Mathematics: Find the measure of each acute angle. The measure of the top-left angle in the triangle is__º and the measure of the bottom-right angle is__º.

Find the measure of each acute angle. The measure

Ответ:

livirose1514

11.06.2021

Mathematics

the top angle is 48

the bottom right angle is 42

Step-by-step explanation:

The sum of the angles of a triangle is 180

90 + 8x+7x = 180

Combine like terms

90 +15x = 180

Subtract 90 from each side

15x = 180-90

15x = 90

Divide by 15

15x/15 =90/15

x = 6

8x = 8*6 = 48 so the top angle is 48

7x = 7*6 = 42 so the bottom angle is 42

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Ответ:

martinbricein10

11.06.2021

Mathematics

Solution given:

8x°+7x°+90°=180°[sum of interior angle of a triangle is 180°]

15x=180°-90°

x=90/15

x=6°

8x=8*6=48°

7x=7*6=42°

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Ответ:

elijaahstepp041

31.03.2020

Mathematics

Explicit formula for

1) a_n=17-3(n-1)

2) $a_n=(\frac{1}{2} )^{n-1}\cdot 20$

Step-by-step explanation:

W have to find the first few terms of the given sequence and then find an explicit equation

Explicit formula = f(n) + d(n-1) , where,

f(n) is first term,

d is common difference,

n-1 is one term less than the term number.

Given : $a_1=17\\\\a_{n+1}=a_n-3$

$\text{We put n =1, we get,}\\\\a_2=a_1-3=17-3=11=17-3(2-1)\\\\\\\text{We put n =2, we get,}\\\\\\a_3=a_2-3=11-3=8=17-3-3=17-3\cdot 2=17-3(3-1)\\\\\text{We put n =3, we get,}\\\\\\a_4=a_3-3=8-3=5=17-3-3-3=17-3\cdot 3=17-3(4-1)\\$

Thus, we obtained an explicit formula,

a_n=17-3(n-1)

Given : $a_1=20\\\\a_{n+1}=\frac{1}{2} \cdot a_n$

Thus, we obtained an explicit formula,

a_n=17-3(n-1)

$\text{We put n =1, we get,}\\\\\\a_2=\frac{1}{2} \cdot a_1=\frac{1}{2} \cdot 20=10=(\frac{1}{2} )^{2-1}\cdot 20\\\\\\\text{We put n =3, we get,}\\\\\\a_4=\frac{1}{2} \cdot a_3=\frac{1}{2} \cdot 5=\frac{5}{2}=(\frac{1}{2} )^{4-1}\cdot 20\\\\\\$

Thus, we obtained an explicit formula,

$a_n=(\frac{1}{2} )^{n-1}\cdot 20$

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Find the measure of each acute angle. The measure of the top-left angle in the triangle isº and the measure of the bottom-right angle isº.

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