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Brut [27]
3 years ago
8

What is the constant in -4×_9

Mathematics
1 answer:
Mamont248 [21]3 years ago
6 0

Answer:

Do u have like a worksheet for the problem because that kinda confusing , but I would love to help if you could screens shot or post the pic.

Step-by-step explanation:

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5/6 but if i mulsiply by 4 it gives me whole diff
miskamm [114]

Answer:

3.33 or 3 1/3

Step-by-step explanation:

You're correct to multiply by 4.

Your answer will be 10/3

Just make it a mixed number: 3 1/3

3.33 is 10/3 as a decimal.

4 0
2 years ago
What is SOH CAH TOA?
vitfil [10]

Answer:

Soh= sin= opp/hyp cah= cos=adj/hyp

toa= tan= opp/adj

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Find the distance between the two points rounding to the nearest tenth (if necessary).
leonid [27]

Answer:

\boxed {\boxed {\sf d\approx 4.2}}

Step-by-step explanation:

The formula for distance is:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

Where (x₁, y) and (x₂, y₂) are the points.

We are given (-6, 6) and (-3, 3). If we match the value and its corresponding variable, we see that:

  • x₁= -6
  • y₁ = 6
  • x₂ = -3
  • y₂ = 3

Substitute the values into the formula.

d= \sqrt{ (-3 - -6)^2+(3-6)^2

Solve inside the parentheses.

  • -3 --6 = -3+6 = 3
  • 3-6 = -3

d= \sqrt{(3)^2+ (-3)^2

Solve the exponents.

  • (3)²= 3*3= 9
  • (-3)²= -3*-3 =9

d= \sqrt{9+9

Add.

d= \sqrt18

d= 4.24264069

Round to the nearest tenth. The 4 in the hundredth place tells us to leave the 2 in the tenth place.

d \approx 4.2

The distance between the two points is apprximately <u>4.2</u>

4 0
3 years ago
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
What 2 numbers multiply to -25 and add to -24​
S_A_V [24]

Answer: 1 and -25

Step-by-step explanation:

1 x -25 = -25

1 + -25 = -24

7 0
3 years ago
Read 2 more answers
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