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Rashid [163]
3 years ago
10

For a basic subscription, a cable television provider charges an activation fee of $60, plus $125 per month. What linear functio

n represents the total cost of a basic cable subscription for t months? What is the total cost for two years of service?
pls will give brainlyiest
Mathematics
1 answer:
kap26 [50]3 years ago
8 0

Answer:

When you add 60 plus 125 dived the days in the month I think you have your awnser!

Step-by-step explanation:

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How do you solve z= - 2/3a, for a
gogolik [260]
Z = -2/3a
divide by -2/3 on both sides
a = z divided by -2/3
a = -3/2z
6 0
4 years ago
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What do you think about geometry?
Brums [2.3K]

Answer:

I think its better than other types of math.

Step-by-step explanation:

7 0
3 years ago
From a point A that is 8.20 m above level ground, the angle of elevation of the top of a building s 31 deg 20 min and the angle
Mariulka [41]

Answer:

30.11 meters ( approx )

Step-by-step explanation:

Let x be the distance of a point P ( lies on the building ) from the top of the building such that AP is perpendicular to the building and y be the distance of the building from point A, ( shown in the below diagram )

Given,

Point A is 8.20 m above level ground,

So, the height of the building = ( x + 8.20 ) meters,

Now, 1 degree = 60 minutes,

⇒ 1\text{ minute } =\frac{1}{60}\text{ degree }

20\text{ minutes }=\frac{20}{60}=\frac{1}{3}\text{ degree}

50\text{ minutes }=\frac{50}{60}=\frac{5}{6}\text{ degree}

By the below diagram,

tan ( 12^{\circ} 50') = \frac{8.20}{y}

tan(12+\frac{5}{6})^{\circ}=\frac{8.20}{y}

tan (\frac{77}{6})^{\circ}=\frac{8.20}{y}

\implies y=\frac{8.20}{tan (\frac{77}{6})^{\circ}}

Now, again by the below diagram,

tan (31^{\circ}20')=\frac{x}{y}

tan(31+\frac{1}{3})=\frac{x}{y}

\implies x=y\times tan(\frac{94}{3})=\frac{8.20}{tan (\frac{77}{6})^{\circ}}\times tan(\frac{94}{3})^{\circ}=21.9142943216\approx 21.91

Hence, the height of the building = x + 8.20 = 30.11 meters (approx)

8 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
How to solve 2d-e=7 d+e=5
loris [4]
<span>2d – e = 7 </span>
<span>d + e = 5 </span>

<span>3d=12 </span>

<span>d=4 </span>


8 0
4 years ago
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