Ask question

Ask question

All categories

1 year ago

13

Please help quickly. Find the measure of the unknown angle

2 answers:

Vladimir [108]1 year ago

5 0

Answer:

65

Step-by-step explanation:

yessss

Natalija [7]1 year ago

4 0

Answer:

the measure of the unknown angle is 31 degrees. 180 - 149 = 31.

Step-by-step explanation:

hope this helps

You might be interested in

"Three times a number increased by eleven is seventeen" translates to which of the following equations?

ololo11 [35]

3x + 11 = 17 i think

7 0

2 years ago

Read 2 more answers

Theresa Morgan, a chemist, has a 30% hydrochloric acid solution and a 50% hydrochloric acid solution. How many liters of each sh

Bogdan [553]

Let x represent the number of liters of 50% acid Theresa puts into the mix. The the number of liters of 30% acid will be (420-x). The total amount of acid in the final solution will be ...
0.50x + 0.30(420-x) = 0.45(420)
0.20x + 126 = 189 . . . . . . . . . . . . . . . simplify
0.20x = 63 . . . . . . . . . . . . . . . . . . . . . subtract 126
x = 63/0.20 = 315 . . . . . . . . . . . . . . . liters of 50% solution
(420-x) = 420-315 = 105 . . . . . . . . . liters of 30% solution

Theresa should mix ...
105 liters of 30% solution
315 liters of 50% solution

7 0

2 years ago

$35 to spend cake is $18 balloons are $4

Lera25 [3.4K]

Answer:

he spends $22 and has $13left

Step-by-step explanation:

3 0

2 years ago

Find the measure of each angle indicated. Round to the nearest tenth

MArishka [77]

Answer:

1. A 2.C

Step-by-step explanation:

8 0

1 year ago

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference

allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

$\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}$

so that

$a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d$

$a_n=a_1+(n-1)d$

$33=8+(n-1)d$

$21=(n-1)d$

$n$ has to be an integer, which means there are 4 possible cases.

Case 1: $n-1=1$ and $d=21$ . But

$\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123$

Case 2: $n-1=21$ and $d=1$ . But

$\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123$

Case 3: $n-1=3$ and $d=7$ . But

$\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123$

Case 4: $n-1=7$ and $d=3$ . But

$\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123$

8 0

2 years ago