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castortr0y [4]
3 years ago
9

What’s |z-15|+5c+j z=10 c=-2 j=-4

Mathematics
1 answer:
Mashutka [201]3 years ago
4 0

Answer:

-9

Step-by-step explanation:

10-15 = -5 I-5I= 5

5*-2= -10

5+-10+-4 = -9

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4=32b<br> solve for b WILL GIVE BRAINLIEST
Makovka662 [10]

Answer:

1/8

Step-by-step explanation:

32b \div 32 = 4 \div 32 \\

5 0
2 years ago
Use the function f(x) = -3x^2+13x+3 to find x when f(x) = -7
Bond [772]

Answer: - 235

<u>Step-by-step explanation:</u>

f(x) = -3x² + 13x + 3

f(-7) = -3(-7)² + 13(-7) + 3

      = -3(49) - 91 + 3

      = -147 - 88

      = -235  

7 0
3 years ago
Read 2 more answers
Write the equation for a parabola with the focus at
katen-ka-za [31]
The answer is (y - 4)2 = -12(x - 2)
Since the graph is facing the left, the starting equation would be (y - k)2 = -4p(x - h).
(h, k) is the vertex of the graph. If you plot the focus and the directrix, you can see that the distance is 6. Divide 6 by 2, thus the vertex should be 3 squares away from the focus and 3 squares away from the directrix. The vertex is (2, 4). And since the distance from the focus to the vertex and the distance from the directrix to the vertex is 3, P = 3.
Insert the numbers in and you should get the last choice.
5 0
3 years ago
Read 2 more answers
What will be the unit digit of the cube of a number ending with 9 ?
Bess [88]

Answer:

9

Step-by-step explanation:

If I’m understanding you correctly, the answer would be nine.  Any number with a 9 in the one’s place (unit digit) would still have a nine in that place after being cubed.

29³=24,389

69³=328,509

3 0
2 years ago
Read 2 more answers
Use synthetic division to determine whether the number k is an upper or lower bound (as specified for the real zeros of the func
mote1985 [20]

Solution:

The given Polynomial is :

f(x) = 5x^4 + 4x^3 - 2x^2 + 2x + 4=5( x^4 + \frac{4}{5}x^3 - \frac{2}{5}x^2 + \frac{2}{5}x + \frac{4}{5})

By Rational Root theorem the  of Zeroes of the Polynopmial are:

\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5},\pm1,\pm2,\pm4

But , f(\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5},\pm1,\pm2,\pm4)\neq 0

So, no root of this polynomial is real.

Therefore, All the four roots of Polynomial are imaginary.

So, we can't say whether the number k=2, is an upper or lower bound of the polynomial f(x) = 5x^4 + 4x^3 - 2x^2 + 2x + 4=5( x^4 + \frac{4}{5}x^3 - \frac{2}{5}x^2 + \frac{2}{5}x + \frac{4}{5}).

8 0
3 years ago
Read 2 more answers
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