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3 years ago

A rectangle has vertices a(1,2), B(1,7), c(4,7), d(4,2). what is the perimiter of the rectangle

Mathematics

2 answers:

aliina [53]3 years ago

7 0

Answer:

Step-by-step explanation:

You see, if you were to plot these points, (1, 2) and (1, 7) would be 5 points away from each other. If you were to plot the points (4, 2) and (4, 7) those points would also be 5 points away from each other. Now, (1, 2) and (4, 2) are 3 points apart as well as (1, 7) and (4, 7). Therefore, 5 + 5 = 10, 3 + 3 = 6, and finally, 10 + 6 = 16!

mote1985 [20]3 years ago

6 0

Answer:

Step-by-step explanation:

as you can see on the picture, perimeter of the rectangle is 5 + 5 + 3 + 3 = 16

or second practical solving way ;

B(1,7) - A(1,2) = 5

C(4,7) - D(4,2) = 5

C(4,7) - B(1,7) = 3

D(4,2) - A(1,2) = 3

5 + 5 + 3 + 3 = 16

Hope this helps ^-^

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In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.

konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

$BC^2=BD\cdot AB.$

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

$x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.$

Take positive value x. You get

$AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.$

2. According to the previous theorem,

$AC^2=AD\cdot AB.$

Then

$AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

Answer: $AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.$

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

$CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D$

This means that you cannot find solutions of this equation. Then CD≠2 cm.

8 0

3 years ago

Read 2 more answers

Suppose f(x) = x². Find the graph of f(x+3). ???

alexdok [17]

Answer: x^2+3

Step-by-step explanation:

If regular f(x) is equal to x^2, then if you just add 3 to the original, you add 3 to the same x^2. The three becomes a shift upwards on the graph. You put your vertex point up 3 on the vertical line then make your parabola

4 0

3 years ago

PLEASE HELP ASAP!!! (ii only)

mixer [17]

The equation of the tangent line can be found using slope formula and fact that slope = dy/dx.

$\frac{dy}{dx} = \frac{y - y_0}{x - x_0}$

where (x0,y0) is point A ---> (0,-1)

Then rearrange equation into slope-intercept form:
$y = \frac{dy}{dx}(x - 0) -1$

Finally sub in x=0 into dy/dx to get slope at the point (0,-1)

$y = 3x -1$