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2 years ago

A rectangle that is 30 inches long by 10 inched wide can be divided into three equal squares.

Mathematics

1 answer:

Darya [45]2 years ago

6 0

Yes k. And? I don’t get how this is a question.

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Base area:12.5m^2;height:1.2m

lilavasa [31]

? Ummm What is the question???

8 0

3 years ago

? the coordinates of a, b, c, and d are a (-6, 1), b (-9, 4), c (-1, 1), and d (-7, 6). how are ab←→ and cd←→ related? they are

balandron [24]

AB :
(-6,1)(-9,4)
slope = (4 - 1) / (-9 - (-6) = 3 / -3 = - 1

CD:
(-1,1)(-7,6)
slope = (6 - 1) / (-7 - (-1) = -5/6

different slopes....slopes are not negative reciprocals...means they are not perpendicular, parallel, or coincident. These lines will intersect at one point.

4 0

3 years ago

Volume question

ioda

Answer:

The answer is 61.4 L

Step-by-step explanation:

Using the formula of cuboid, V = l×b×h, to find the total amount of water contained :

l = 120cm

b = 40cm

h = 15cm

V = 15×120×40

= 72 000cm³

= 72 L

It is given that she uses 10.6L for shower. Then calculate the remaining water by substracting 10.6L from the original amount of water :

remaining = 72 - 10.6

= 61.4 L

8 0

3 years ago

Which point is located at (-6, - 3) a b c or d

Kamila [148]

D is located at (-6, -3)!

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

5 0

3 years ago

Identify the standard form of the equation by completing the square.

OLEGan [10]

Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

Given equation:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

To complete the square for a hyperbola

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$

4 0

1 year ago