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alexgriva [62]
3 years ago
6

Q6 (i) please i don’t know what i’m doing wrong

Mathematics
2 answers:
UkoKoshka [18]3 years ago
6 0

Answer:

x = 4

Step-by-step explanation:

Square both sides

x+5 = 25-10sqrt(x)+x

20 = 10sqrt(x)

sqrt(x) = 2

Square both sides again

x=4

Verify sol

x = 4

Dennis_Churaev [7]3 years ago
5 0
Start by squaring the numbers on both sides to get rid of the square root on both the left and right sides to get:

x + 5 = 5^2 – x

x + 5 = 25 – x

Then rearrange to make x the subject.

x + 5 = 25 – x

2x + 5 = 25

2x = 20

x = 10
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Answer:

Black = 84 cm^2

Blue = 84cm^2

gold = 60.2cm^2

1 ornament = 228.2 cm^2

100 ornaments - 22820 cm^2

Step-by-step explanation:

Black is a rectangle = A = l*w = 7*12 = 84 cm^2

We have 2 blue   This is a triangle  so multiply by 2  the area of a triangle

2 * A = 2(1/2 b*h) = bh = 12*7 = 84 cm^2

We have 2 gold   This is a triangle  so multiply by 2  the area of a triangle

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For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant
nirvana33 [79]

Answer:

The instantaneous rate of change of y with respect to x at the value x = 3 is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of y with respect to x over the interval by definition of secant line:

r = \frac{y(b) -y(a)}{b-a} (1)

Where:

a, b - Lower and upper bounds of the interval.

y(a), y(b) - Function exaluated at lower and upper bounds of the interval.

If we know that y = 3\cdot x^{2}, a = 3 and b = 6, then the average rate of change of y with respect to x over the interval is:

r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}

r = 27

The average rate of change of y with respect to x over the interval [3,6] is 27.

b) The instantaneous rate of change can be determined by the following definition:

y' =  \lim_{h \to 0}\frac{y(x+h)-y(x)}{h} (2)

Where:

h - Change rate.

y(x), y(x+h) - Function evaluated at x and x+h.

If we know that x = 3 and y = 3\cdot x^{2}, then the instantaneous rate of change of y with respect to x is:

y' =  \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}

y' =  3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}

y' = 3\cdot  \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}

y' = 6\cdot  \lim_{h \to 0} x +3\cdot  \lim_{h \to 0} h

y' = 6\cdot x

y' = 6\cdot (3)

y' = 18

The instantaneous rate of change of y with respect to x at the value x = 3 is 18.

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Whats the answer and show work <br> -7x-2=24-9x
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