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3 years ago

Given ABCD and D = 145, what is the measure of B? A. 145 B.35 C. 90 D. 235 E. 10 F. 55 DEGREES

Mathematics

2 answers:

kari74 [83]3 years ago

5 0

Don't think this can be answered without further info
Sorry bud

BartSMP [9]3 years ago

5 0

Probably, the answer is A. 145

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3 years ago

The area of a triangle is found by multiplying half of the measure of its base by its height. The area of the following right tr

GarryVolchara [31]

Answer:

$\sf Area\:of\:a\:triangle=\dfrac12 \times base \times height$

Given:

area = 18 cm²
base = 6 cm
height = (h + 1) cm

Substitute the given values into the equation and solve for h:

$\sf \implies 18 = \sf \dfrac12 \times 6 \times (h+1)$

$\sf \implies 18 = \sf \dfrac62 (h+1)$

$\sf \implies 18 =3(h+1)$

expand using distributive property of addition $a(b+c)=ab+ac$ :

$\sf \implies 18=3h+3$

subtract 3 from both sides:

$\sf \implies 3h=15$

divide both sides by 3:

$\sf \implies h=5$

<u>To verify</u>:

Half of the base is 3 cm.

If h=5, then the height of the triangle is 6 cm.

Multiplying 3 by 6 is 18.

This matches the given area, so we can verify that h = 5.

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2 years ago

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STALIN [3.7K]

Answer:

Step-by-step explanation:

$\dfrac{5+2\sqrt3}{7+4\sqrt3}\qquad\text{use}\ (a+b)(a-b)=a^2-b^2\\\\=\dfrac{5+2\sqrt3}{7+4\sqrt3}\cdot\dfrac{7-4\sqrt3}{7-4\sqrt3}=\dfrac{(5+2\sqrt3)(7-4\sqrt3)}{7^2-(4\sqrt3)^2}\qquad\text{use FOIL}\\\\=\dfrac{(5)(7)+(5)(-4\sqrt3)+(2\sqrt3)(7)+(2\sqrt3)(-4\sqrt3)}{49-4^2(\sqrt3)^2}\\\\=\dfrac{35-20\sqrt3+14\sqrt3-8(3)}{49-(16)(3)}=\dfrac{(35-24)+(-20\sqrt3+14\sqrt3)}{49-48}\\\\=\dfrac{11-6\sqrt3}{1}=11-6\sqrt3$

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3 years ago

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NARA [144]

13 m is the volume of the prism

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3 years ago

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olga2289 [7]

Answer:

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reason:

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3 years ago