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sattari [20]
3 years ago
11

Please help :(( (will mark brainliest!!!!)

Mathematics
1 answer:
garri49 [273]3 years ago
6 0

Answer:

(-8,0) (0,-4) (4,-6)

Step-by-step explanation:

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Fill in the table using this function rule. y=-3x-5
Gelneren [198K]

Answer:

1, -2, -5, -8

Step-by-step explanation:

(-3*x)-5

just do that i swear

7 0
3 years ago
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Can someone answer this question?
KonstantinChe [14]

Answer:

No solution.

Step-by-step explanation:

(1/12)^-2b * 12^(-2b + 2) = 12

1/12 is basically the same thing as 12^-1.

12^(-1 * -2b) * 12^(-2b + 2) = 12

12^2b * 12^(-2b + 2) = 12

12^(2b - 2b + 2) = 12

12^2 = 12

Since the answers don't match up, the answer will be no solution.

Hope this helps!

5 0
4 years ago
Read 2 more answers
The question is ab+c=d solve for b
steposvetlana [31]

Answer:

b = (d - c)/a

Step-by-step explanation:

ab + c = d

ab = d - c

b = (d - c)/a

7 0
4 years ago
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If a ball is thrown into the air with a velocity of 34 ft/s, its height (in feet) after t seconds is given by y = 34t − 16t2. Fi
Finger [1]

<u>ANSWER: </u>

If a ball is thrown into the air with a velocity of 34 feet per second, then velocity of the ball after 1 second is 2 feet per second

<u>SOLUTION: </u>

Given, a ball is thrown into the air with a velocity of 34 feet per second

Initial velocity (u) = 34 feet per second

And also given a relation between displacement and time = \mathrm{y}=34 \mathrm{t}-16 \mathrm{t}^{2} --- eqn 1

We need to find the velocity when t = 1 ; v = ?

We know that, v = u + at and \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}

where v is instantaneous velocity and u is initial velocity

a is acceleration

t is time interval  

s is displacement

using the displacement and time relation eqn (1) we get

Now, when t = 1, displacement s = 34(1) – 16(1)

\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=34-16

34 \times 1+\frac{1}{2} \times a \times 1^{2}=18

34+\frac{a}{2}=18

\begin{array}{l}{\frac{a}{2}=18-34} \\\\ {\frac{a}{2}=-16} \\ {a=-16 \times 2} \\ {a=-32}\end{array}

here, -ve sign indicates that object is in deceleration . so acceleration is -32 ft/s

now put a value in v = u + at

v = 34 + (-32)(1)

v = 34 – 32

v = 2 ft/s

Hence, velocity of the ball after 1 second is 2 ft/s

6 0
3 years ago
-2/3p - 4 &lt; 8<br><br>help for 15 points.​
wel

Answer:

p > -18

Step-by-step explanation:

-2/3p - 4 < 8

-2/3p < 12

p > 12 x -3/2

p > -18

3 0
3 years ago
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