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Aleks04 [339]
3 years ago
14

I need help with this.​

Mathematics
1 answer:
SOVA2 [1]3 years ago
6 0

Answer:

x + (3x - 9) = 5x - 30 \\ x + 3x - 9 = 5x - 30 \\ 4x - 9 = 5 x - 30 \\  - 9 + 30 = 5x - 4x \\ 21 = x \\ so..value \: of \: x \: is \: 21 \\ so. \\

<ACB = 105°

<h2>it is correct✅ ans. </h2>

❤❤❤❤❤❤❤❤❤

<h3>plazzz mark is as a brilliant ans</h3>
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What is the range of f(x) = 3x + 9?
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Anarel [89]
The range of a function is the value of y where it is defined
 Note the coordinates of the 3 points:

A(-2,0)
B(2,4)
C(3,-3)
As described the range (values of y) are 0,4 & -3
Hence Range ={ y ∈ -3 ≤ y ≤ 4}

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Semenov [28]

To find the value of q, we need to find d(-8). Put another way, we need to find the value of d(x) when x = -8

d(x) = -\sqrt{\frac{1}{2}x+4}

d(-8) = -\sqrt{\frac{1}{2}(-8)+4}

d(-8) = -\sqrt{-4+4}

d(-8) = -\sqrt{0}

d(-8) = 0

So this means q = 0. Note that -0 is just 0.

===========================================

The value of r will be a similar, but now we use f(x) this time.

Plug in x = 0

f(x) = \sqrt{\frac{1}{2}x+4}

f(0) = \sqrt{\frac{1}{2}*0+4}

f(0) = \sqrt{0+4}

f(0) = \sqrt{4}

f(0) = 2

Therefore, r = 2.

===========================================

For s, we plug x = 10 into f(x)

f(x) = \sqrt{\frac{1}{2}x+4}

f(10) = \sqrt{\frac{1}{2}*10+4}

f(10) = \sqrt{5+4}

f(10) = \sqrt{9}

f(10) = 3

So s = 3.

===========================================

Finally, plug x = 10 into d(x) to find the value of t

d(x) = -\sqrt{\frac{1}{2}x+4}

d(10) = -\sqrt{\frac{1}{2}(10)+4}

d(10) = -\sqrt{5+4}

d(10) = -\sqrt{9}

d(10) = -3

A shortcut you could have taken is to note how d(x) = -f(x), so this means

d(10) = -f(10) = -9 since f(10) = 9 was found in the previous section above.

Whichever method you use, you should find that t = -3.

===========================================

<h3>In summary:</h3><h3>q = 0</h3><h3>r = 2</h3><h3>s = 3</h3><h3>t = -3</h3>
8 0
3 years ago
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