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4 years ago

Between which two numbers is Negative two-thirds located on a number line?

Mathematics

1 answer:

klemol [59]4 years ago

5 0

Answer:-1 and 0

Step-by-step explanation:

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Fill in the blank to make equivalent rational expressions?

laiz [17]

The answer is -49 because you multiply by 7.

7 0

3 years ago

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Trigonometry

schepotkina [342]

The area of the triangle ABC is 207.5 square units.

Explanation:

The measurements of the sides of the triangle are $AB=19$ , $AC=24$ and $m\angle A=65^{\circ}$

We need to determine the area of the triangle ABC.

<u>Area of the triangle:</u>

The area of the triangle can be determined using the formula,

${Area}=\frac{1}{2} b c \sin A$

where $b=19$ , $c=24$ and $m\angle A=65^{\circ}$

Substituting these values in the above formula, we get,

${Area}=\frac{1}{2}(19)(24) \sin 65^{\circ}$

Simplifying the values, we get,

${Area}=\frac{1}{2}(456) (0.91)$

${Area}=\frac{1}{2}(414.96)$

${Area}=207.48$

Rounding off to the nearest tenth, we get,

${Area}=207.5$

Thus, the area of the triangle ABC is 207.5 square units.

5 0

3 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:

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