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2 years ago

You invested $100 at 8.2% which is compounded annually for 7 years. How much will your $100 be worth in 7 years?

Mathematics

1 answer:

____ [38]2 years ago

7 0

Answer:

57.4%

Step-by-step explanation:

8.2 × 7 = 57.4

hope this helps!!

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lyudmila [28]

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

★ ∆ ABC is similar to ∆DEF

★ Area of triangle ABC = 64cm²

★ Area of triangle DEF = 121cm²

★ Side EF = 15.4 cm

${\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}$

★ Side BC

${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

Since, ∆ ABC is similar to ∆DEF

[ Whenever two traingles are similar, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. ]

$\therefore \tt \boxed{ \tt \dfrac{area( \triangle \: ABC )}{area( \triangle \: DEF)} = { \bigg(\frac{BC}{EF} \bigg)}^{2} }$

❍ Putting the values, [Given by the question]

• Area of triangle ABC = 64cm²

• Area of triangle DEF = 121cm²

• Side EF = 15.4 cm

$\implies \tt \dfrac{64 \: {cm}^{2} }{12 \: {cm}^{2} } = { \bigg( \dfrac{BC}{15.4 \: cm} \bigg) }^{2}$

❍ By solving we get,

$\implies \tt \sqrt{\dfrac{{64 \: cm}^{2} }{ 121 \: {cm}^{2} }} = \bigg( \dfrac{BC}{15.4 \: cm} \bigg)$

$\implies \tt \sqrt{\dfrac{{(8 \: cm)}^{2} }{ {(11 \: cm)}^{2} }} = \bigg( \dfrac{BC}{15.4 \: cm} \bigg)$

$\implies \tt \dfrac{8 \: cm}{11 \: cm} = \dfrac{BC}{15.4 \: cm}$

$\implies \tt \dfrac{8 \: cm \times 15.4 \: cm}{11 \: cm} = BC$

$\implies \tt \dfrac{123.2 }{11 } cm = BC$

$\implies \tt \purple{ 11.2 \: cm} = BC$

Hence, BC = 11.2 cm.

${\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}$

★ Figure in attachment.

$\rule{280pt}{2pt}$

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Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F

Svet_ta [14]

Answer:

$A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36$

Step-by-step explanation:

The equations are:

$y = x^{2} + 2x + 3$

$y = 2x + 12$

The two graphs intersect when:

$x^{2} + 2x + 3 = 2x + 12$

$x^{2} = 0$

$x_{1} = 3\\x_{2} = -3$

To find the area under the curve for the first equation:

$A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

To find the area under the curve for the second equation:

$A_{2} = \int\limits^3__-3}{2x + 12} \, dx$

To find the total area:

$A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

Simplifying the equation:

$A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx$

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).

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