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lesya [120]
2 years ago
11

Can someone help pls.

Mathematics
2 answers:
kap26 [50]2 years ago
8 0

<em>My goodness, this is rather confusing in the way it is worded. Nevertheless, I will attempt to do what I can. Just please keep in mind that this is my own interpretation of the problem, and therefore could be... incorrect.</em>

<em>I think, to start out, we could set up the problem like so</em>

<em>15 + t ≥ 26</em>

<em>because t is not a set number. </em>

<em>Then all that is needed is to subtract 15 from both sides, and the equation becomes</em>

<em>t ≥ 11</em>

<em>So the resulting answer is t ≥ 11.</em>

<em />

<em>I hope that my interpretation helps.</em>

<em>-Toremi</em>

<em />

LekaFEV [45]2 years ago
3 0

Answer:

Step-by-step explanation:

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Use the quadratic model y = −4x2 − 3x + 4 to predict y if x equals 5.
REY [17]

Answer:

-111

Step-by-step explanation:

substitute 5 in place of each x

    -4(5)^2 - 3(5) + 4

then evaluate or type it into your calculator

    -100 - 15 + 4

7 0
3 years ago
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Barty has 70 balloons, and he will pop four of them every second. Fill in the missing information on the table, it relates b, th
Vedmedyk [2.9K]
T b
0 70
1 66
2 62
3 58
4 54

This is not a proportional relationship because the ratios are not equivalent.
1/66 (t/b) is not equivalent to 2/62 (t/b).
4 0
2 years ago
What is y 12y - 19= 6y+1=
nadezda [96]

Answer:

<h2><u><em>Y= 10/3 or 3.333....</em></u></h2>

Step-by-step explanation:

12y-19= 6y+1

12y-6y = 1 +19

6y = 20

y = 20/6

y= 10/3 or 3.33..

 

<h2><u><em>HAVE A NICE TIME :)</em></u></h2><h2><em>pls give a </em><u><em>brainliest</em></u><em> if it is CORRECT</em></h2>
3 0
1 year ago
Write the equation of the ellipse with foci at (-2,2) and (4,2) and major axis of length 10 *
Luda [366]

Answer:

The answer to your question is   \frac{(x - 1)^{2}}{25} + \frac{(y - 2)^{2}}{16} = 1          

Step-by-step explanation:

Data

Foci  (-2, 2)  (4, 2)

Major axis = 10

Process

1.- Plot the foci to determine if the ellipse is vertical or horizontal. See the picture below.

From the graph we conclude that it is a horizontal ellipse.

2.- Determine the foci axis (distance between the foci)

                  2c = 6

                    c = 6/2

                    c = 3

3.- Determine a

                  2a = 10

                    a = 10/2

                    a = 5

4.- Determine b using the Pythagorean theorem

                   a² = b² + c²

-Solve for b

                   b² = a² - c²

                   b² = 5² - 3²

                   b² = 25 - 9

                   b² = 16

                   b = 4

5.- Find the center (1, 2)  From the graph, it is in the middle of the foci

6.- Find the equation of the ellipse

                      \frac{(x - 1)^{2}}{5^{2}} + \frac{(y - 2)^{2}}{4^{2}} = 1

                       \frac{(x - 1)^{2}}{25} + \frac{(y - 2)^{2}}{16} = 1                

3 0
2 years ago
The harbormaster wants to place buoys where the river bottom is 20 feet below the surface of the water. Complete the absolute va
Natalija [7]

Answer:

The answer is below

Step-by-step explanation:

The bottom of a river makes a V-shape that can be modeled with the absolute value function, d(h) = ⅕ ⎜h − 240⎟ − 48, where d is the depth of the river bottom (in feet) and h is the horizontal distance to the left-hand shore (in feet). A ship risks running aground if the bottom of its keel (its lowest point under the water) reaches down to the river bottom. Suppose you are the harbormaster and you want to place buoys where the river bottom is 20 feet below the surface. Complete the absolute value equation to find the horizontal distance from the left shore at which the buoys should be placed

Answer:

To solve the problem, the depth of the water would be equated to the position of the river bottom.

d(h)=river \ bottom\\\\The \ river \ bottom=-20\ feet(below)\\\\d(h) = -20\\\\\frac{1}{5}|h-240|-48=-20\\ \\\frac{1}{5}|h-240|=-20+48\\\\\frac{1}{5}|h-240|=28\\\\|h-240|=28*5\\\\|h-240|=140\\\\h-240=140\ or\ h-240=-140\\\\h=140+240\ or\ h=-140+240\\\\h=380\ or\ h=100\\\\The\ buoys\ should\ be\ placed\ at\ 100\ feet\ and\ 380\ feet\ from\ left-hand\ shore

4 0
3 years ago
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