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3 years ago

8

Which number goes in the box to make this a true sentence? 65 × 32 = (65 × 30) + (65 × ) PLZZ HELP!

1 answer:

lutik1710 [3]3 years ago

7 0

The answer would be 2.

65 x 32 = 2080.

(65 × 30) = <span>1950</span><span>
(65 × 2) = 130

1950 + 130 = 2080

To make this easier, you can 30 from 32.</span>

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Tamira invests $5,000 in an account that pays 4% annual interest. How much will there be in the account after 3 years if the int

Talja [164]

Answer:

There will be $5624.32 in the account after 3 years if the interest is compounded annually.

There will be $5630.812 in the account after 3 years if the interest is compounded semi-annually.

There will be $5634.125 in the account after 3 years if the interest is compounded quarterly.

There will be $5636.359 in the account after 3 years if the interest is compounded monthly

Step-by-step explanation:

Tamira invests $5,000 in an account

Rate of interest = 4%

Time = 3 years

Case 1:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 1

Formula : $A=P(1+r)^t$

$A=5000(1+0.04)^3$

A=5624.32

There will be $5624.32 in the account after 3 years if the interest is compounded annually.

Case 2:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 2

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

$A=5000(1+\frac{0.04}{2})^{2 \times 3}$

A=5630.812

There will be $5630.812 in the account after 3 years if the interest is compounded semi-annually.

Case 3:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 4

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

$A=5000(1+\frac{0.04}{4})^{4 \times 3}$

A=5634.125

There will be $5634.125 in the account after 3 years if the interest is compounded quarterly.

Case 4:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 4

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

$A=5000(1+\frac{0.04}{12})^{12 \times 3}$

A=5636.359

There will be $5636.359 in the account after 3 years if the interest is compounded monthly

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2 years ago

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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

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3 years ago