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3 years ago

1. Find the simple interest.

Mathematics

2 answers:

telo118 [61]3 years ago

5 0

Correct this is correct

lions [1.4K]3 years ago

4 0

Answer:

Step-by-step explanation:

2 + (9.2)−8x = 2.32. Solution: ... 3. Which will be worth more in 10 years: $10,000 invested at 8.2% simple ... For an account with an annual interest rate of 6%, find the annual percentage yield ... Solution: This is the “remaining balance” entry that would be at the end of the.

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I need help on this question <br> Factor by grouping.<br> 2u^3+3u^2+14u+21

Andru [333]

U^2(2u+3)+7(2u+3)

(u^2+7)(2u+3)

6 0

3 years ago

Expand and simplify 9a+3(8-2a).

olchik [2.2K]

Hi there

9a + 3(8-2a)
9a + 24 - 6a
(9a -6a) +24
3x + 24

I hope that's help !

3 0

3 years ago

Read 2 more answers

What is the simplified form of the following expression? 2√27+√12-3√3-2√12

LiRa [457]

Answer:

√3

Step-by-step explanation:

The given expression to be simplified is

$2 \sqrt{27}+\sqrt{12}-3\sqrt{3}-2 \sqrt{12}$

but

$2 \sqrt{27}=2\sqrt{9 \times 3}=2 \times \sqrt{9}\times \sqrt{3}=2 \times 3 \times\sqrt{3}=6 \sqrt{3}$

$\sqrt{12}=\sqrt{4\times3}= \sqrt{4}\times \sqrt{3}=2\sqrt{3}$

Since √12=2√3,this implies that,

$2\sqrt{12}=2\times2\sqrt{3}=4 \sqrt{3}$

Therefore,

$2 \sqrt{27}+\sqrt{12}-3\sqrt{3}-2 \sqrt{12}=6\sqrt{3}+2\sqrt{3}-3 \sqrt{3} -4\sqrt{3}$

$=(6+2-3-4)\sqrt{3}$

$=\sqrt{3}$

The simplified form of ,

$2 \sqrt{27}+\sqrt{12}-3\sqrt{3}-2 \sqrt{12}$ is √3

8 0

3 years ago

Identify the point from the equation y+2=5(x-5)

Oxana [17]

Answer:

use desmos :)) it helps

Step-by-step explanation:

8 0

3 years ago

Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).

andre [41]

Check the picture below. So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}$

7 0

4 years ago