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ch4aika [34]
3 years ago
9

Please help me with that

Mathematics
1 answer:
PtichkaEL [24]3 years ago
7 0
Common solution for each problem:
let
1st point=(x1,y1)
&,
2nd point=(x2,y2),then,

distance between 1st and 2nd point is:
√[(x2-x1)²+(y2-y1)²]

Just replace these x2,y2,x1,y1 with given points and you will get required answer..
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A) Find a polynomial, which, when added to the polynomial 5x2–3x–9, is equivalent to: 18
VMariaS [17]

Answer:

a) -5x^{2}+3x+27

b) -5x^{2}+3x+9

Step-by-step explanation:

a) Let the required polynomial be p(x).

We have the relation, 5x^{2}-3x-9 + p(x) = 18

i.e. p(x) = 18 -5x^{2}+3x+9

i.e. p(x) = -5x^{2}+3x+27


b) Let the required polynomial be q(x).

We have the relation, 5x^{2}-3x-9 + q(x) = 0

i.e. q(x) = 0 -5x^{2}+3x+9

i.e. q(x) = -5x^{2}+3x+9

3 0
3 years ago
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If a=5, the value of 2a+5 is _____​
slavikrds [6]

Answer:

15

Step-by-step explanation:

2(5) +5

10 + 5 = 15

8 0
3 years ago
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For the function given, state the middle line; f(t) = 40cos (80t + 20)
Digiron [165]

In the standard form of the equation

\\ \ f(t)=Acos[b(t\pm c)]+k\\ \\

The middle line =k

For our given problem

f(t) = 40cos (80t + 20)

On comparison we get k=0

Hence middle line=0

8 0
3 years ago
Sophia is dilating a figure. One of her pre-image points with coordinates (-3, 5) has an image with coordinates (-9, 15). If one
Galina-37 [17]
The dilation factor Sophia is using appears to be
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Then the problem tells us
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so we conclude
  (pre-image point) = (3, 1)/3 = (1, 1/3)
8 0
3 years ago
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Use addition to solve the linear system of equations.
Degger [83]

Answer:

(4,-5)

Step-by-step explanation:

Y = 3-2X

8X +5(3-2X) =7

8x + 15 - 10x = 7

-2x = -8

x = 4

y = 3 - 2(4)

y = -5

8 0
2 years ago
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