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Elenna [48]
3 years ago
6

What is the area of a parallelogram that has the length of 3.5cm and the height of 2.3cm?

Mathematics
1 answer:
lianna [129]3 years ago
6 0
Area of parallelogram = bh

3.5 x 2.3
= 8.05cm
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Please help me..............................
myrzilka [38]

Answer:

y = 12

Step-by-step explanation:

∵ m∠ABC = 40

∵ BD ray is the bisector of ∠ABC

∴ m∠ABD = m∠CBD = 20

In 2 Δ BAD and BCD:

∵ m∠BAD = m∠BCD = 90°

∵ m∠ABD = m∠CBD = 20°

∵ BD is a common side in the two triangles

∴ ΔABD congruent to ΔCBD⇒(AAS)

∴ AD = CD

∴ 3y + 6 = 5y - 18

∴ 5y - 3y = 6 + 18

∴ 2y = 24

∴ y = 24/2 = 12

6 0
3 years ago
Suppose the base and height of a triangle sum to 18 cm, and the area of the triangle is 36 square cm. if the base is larger than
pentagon [3]
<span>15-h)*h=40
15h-h^2=40
h^2-15h+40=0
solve for h by quadratic formula:

a=1, b=-15, c=20
ans:
h=3.47 or 11.53 cm (height)
b=15-h=11.53 or 3.47 cm (base)</span>
7 0
3 years ago
20-10 write four equivalent expressions
igor_vitrenko [27]
Here's two. I can't think of any more.


20/10
20-(-10)
5 0
3 years ago
Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y
katrin2010 [14]

Answer:

The vertices are (3 , -5) , (-5 , -5)

The foci are (4 , -5) , (-6 , -5)

Step-by-step explanation:

* Lets study the equation of the hyperbola

- The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the x-axis is

  (x - h)²/a² - (y - k)²/b² = 1

- The length of the transverse axis is 2 a

- The coordinates of the vertices are  (h  ±  a  ,  k)

- The coordinates of the foci are (h ± c , k), where c² = a² + b²

- The distance between the foci is  2c

* Now lets solve the problem

- The equation of the hyperbola is (x + 1)²/16 - (y + 5)²/9 = 1

* From the equation

# a² = 16 ⇒ a = ± 4

# b² = 9 ⇒ b = ± 3

# h = -1

# k = -5

∵ The vertices are (h + a , k) , (h - a , k)

∴ The vertices are (-1 + 4 , -5) , (-1 - 4 , -5)

* The vertices are (3 , -5) , (-5 , -5)

∵ c² = a² + b²

∴ c² = 16 + 9 = 25

∴ c = ± 5

∵ The foci are (h ± c , k)

∴ The foci are (-1 + 5 , -5) , (-1 - 5 , -5)

* The foci are (4 , -5) , (-6 , -5)

4 0
3 years ago
Read 2 more answers
Determine the equation of the line that passes through the given points. (If you have a graphing calculator, you can use the tab
expeople1 [14]

<span>
You can write the equation in point-slope form, which has the format <em>y-y</em>subscript1=<em>m</em>(<em>x-x</em>subscript1), with <em>y</em>subscript1 and <em>x</em>subscript1 being the y and x coordinates for a point on the line, and <em>m</em> being the slope. </span>

<span /><span>Substitute a y and x coordinate into the equation so you have <em>y</em>-6=<em>m</em>(<em>x</em>-2)</span>

<span /><span><span>Then find the slope so you can replace <em>m</em>. The slope formula is <em />(<em>y</em>subscript2-<em>y</em>subscript1)/(<em>x</em>subscript2-<em>x</em>subscript1). </span><span>Substitute the coordinates in so you have <em>m</em>=(16-6)/(4-2), which simplifies to 10/2 and then 5.</span></span>

<span><span /></span><span>Now the equation is <em>y</em>-6=5(<em>x</em>-2)</span>

<span />If you want a different form, for example slope-intercept form, you can change it to that:

<span><em>y</em>-6=5(<em>x</em>-2)</span>

<span><em>y</em>=5x-4</span>

6 0
3 years ago
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