The answer is _________
9*4=36
9*7=63
7*4=28
9*4=36
9*7=63
7*4=28
Now add the numbers all up. You get <span>254, </span>
Answer:
B. $2862
Step-by-step explanation:
Using n=5 in the given equation, we get ...
A(5) = 2700 + (5-1)(.015·2700) = 2700 +4(40.50)
A(5) = 2862.00
In year 5, you will have $2862 in the account.
_____
<em>Comment on the given equation</em>
The given equation tells you the amount in the account at the <em>beginning</em> of the year, before it earns any interest. Since that is the equation given, we presume that is the answer desired. In most "account balance" problems, you are interested in the amount at the <em>end</em> of the interest-earning period.
Answer:
Rhombus
Step-by-step explanation:
Mark as Brainllest
Ok , lots of questions=lower standard of explanation
just answers
5. 8.5 times 10^12
6. 0.001260-7,003,000
7.3.843 times 10^4
9. x=1
10. 7^(x+5)=49^(x+3)
11. 1.03
12. 25,700(0.85)^x
13. 10a√a
14.
![2 x^{2} y^{4} \sqrt[3]{5x^{2}}](https://tex.z-dn.net/?f=2%20x%5E%7B2%7D%20%20y%5E%7B4%7D%20%20%5Csqrt%5B3%5D%7B5x%5E%7B2%7D%7D%20)
15.6√3
16.

17.
![4 \sqrt[3]{x^{2}}](https://tex.z-dn.net/?f=%204%20%5Csqrt%5B3%5D%7Bx%5E%7B2%7D%7D%20)
18.

19.√109
20. (8,(3/2))
21. 5 units
Answer:
Step-by-step explanation:
The diagram of the triangles are shown in the attached photo.
1) Looking at ∆AOL, to determine AL, we would apply the sine rule
a/SinA = b/SinB = c/SinC
21/Sin25 = AL/Sin 105
21Sin105 = ALSin25
21 × 0.9659 = 0.4226AL
AL = 20.2839/0.4226
AL = 50
Looking at ∆KAL,
AL/Sin55 = KL/Sin100
50/0.8192 = KL/0.9848
50 × 0.9848 = KL × 0.8192
KL = 49.24/0.8192
KL = 60
AK/Sin25 = AL/Sin 55
AKSin55 = ALSin25
AK × 0.8192 = 0.4226 × 50
AK = 21.13/0.8192
AK = 25.8
2) looking at ∆AOC,
Sin 18 = AD/AC = 18/AC
AC = 18/Sin18 = 18/0.3090
AC = 58.25
Sin 85 = AD/AB = 18/AB
AB = 18/Sin85 = 18/0.9962
AB = 18.1
To determine BC, we would apply Sine rule.
BC/Sin77 = 58.25/Sin85
BCSin85 = 58.25Sin77
BC = 58.25Sin77/Sin85
BC = 58.25 × 0.9744/0.9962
BC = 56.98