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3 years ago

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Find the values of x and y in the parallelogram when AB=x^2-18, CD=7x, mD=(5y+31), mB=96

1 answer:

Fynjy0 [20]3 years ago

8 0

Greetings, I Am BrotherEye

Answer:

AP = CP, BP = DP; sample answer: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram, so if AP = CP and BP = DP, then the string forms a parallelogram. ALGEBRA Find x and y so that the quadrilateral is a parallelogram. BP = DP, then the string forms a parallelogram.

Step-by-step explanation:

Answer: B

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I NEED HELP ASAP<br> QUESTION BELOW

Gekata [30.6K]

Answer:

Circumference ≈ 50.27

Area ≈ 201.06

Step-by-step explanation:

The circumference of a circle is $2\pi r$ while the area of the circle $\pi r^{2}$

Since the diameter is the radius multiplied by 2, the radius of the circle is 8. So substituting 8 for r, the circumference is 2 × $\pi$ × 8, which is approximately 50.27 (rounded to the nearest hundredth) and the area of the circle would be $\pi$ × 8². 8² is 64 so 64 times pi is approximately 201.06 (rounded to the nearest hundredth)

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1 year ago

Given 2x - 9y - 12 = 0, express 5x - 4y + 7 = 0 in terms of x only

olga2289 [7]

Answer:

$4 \frac{1}{9}x$ + $12\frac{1}{3}$ = 0

Step-by-step explanation:

Based on 2x - 9y - 12 = 0

-9y = 12 - 2x

9y = -12 + 2x

y = $-1\frac{1}{3} + \frac{2}{9} x$

Therefore...

5x - 4y + 7 = 0

5x - 4( $-1\frac{1}{3} + \frac{2}{9} x$ ) + 7 = 0

5x + $5\frac{1}{3}$ - $\frac{8}{9} x$ + 7 = 0

$4 \frac{1}{9}x$ + $12\frac{1}{3}$ = 0

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2 years ago

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grandymaker [24]

Answer:

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Step-by-step explanation:

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2 years ago

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Find the total area the regular pyramid.<br><br> T. A. =

lorasvet [3.4K]

Answer:

T.A. = 16√3 units²

Step-by-step explanation:

∵ The total area = the area of the four faces

∵ The four faces are equilateral triangles with side length 4

∵ Area of the equilateral Δ = 1/4 s² √3

∴ T.A. = 4 × 1/4 × 4² × √3 = 16√3 units²

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2 years ago

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ValentinkaMS [17]

Answer:

Step-by-step explanation:

Given

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i.e. distance from x axis is $b\sin (at)---1$

Distance from y axis at

velocity is given by $v=\frac{\mathrm{d} r}{\mathrm{d} t}$

$v=a\hat{i}+ba\cos (at)\hat{j}$

Similarly acceleration is given by

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

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Magnitude of acceleration is $=\sqrt{(-a^2b\sin (at))^2}$

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From 1 and 2 we can see that

Magnitude of acceleration is proportional to distance from x axis

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