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Irina-Kira [14]
3 years ago
14

HELP ME PLZ! I will not accept non sense answers. BRAINLIEST will be given to the person who gets it correct with full solutions

.

Mathematics
1 answer:
Mariana [72]3 years ago
7 0
<h3>Answer: Choice B.  </h3>

A = 1500(1.00167)^{12t}

=================================

Explanation:

You could use the formula

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

to plug in P = 1500, r = 0.02 and n = 12 to get the answer above. This is the compound interest formula. The interest rate r is divided into n = 12 to account for monthly compounding. The original exponent t turns into 12t. So for instance, if t = 2 years go by, then 12*t = 12*2 = 24 months have gone by.

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Connor is painting a superhero with this belt buckle shown. What is the area of the belt buckle in square Inches? 9 inches 2.5 i
antoniya [11.8K]

Answer:

87.5 square inches

Step-by-step explanation:

The belt knuckle as shown in the diagram above consist of a 4 rectangles (2 are the of the same dimension, the other two are of the same direction differently) and a rectangle.

Area of the belt knuckle = area of the 4 triangles + area of the rectangle

✔️Area of the 2 rectangles with the following dimensions:

base (b) = 9 inches

height (h) = 2.5 inches

Area of the two triangles = 2(½*b*h)

= 2(½*9*2.5)

= 22.5 inches²

✔️Area of the 2 rectangles with the following dimensions:

base (b) = 4 inches

height (h) = 5 inches

Area of the two triangles = 2(½*b*h)

= 2(½*4*5)

= 20 inches²

✔️Area of the rectangle = l*w

l = 9 inches

w = 5 inches

Area = 9*5 = 45 inches²

✔️Area of the belt knuckle = 22.5 + 20 + 45 = 87.5 square inches

3 0
3 years ago
HELP ME QUICK PLEASEEEE solve for x: 2/x-2+7/x^2-4=5/x
Anna35 [415]

 

\displaystyle\\\\\frac{2}{x-2}+\frac{7}{x^2-4}=\frac{5}{x}\\\\\frac{^{x+2)}2~~~~~}{x-2}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2(x+2)}{(x-2)(x+2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2x+4+7}{x^2-4}=\frac{5}{x}\\\\\frac{2x+11}{x^2-4}=\frac{5}{x}\\\\5(x^2-4)=x(2x+11)\\\\5x^2-20=2x^2+11x\\\\5x^2-2x^2 -11x-20=0


\displaystyle\\3x^2-11x-20=0\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+4\cdot3\cdot20}}{2\cdot3}=\\\\=\frac{11\pm\sqrt{121+240}}{6}=\frac{11\pm\sqrt{361 }}{6}=\frac{11\pm19}{6}\\\\x_1=\frac{11-19}{6}=\frac{-8}{6}\\\\\boxed{\bf x_1=-\frac{4}{3}}\\\\x_2=\frac{11+19}{6}=\frac{30}{6}\\\\\boxed{\bf x_2=5}




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I need to find what m
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Answer:

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Step-by-step explanation:

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