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inysia [295]
3 years ago
9

A, B, C, or, D please answer correctly ;-;

Mathematics
1 answer:
lisov135 [29]3 years ago
4 0

Answer:

A

Step-by-step explanation:

Graph A has a y intercept of -3 and a slope of 3/4

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What 12r+5s-7t+11r+9s-4r
AysviL [449]
Add the like terms.
12r+11r-4r= 19r
5s+9s=14s
-7t=-7t
19r+14s-7t
4 0
3 years ago
Johnson bought 4 1/4 gallons of ice cream for the celebration the baseball team ate 3/4 of it how many gallons of ice cream did
mezya [45]

Answer:

They ate 0.75. I only wrote it as a decimal.

3÷4= 0.75 Hope it helps :)

Have a good day.

4 0
3 years ago
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Jayden, Maggy, and Ramon shared $49 in the ratio 2:5:7. How much more did Ramon get then Jayden
Drupady [299]
Since the ratio is 2:5:7, you can set up the equation 2x + 5x + 7x = 49. Solving it will give you 14x = 49 or x = 3.5. This means that Jayden got 7(3.5) = $24.50 and Ramon got 2(3.5) = $7. Therefore Ramon got $24.50 - $7 more than Jayden, which is equal to $17.50.
6 0
4 years ago
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How many different 7-digit telephone numbers are possible if the first digit cannot be 0, but the last two digits must be 0?
solniwko [45]
The first digit can only range from 1-9, resulting in 9 possible options. The next four digits can range from 0-9, resulting in 10 options for each. Since the last two digits remain the same, they do not affect the sample size. Using the fundamental counting principle, we can find the amount of telephone numbers possible in the following equation.
9 \times 10 \times 10 \times 10 \times 10 \times 1 \times 1 \\ 90000

4 0
3 years ago
Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




6 0
3 years ago
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