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Aneli [31]
3 years ago
15

Solve,if integral roots cannot be found estimate the roots by stating the consecutive intergers between which the roots lie.

Mathematics
1 answer:
miv72 [106K]3 years ago
8 0
Hey mate 

The answers to this are going to be in the pic down below.
Hope this helped

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What is (16+10)-1+13x5=
Mariana [72]
(16+10)-1+13*5
26-1+65
25+65
90
8 0
4 years ago
Read 2 more answers
Please can someone solve this
Crazy boy [7]

Answer:

200.86

Step-by-step explanation:

Using the formulas:

A=πr2

C=2πr

Solving for A:

A=C2

4π=50.242

4·π≈200.85812

I hope this helps.

4 0
3 years ago
F(x) = 2x^2 – 5x – 3<br> g(x) = 2x^2 + 5x + 2<br> Find: (f/g) (x)
GrogVix [38]

Answer:

Since this means f of g so I will get the function g(x) and replace it with X in the function of f (x)

5 0
3 years ago
I NEED HELP!!!!! PLEASE ​
lys-0071 [83]

Answer:

I don't know

Step-by-step explanation:

your image is blocked on my computer. sorry

7 0
3 years ago
PLEASE NEEED HEELPP
krok68 [10]
There are many systems of equation that will satisfy the requirement for Part A.
an example is y≤(1/4)x-3 and y≥(-1/2)x-6
y≥(-1/2)x-6 goes through the point (0,-6) and (-2, -5), the shaded area is above the line. all the points fall in the shaded area, but
y≤(1/4)x-3 goes through the points (0,-3) and (4,-2), the shaded area is below the line, only A and E are in the shaded area. 
only A and E satisfy both inequality, in the overlapping shaded area.

 
Part B. to verify, put the coordinates of A (-3,-4) and E(5,-4) in both inequalities to see if they will make the inequalities true. 
 for y≤(1/4)x-3: -4≤(1/4)(-3)-3
-4≤-3&3/4 This is valid.
For y≥(-1/2)x-6: -4≥(-1/2)(-3)-6
-4≥-4&1/3 this is valid as well. So Yes, A satisfies both inequalities. 
Do the same for point E (5,-4)

Part C: the line y<-2x+4 is a dotted line going through (0,4) and (-2,0)
the shaded area is below the line
farms A, B, and D are in this shaded area. 
8 0
3 years ago
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