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3 years ago

Express in standard form 15/65

Mathematics

1 answer:

Naily [24]3 years ago

6 0

The correct answer is 3/13. Divided both by 5.
15/65
3/13

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PLS HELP I WILL GIVE BRAINLIEST

BaLLatris [955]

Answer:

81°

Step-by-step explanation:

By the outside angle theroem, we know that 2 * angle APC is differece of arcs DC and AB.

Angle APC = 27° so the difference is 54°

Arc DC = 27°, so Arc AB = 27° + 54° or 81°.

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3 years ago

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IgorC [24]

Answer: x is, 2 6 8 10 20-- i think

Step-by-step explanation: y is, 2 3 4 6 10

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3 years ago

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5(m+2)-m=2m+3<br><br>Please I need help

Blababa [14]

Answer:

m=-7/2

Step-by-step explanation:

(5 • (m + 2) - m) - (2m + 3) = 0

2m + 7 = 0

3.1 Solve : 2m+7 = 0

Subtract 7 from both sides of the equation :

2m = -7

Divide both sides of the equation by 2:

m = -7/2 = -3.500

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2 years ago

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Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

3 years ago

16. Which of the following are equivalent to(x) 16xA) g(r) 8.21B) g(c) 4096.16-3g (x) -4.4xD) g(x) 0.0625-16'+1E) g(x) 32.16E) g

swat32

F(x) = 16ˣ

A. g(x) = 8(2ˣ)
g(x) = (2³)(2ˣ)
g(x) = 2ˣ⁺³
The answer is not A.

B. g(x) = 4096(16ˣ⁻³)
g(x) = (16³)(16ˣ⁻³)
g(x) = 16ˣ
The answer is B.

C. g(x) = 4(4ˣ)
g(x) = 4ˣ⁺¹
The answer is not C.

D. g(x) = 0.0625(16ˣ⁺¹)
g(x) = (16⁻¹)(16ˣ⁺¹)
g(x) = 16ˣ
The answer is D.

E. g(x) = 32(16ˣ⁻²)
g(x) = (2⁵)(2⁴ˣ⁻⁸)
g(x) = 2(⁴ˣ⁻³)
The answer is not E.

F. g(x) = 2(8ˣ)
g(x) = 2(2³ˣ)
g(x) = 2³ˣ⁺¹
The answer is not F.

The answer is B and D.

6 0

3 years ago