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2 years ago

Can someone please help me fill out this table

Mathematics

1 answer:

Amiraneli [1.4K]2 years ago

4 0

I thought I saw this one earlier and there were people answering. Guess not.

h(t) = 5 + 98t - 4.9t²

h(0) = 5 meters

h(5) = 5 + 98(5) - 4.9(5)² = 372.5 meters

h(10) = 5 + 98(10) - 4.9(10)² = 5 + 980 - 490 = 495 meters

h(15) = 5 + 98(15) - 4.9(15²) = 372.5 meters

h(20) = 5 meters

I'll leave the graphing to you, plot those points

(0,5), (5,372,5),(10,495),(15,372.5) (20,5)

and connect the dots

It's coming down pretty fast at 5 feet up 20 seconds. That last bit is 5/495 or 1% of the trip, so probably double the average speed of around 50m/s so 100msec or .05 sec to go 5 meters.

So it's in the ground at about 20.05 seconds after launch.

Answer: 20.05 meters

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12k+5=k+49 what's the answer

dybincka [34]

The answer is k=4 you're welcome

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3 years ago

GI bisects DGH so that m DGI is x-3 and mIGH is 2x-13 find the value of x

baherus [9]

x = 10

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3 years ago

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Find the area of a triangle with a =33, b =51, and c =42.

PolarNik [594]

Answer:

d. 690.1 units²

Step-by-step explanation:

1. You can use the Heron's formula to calculate the area of the triangle, which is:

$K=\sqrt{s(s-a)(s-b)(s-c)}$

Where:

$s=\frac{a+b+c}{2}$

2. Calculate $s$ :

$s=\frac{33+51+42}{2}\\s=63$

3. Substitute values into the formula shown above and calculate the area, as following:

$K=\sqrt{63(63-33)(63-51)(63-42)}\\K=690.1units^{2}$

4. Therefore, the area is 690.1 units².

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3 years ago

Factorise 2x^2 - 3x -2 < 0

Galina-37 [17]

Step 1: Factor $2 x^{2} - 3x -2$
1. Multiply 2 by -2, which is -4.
2. Ask: Which two numbers add up to -3 and multiply to -4?
3. Answer: 1 and -4
4. Rewrite $-3x$ as the sum of $x$ and $-4x$

$2 x^{2} +x-4x-2\ \textless \ 0$

Step 2: Factor out common terms in the first two terms, then in the last two terms.

 $x(2x+1)-2(2x+1)\ \textless \ 0$

Step 3: Factor out the common term $2x+1$

$(2x+1)(x-2)\ \textless \ 0$

Step 4: Solve for $x$

1. Ask: When will $(2x+1)(x-2)$ equal zero?
2. Answer: When $2x + 1 = 0$ or $x-2=0$
3. Solve each of the 2 equations above:

 $x=- \frac{1}{2} ,2$

Step 5: From the values of $x$ above, we have these 3 intervals to test.
x = < -1/2
-1/2 < x < 2
x > 2

Step 6: Pick a test point for each interval

For the interval $x\ \textless \ - \frac{1}{2} 
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Lets pick $x=-1.$ Then, $2(-1) ^{2} -3 * -1 -2 \ \textless \ 0$
After simplifying, we get $3\ \textless \ 0$ , Which is false.
Drop this interval.

For this interval $- \frac{1}{2} \ \textless \ x\ \textless \ 2$
Lets pick $x=0$ . Then, $2* 0^{2} - 3 * 0-2\ \textless \ 0$ . After simplifying, we get $-2\ \textless \ 0,$ which is true. Keep this interval.

For the interval $x\ \textgreater \ 2$

Lets pick $x = 3.$ Then, $2 * 3 ^{2} -3*3-2\ \textless \ 0.$ After simplifying, we get $7\ \textless \ 0$ , Which is false. Drop this interval.

.Step 7: Therefore,
$- \frac{1}{2} \ \textless \ x\ \textless \ 2$

Done! :)

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3 years ago

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charle [14.2K]

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3 years ago

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