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Varvara68 [4.7K]
3 years ago
5

Hey guys help my please !!Q 4,6,7

Mathematics
1 answer:
padilas [110]3 years ago
6 0
Q4, 4y²=60
y=√15 ( i am not sure)
q6, (5+y) ²=81
5+y=9
y=4
q7 (x+4) (x) =77
x²+4x-77=0
(x+11) (x-7) =0
x=-11(rej) or x=7
therefore ,length= 7+4=11cm
breadth=7cm
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Break apart the number you are subtracting write the different 73-7=
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The answer is 66..................................................
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Growth mindset is a belief that intelligence can improve through________
RoseWind [281]

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Step-by-step explanation:

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Does the absolute value graph below flips or not flips? F(x)=2|x-9|+3
True [87]

Answer:

The absolute value graph below does not flip.

Step-by-step explanation:

New graphs are made when transformed from their parents graphs. The parent graph for an absolute value graph is f(x) = |x|.

The equation used for a new graph transformed from the parent graph is in the form f(x) = a |k(x - d)| + c.

"a" shows vertical stretch (a>1) or vertical compression (0<a<1), and <u>flip across the x-axis if "a" is negative</u>.

"k" shows horizontal stretch (0<k<1) or horizontal compression (k>1), and <u>flip across the y-axis if "k" is negative</u>.

"d" shows horizontal shifts left (positive number) or right (negative number).

"c" shows vertical shifts up (positive) or down (negative).

The function f(x)=2|x-9|+3 has these transformations from the parent graph:

a = 2; Vertical stretch by a factor of 2

k = 1; No change

d = 9; Horizontal shift right 9 units

c = 3; Vertical shift up 3 units

Since neither "a" nor "k" was negative, there were no flips, <u>also known as reflections</u>.

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N=4; 2i and 3i are zeros <br> f(-1)=50
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Solution:- \text{Let f(x) be any nth degree polynomial with n=4}\\

\text{Given that 2i and 3i are the zeroes of f(x)}

\text{so (x-2i) and (x-3i) are factors of f(x)}

\text{Since 2i is a zero of f(x) then its conjugate -2i is also a zero of f(x)}

\Rightarrow(x+2i) \text{is a factor}\\\text{Similarly, conjugate of 3i is -3i is also a zero of f(x) }\\\Rightarrow(x+3i)\text{is a factor}\\\text{So , }\\f(x)=k(x-2i)(x+2i)(x-3i)(x+3i)\\=k(x^2-(2i)^2)(x^2-(3i)^2)\\=k(x^2-4i^2)(x^2-9i^2)\\=k(x^2+4)(x^2+9)\\=k(x^4+13x^2+36)\\\text{As given}\\f(-1)=50\\\Rightarrow k((-1)^4+13(-1)^2+36)=50\\\Rightarrow k(1+13+36)=50\\\Rightarrow k(50)=50\\\Rightarrow k=1\\\text{So by substituting k=1 in f(x) we get ,}\\f(x)=(x^4+13x^2+36)

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