9 pg's. So 9 pg-1 pg= 8 pg

Hope this helps!

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Write a rule in coordinate form for the translation from Triangle EFG to Triangle
DanielleElmas [232]

Answer:

1st option

Step-by-step explanation:

Consider the coordinates of F and F'

F (- 3, 4 ) and F' (- 1, 3 )

To go from - 3 → - 1 in the x- directions , means adding 2

To go from 4 → 3 in the y- direction means subtracting 1

Then translation rule is

(x, y ) → (x + 2, y - 1 )

7 0
3 years ago
X=-4+y and -2x-5y=15
Scrat [10]

Answer:

x=-5, y=-1

Step-by-step explanation:

Substitute x=-4+y into -2x-5y=15. So we have

-2(-4+y)-5y=15

8-2y-5y=15

-7y=7

y=-1

And x is -4+y so x= -4+(-1)= -5.

7 0
3 years ago
Someone help with this, please. It's for my test ;(
Gnom [1K]

Answer:

y=4

Step-by-step explanation:

90 - 66 = 24

therefore, 8y-8 = 24

8y = 32

divide both sides by 8

y = 4

8 0
3 years ago
PLEASE HELP NTH TERM MATHS
fredd [130]

Answer:

odd numbers

Step-by-step explanation:

4 - 7 is 3, 7 - 12 is 5, 12 - 19 is 7, 19 to 28 is 9

So just keep counting

4 0
2 years ago
Read 2 more answers
PRECAL:<br> Having trouble on this review, need some help.
ra1l [238]

1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}

8. Divide through by x² again:

\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}

10. Factorize the numerator and simplify:

\dfrac{-2x^2+2}{x+1} = -2 \times \dfrac{x^2-1}{x+1} = -2 \times \dfrac{(x+1)(x-1)}{x+1} = -2(x-1) = -2x+2

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

\displaystyle \lim_{x\to\infty} \frac{-2x^2+2}{x+1} = \lim_{x\to\infty} (-2x+2) = \boxed{-\infty}

6 0
2 years ago
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