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3 years ago

11

I need an answer for this A.S.A.P if the answer is correct I will mark it brainliest, If you don't know it don't bother answerin

g thanks!

2 answers:

ki77a [65]3 years ago

6 0

I can see that this set of lines were perpendicular until that last line that split 4 and 3. We know that a combination of 4 and 3 is equal to 90°. Plus the angle next to 1 has a measure of 57°. Those angles have the same measure. So now we know what equation to solve.

$57 = 2x - 7\\57 + 7 = 2x - 7 + 7\\64 = 2x\\\frac{64}{2} = \frac{2}{2}x\\32 = x$

So now we know that x = 32.

natita [175]3 years ago

3 0

Answer:

x = 32.

Step-by-step explanation:

To solve this correctionly, I like use to make some points clear.

Opposite angles are equal. That is, angle 4 and 57° are the same.

If that be the case:

(2x - 7)° = 57°

2x = 57 + 7

2x = 64

x = 64/2

x = 32°

We can also solve it with another approach.

We know that angles on a straight line equals 180°

If that be the case, then:

Angles 3,4 and 5 are on a straight line.

Angle 3 = 33°

Angle 4 = (2x - 7)°

Angle 5 = 90°

Therefore,

90° + 33° + (2x - 7)° = 180°

123° + 2x - 7 = 180°

116° + 2x = 180

2x = 180° - 116°

2x = 64°

Divide through by 2

x = 64/2

x = 32°

Enjoy Maths.

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Basile [38]

Arithmetic sequences have a common difference between consecutive terms.

Geometric sequences have a common ratio between consecutive terms.

Let's compute the differences and ratios between consecutive terms:

Differences:

$0.4-0.2 = 0.2,\quad 0.6-0.4=0.2,\quad 0.8-0.6=0.2,\quad 1-0.8=0.2$

Ratios:

$\dfrac{0.4}{0.2}=2,\quad \dfrac{0.6}{0.4} = 1.5,\quad \dfrac{0.8}{0.6} = 1.33\ldots, \quad\dfrac{1}{0.8}=1.25$

So, as you can see, the differences between consecutive terms are constant, whereas ratios vary.

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3 years ago

ΔCDE is reflected over the x-axis. What are the vertices of ΔC'D'E' ?

sattari [20]

Answer:

B.

Step-by-step explanation:

When reflecting over the x-axis:

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7 0

2 years ago

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

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