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4 years ago

Please anser. please answer. please answer. thank you.

Mathematics

1 answer:

Agata [3.3K]4 years ago

8 0

A) after 5 years she will pay $7,200
b) she will save $2880

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Use a number line to add 3 + (–5).

Sonja [21]

Answer:

3 plus -5 means you go backwards from +3 to -2 thats your answer

Step-by-step explanation:

8 0

3 years ago

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You roll a standard number cube. Find p(number greater than 2)

Anna35 [415]

A number cube contains the numbers 1-6. Without counting the numbers that are less than two, you will get 3, 4, 5, and 6.

7 0

3 years ago

Analyze the diagram below and complete the instructions that follow.

Jlenok [28]

Step-by-step explanation:

$scale \: factor = \frac{7}{28} = \frac{1}{4} \\$

$ratio \:of \:surface \:areas\\=\frac{2[5\times 7 + 7\times 3+3\times 5]}{2[20\times 28+ 28\times 12+12\times 20] }\\\\=\frac{[35 + 21+15]}{[560+ 336+240] }\\\\=\frac{71}{1136}\\\\=\frac{1}{16}\\\\= 1\: :\: 16$

3 0

3 years ago

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)

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3 years ago

A sea turtle is 3 feet below the surface of the sea. If its position can be recorded as - 3 feet, what would the position 0 repr

svp [43]

I am thinking it’s 1

5 0

3 years ago

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