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blagie [28]
3 years ago
8

The area of a rectangle with base b and height h is bh. A rectangular sports goal is 10 ft wide and 8 ft high. Find the area of

the opening.
Mathematics
1 answer:
Naya [18.7K]3 years ago
5 0

Answer:

80 (ft)^{2}

Step-by-step explanation:

As stated in the question, the area of a rectugle is its base by its height.

A = b*h = 10ft * 8 ft = 80 (ft)^2

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The base edge of an oblique square pyramid is represented as x cm. If the height is 9cm, what is the volume of the pyramid in te
Tom [10]
A is correct:

Let volume v, length l, and h height.

v= (l²h)/3

v=(x²9)/3

v=3x²
5 0
3 years ago
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SOS if anyone could help that would be appreciated!!!❤️
Anastaziya [24]

Answer:

See below

Step-by-step explanation:

2e)15-4x=10x+45

Solve:

15-4x=10x+45\\-4x-10x=45-15\\-14x=30\\ \boxed {x=-\dfrac{15}{7}}3(x+1)=21

Check:

15-4 \left(-\dfrac{15}{7} \right)=10\left(-\dfrac{15}{7} \right)+45

15+\dfrac{60}{7} \right)=-\dfrac{150}{7} +45

\dfrac{210}{7}  =30

30=30

3a) 3(x+1)=21

Solve:

3(x+1)=21\\3x+3=21\\3x=18\\x=6

3b) 2+3(x+1)=6x

2+3x+3=6x\\5=3x\\

x=\dfrac{5}{3}

3c) -2(2x-3)=-2

Solve:

-2(2x-3)=-2\\-4x+6=-2\\-4x=-8\\x=2

3 0
3 years ago
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.

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Answer:

231.12

Step-by-step explanation:

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42 can be

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