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tankabanditka [31]
1 year ago
5

Is it possible to draw a quaderilateral that is not a trapezoid and not a parallelogram

Mathematics
1 answer:
Scrat [10]1 year ago
5 0

Answer:

Yes

Step-by-step explanation:

A square is a quadlirateral.

A rectangle is a quadrilateral.

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Plz help me I really need it ASAP
vekshin1

Answer:

32

Step-by-step explanation:

\frac{4ab}{8c}

\frac{4(8)(4)}{8(\frac{1}{2})}

\frac{4(32)}{4}

32

8 0
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Round 3.255 to the nearest hundredth.
Oduvanchick [21]

Answer:

3.26

Step-by-step explanation:

if there is a five you round it up too 3.26

3 0
3 years ago
Please help me I don't get it 7 1/5−(−3/5)
Fudgin [204]

Answer:

39/5

Step-by-step explanation:

8 0
3 years ago
10 points, doing this one again because I got a wrong answer last time but regardless thank you for trying to help​
Alik [6]

Hey ! there

Answer:

  • Value of missing side i.e. TE is <u>1</u><u>2</u><u> </u><u>feet</u>

Step-by-step explanation:

In this question we are provided with a <u>right</u><u> </u><u>angle </u><u>triangle</u> having <u>TS </u><u>-</u><u> </u><u>35</u><u> </u><u>ft </u><u>and</u><u> </u><u>SE </u><u>-</u><u> </u><u>37</u><u> </u><u>ft </u>. And we are asked to find the missing side that is <u>TE </u>using Pythagorean Theorem .

<u>Pythagorean Theorem :</u> -

According to Pythagorean Theorem sum of squares of perpendicular and base is equal to square of hypotenuse in a right angle triangle i.e.

  • H² = P² + B²

<u>Where </u><u>,</u>

  • H refers to <u>Hypotenuse</u>

  • P refers to <u>Perpendicular</u>

  • B refers to <u>Base</u>

<u>Solution</u><u> </u><u>:</u><u> </u><u>-</u>

In the given triangle ,

  • Base = <u>TE </u>

  • Perpendicular = <u>TS </u><u>(</u><u> </u><u>35</u><u> </u><u>feet </u><u>)</u>

  • Hypotenuse = <u>SE </u><u>(</u><u> </u><u>37</u><u> </u><u>feet </u><u>)</u>

Now applying Pythagorean Theorem :

\quad \longmapsto \qquad \:SE {}^{2}  = TS {}^{2}  + TE {}^{2}

Substituting values :

\quad \longmapsto \qquad \:37 {}^{2}  = 35 {}^{2}  + TE {}^{2}

Simplifying it ,

\quad \longmapsto \qquad \:1369 = 1225  + TE {}^{2}

Subtracting 1225 on both sides :

\quad \longmapsto \qquad \:1369 - 1225  = \cancel{1225}  + TE {}^{2}  -  \cancel{1225}

We get ,

\quad \longmapsto \qquad \:144 = TE {}^{2}

Applying square root to both sides :

\quad \longmapsto \qquad \ \sqrt{ 144} =  \sqrt{TE {}^{2}}

We get ,

\quad \longmapsto \qquad \:     \red{\underline{\boxed{\frak{TE  = 12 \: feet}}}} \quad \bigstar

  • <u>Henceforth</u><u> </u><u>,</u><u> </u><u>value </u><u>of </u><u>missing </u><u>side </u><u>is </u><em><u>1</u></em><em><u>2</u></em><em><u> </u></em><em><u>feet </u></em><em><u>.</u></em>

<u>Verifying</u><u> </u><u>:</u><u> </u><u>-</u>

Now we are verifying our answer using Pythagorean Theorem . We know that according to Pythagorean Theorem ,

  • SE² = TS² + TE²

Substituting value of SE , TS and TE :

  • 37² = 35² + <u>1</u><u>2</u><u>²</u>

  • 1369 = 1225 + 144

  • 1369 = 1369

  • L.H.S = R.H.S

  • Hence , Verified .

<u>Therefore</u><u> </u><u>,</u><u> </u><u>our</u><u> answer</u><u> is</u><u> correct</u><u> </u><u>.</u>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
6 0
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I need help on number 17 and 18 on how to find the area of the figure
JulsSmile [24]
17. Area of triangle: 1/2bh
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18. Area of trapezoid: 1/2 h (b1 + b2)
1/2 * 8 ( 12 + 15.4)
4 * 27.4
= 109.6 cm^2
3 0
3 years ago
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