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Lera25 [3.4K]
3 years ago
12

On a boat, a cabin's window is in the shape of an isosceles trapezoid, as shown below. What is the area of the window?

Mathematics
1 answer:
brilliants [131]3 years ago
6 0

Answer:

195

Step-by-step explanation:

½(B¹ + B²) h is the formula for a trapezoid so you would do

1/2(10+16)15 which gives you 195

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Determine the slope of the line that passes through the points (8, 0) and (-3, 9)
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Answer:

-9/11

Step-by-step explanation:

The slope of the line is given by

m = (y2-y1)/(x2-x1)

   = (9-0)/(-3-8)

  = 9/-11

  = -9/11

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3 years ago
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Without using a calculator, fill in the blanks with two consecutive integers to complete the following inequality.
Keith_Richards [23]

Answer:

impossible for me

Step-by-step explanation:

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What will be the linear equation for the shown table?<br><br>​
vladimir1956 [14]

Answer:

y=2x+5

Step-by-step explanation:

To find a linear equation you must find the slope and y intercept to create an equation in y=mx+b form (slope-intercept form)

So slope is = y2-y1 / x2-x1 [2 and 1 are subscripts] so 11-7/3-1 so m=4/2 = 2

Slope = 2, now find b

So using y = mx + b, substitute what you know so 7 = 2(1) + b so b=7-2 = 5

Put it all together y = 2x + 5

3 0
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Help with math pls Question: Which ratio tables are correct? (Also: Select *all* that apply) Which means there is more than one
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3 years ago
Two isosceles triangles share the same base. Prove that the medians to this base are collinear. (There are two cases in this pro
Anna71 [15]

Answer:

Given: Two Isosceles triangle ABC and Δ PBC having same base BC. AD is the median of Δ ABC and PD is the median of Δ PBC.

To prove: Point A,D,P are collinear.

Proof:

→Case 1.  When vertices A and P are opposite side of Base BC.

In Δ ABD and Δ ACD

AB= AC   [Given]

AD is common.

BD=DC  [median of a triangle divides the side in two equal parts]

Δ ABD ≅Δ ACD [SSS]

∠1=∠2 [CPCT].........................(1)

Similarly, Δ PBD ≅ Δ PCD [By SSS]

 ∠ 3 = ∠4 [CPCT].................(2)

But,  ∠1+∠2+ ∠ 3 + ∠4 =360° [At a point angle formed is 360°]

2 ∠2 + 2∠ 4=360° [using (1) and (2)]

∠2 + ∠ 4=180°

But ,∠2 and ∠ 4 forms a linear pair i.e Point D is common point of intersection of median AD and PD of ΔABC and ΔPBC respectively.

So, point A, D, P lies on a line.

CASE 2.

When ΔABC and ΔPBC lie on same side of Base BC.

In ΔPBD and ΔPCD

PB=PC[given]

PD is common.

BD =DC [Median of a triangle divides the side in two equal parts]

ΔPBD ≅ ΔPCD  [SSS]

∠PDB=∠PDC [CPCT]

Similarly, By proving ΔADB≅ΔADC we will get,  ∠ADB=∠ADC[CPCT]

As PD and AD are medians to same base BC of ΔPBC and ΔABC.

∴ P,A,D lie on a line i.e they are collinear.




3 0
3 years ago
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