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valkas [14]
3 years ago
13

The equation of line s,line t, and point p

Mathematics
1 answer:
Veronika [31]3 years ago
3 0

Answer:

T

Step-by-step explanation:

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The segment from (-1,4) to (2,-2) extended three times then terminal point of segment is
ziro4ka [17]
There are various ways in which you could do this problem.  I'm going to share what I think is one of the faster ways.

Instead of thinking of jumping from (-1,4) TO (2,-2),  Consider the horiz. jump separately and the vertical jump separately.  From -1 to 2 is 3 units.  Three times that is 9 units.  Add 9 to -1, obtaining 8.  That's the horiz. component of the terminal point.

From 4 to -2 is -6 units.  Mult. that by 3.  The result is the vert. comp of the terminal point. 
5 0
3 years ago
PLZZZ help with this its pretty simple!
zubka84 [21]

Answer:

2x^3−7x^2+16x−15

Step-by-step explanation:

(2x−3)(x^2−2x+5)

=(2x+−3)(x^2+−2x+5)

=(2x)(x^2)+(2x)(−2x)+(2x)(5)+(−3)(x^2)+(−3)(−2x)+(−3)(5)

=2x^3−4x^2+10x−3x^2+6x−15

=2x3−7x2+16x−15

6 0
3 years ago
Read 2 more answers
A family goes out to eat. The meal costs $44. The sales tax rate is 9%. They leave a 20% tip. How much did they spend?
REY [17]

Answer:

$56.76

Step-by-step explanation:

$56.76 because 20% of 44 is 8.80 which is the tip and 9% of 44 is 3.96 which is the tax rate giving you an answer of $56.76. Hope this helps!

6 0
3 years ago
Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

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

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
5x=15 y 2x=6 <br> Son ecuaciones equivalentes?<br><br> PORFAVOR AYÚDENME ES EMERGENCIA
mihalych1998 [28]
Si, los dos son iguales

Cuando haces 5x=15, la división dice que x = 3. Para 2x=6 puedes usar el resultado de x=3 de la problema de antes, y cuando te usas esto va a ver que 2 de x cuando x=3 es igual a 6, y por eso 5x=15 y 2x=6 son equivalentes.
5 0
3 years ago
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