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Fofino [41]
3 years ago
5

Please help me with this

Mathematics
1 answer:
masya89 [10]3 years ago
4 0
X= 30

That’s the answer I think
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If anyone could help answer this question that would be great.
lys-0071 [83]
I believe the answer is 64
First break it off into 2 rectangles on is 4 width and 10 length other is 8 length and 10 width. Then multiple to find area 10•4=40 8•3=24 then add 40+24= 64
3 0
3 years ago
You have $5600. The best interest rate you can find is 3%
EleoNora [17]

Answer:

18 years

Step-by-step explanation:

The formula for computing accrued amount A for a principal of P at an interest rate of r(in decimal) compounded n times in a year for t years is given by

A = P(1 + \frac{r}{n})^{nt}

Note that r is percentage converted to decimal. So 3% = 3/100 = 0.03

We can rearrange the above equation to:

\frac{A}{P} = (1 + \frac{r}{n})^{nt}

Taking logs on both sides

log(\frac{A}{P}) = log(1 + \frac{r}{n})^{nt}

This gives

log(\frac{A}{P}) =nt \times log(1 + \frac{r}{n})\\So,\\nt = \frac{log(\frac{A}{P})}{ log(1 + \frac{r}{n})}

In this particular problem, n = 4, , A= 9600, P = 5600, r =0.03, so r/n = 0.03/4 = 0.0075

1 + r/n = 1+0.0075 = 1.0075

4t = log(9600/5600)/log(1.0075) = log(1.714) / log(1.0075) = 0.234 /0.00325 = 72

t = 72/4 = 18 years

4 0
2 years ago
How is an equation like a balance scale!?! please make the response short and easy to understand !!
Sauron [17]
You have to equal out each side to fine the value of the variable.

hope this helped!!!
can u mark me brainliest plz?
5 0
3 years ago
Find BC round to the nearest tenth
slega [8]

Answer:

6.7

Step-by-step explanation:

Here, we can apply cosine law, c^2=a^2+b^2-2abcos\theta

We the plug in values: (CB)^2+6^2+5^2-2*6*5*cos74

and solve that CB is around 6.7

3 0
3 years ago
An empty 5-gal water jug weighs 0.75 lb. With 3 c of water inside, the jug weighs 2.25 lb. Which equation models the jug's weigh
Deffense [45]
Weight of 1 c
3 c = 2.25 - empty bucket
3 c = 2.25 - 0.75
3 c = 1.5
c = 1.5/3 =1/2

Jug Weigh = (1/2)x +0.75
8 0
3 years ago
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