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Artemon [7]
3 years ago
11

Can someone please help me with this question ?

Mathematics
1 answer:
spayn [35]3 years ago
6 0

Answer:

  1350 square inches

Step-by-step explanation:

The area of the end trapezoid is ...

  A = (1/2)(b1 +b2)h = (1/2)(12 +36)(5) = 120 . . . . . square inches

The perimeter of the end trapezoid is ...

  P = 13 +12 +13 + 3×12 = 74 . . . . inches

so the total area of the rectangular surfaces (including the bottom) is ...

  (74 in)(15 in) = 1110 in²

The total area of the ramp is this rectangular area plus the two trapezoidal ends:

  total area = rectangle area + 2×trapezoid area

  = 1110 in² +2×120 in²

  total area = 1350 square inches

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15. Ten less the quotient of a number and 3 is 6.
coldgirl [10]

Answer:

(n/3)-10=6

Step-by-step explanation:

Let n be the number

Quotient of n and 3 is n/3

10 less that this is (n/3)-10=6

7 0
3 years ago
Which ordered pair is a solution of the equation y = x - 2?
Liula [17]

Answer:

One order pair is  

( 2,0)

Step-by-step explanation:

What ordered pairs are the options?

Pick a value for  

x

and solve for  

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If  

x

=

2

, then:

y

=

2

−

2

⇒

y

=

0

So we have  

(

2

,

0

)

If  

x

=

0

, then:

y

=

0

−

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⇒

y

=

−

2

Here we have  

(

0

,

−

2

)

You can simply use  

0

for both  

x

and  

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(intercept) to get the same answer.

8 0
3 years ago
A triangle has angles measuring 23 and 48 degrees. What is the measure of the third angle?
icang [17]

Answer:

109

Step-by-step explanation:

Hope this helped :)

6 0
3 years ago
Read 2 more answers
Please help me with this question.
Amiraneli [1.4K]
Newton is 762 Km east of Mayfield. 

I send you a file wich the answer. 

4 0
3 years ago
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In ΔABC, m∠B = m∠C. The angle bisector of ∠B meets AC at point H and the angle bisector of ∠C meets AB at point K. Prove that BH
solniwko [45]

Answer:

See explanation

Step-by-step explanation:

In ΔABC, m∠B = m∠C.

BH is angle B bisector, then by definition of angle bisector

∠CBH ≅ ∠HBK

m∠CBH = m∠HBK = 1/2m∠B

CK is angle C bisector, then by definition of angle bisector

∠BCK ≅ ∠KCH

m∠BCK = m∠KCH = 1/2m∠C

Since m∠B = m∠C, then

m∠CBH = m∠HBK = 1/2m∠B = 1/2m∠C = m∠BCK = m∠KCH   (*)

Consider triangles CBH and BCK. In these triangles,

  • ∠CBH ≅ ∠BCK (from equality (*));
  • ∠HCB ≅ ∠KBC, because m∠B = m∠C;
  • BC ≅CB by reflexive property.

So, triangles CBH and BCK are congruent by ASA postulate.

Congruent triangles have congruent corresponding sides, hence

BH ≅ CK.

5 0
3 years ago
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