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3 years ago

8 1/7 = 2 2/5 + m M=

Mathematics

1 answer:

DIA [1.3K]3 years ago

3 0

Step-by-step explanation:

multiply them

take m on one side

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!!!*BRAINLIEST*!!!

balu736 [363]

Answer:

B. 10,368 square yards

Step-by-step explanation:

120x90= 10800

24x18=432

10800-432= 10,368 square yards

7 0

2 years ago

Read 2 more answers

Write the slope-intercept form of the equation through (4, 2) perpendicular to LaTeX: y=4x-4

marin [14]

Answer:

y=4x-4

slope of the line=-4/-1

4/1

since perpendicular,

m1.m2=-1

4.m2=-1

m2=-1/4

The equation of line passing through the point (4,2)is

y-y1=m2(x-x1)

y-2=-1/4(x-4)

Step-by-step explanation:

i don't say that you have to mark my ans as brainliest but my friend if it has helped you a bit also don't forget to thank me.....

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3 years ago

Can someone help? If b/a=c/b=d/c=2/3, find a:b:c:d.

blagie [28]

Answer: a:b:c:d = 16:24:30:40

You can also divide the ratio by 2 because all terms have 2 in common. Which would be a:b:c:d = 8:12:15:20

8 0

2 years ago

Sally is baking cupcakes for her birthday party.

Katena32 [7]

Answer:

Where's the rest of the question?

5 0

2 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)

8 0

3 years ago