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expeople1 [14]
2 years ago
12

Usha deposited ₹1200 in a finance company which pays 3

title="\frac{1}{2}" alt="\frac{1}{2}" align="absmiddle" class="latex-formula"> % interest per year.
Find the amount she is expected to get after 4 years.
Mathematics
1 answer:
arsen [322]2 years ago
6 0

Answer:

4968

Step-by-step explanation:

using simple interest method

A=P(1+i)n

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16. Peter was told by his teacher that he needed to measure the height of the building in which his
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Answer:

IT'S 20 FEET JUST FOUND OUT

Step-by-step explanation:

6 0
3 years ago
Input the equation of the given line in slope-intercept form. the line with m=2 and b=-4
love history [14]

Answer:

<h2>y = 2x - 4</h2>

Step-by-step explanation:

The slope-intercept form:

y = mx + b

We have m = 2 and b = -4. Substitute:

y = 2x - 4

7 0
3 years ago
In a slope intercept equation , what does “m” represent
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The slope

Your welcome!! Vote me brainliest
6 0
3 years ago
Read 2 more answers
1. Write the equation of the line parallel to y = 7x – 5 and going through the point (4,31).
densk [106]

Answer:

y=7x+3

Step-by-step explanation:

Hi there!

We are given the equation y=7x-5 and we want to find the equation of the line that is parallel to it, and contains the point (4,31)

First, let's find the slope of y=7x-5, since parallel lines have the same slopes

The line y=7x-5 is written slope-intercept form, which is y=mx+b, where m is the slope and b is the y intercept

Since 7 is in the place of where m is, the slope of the line is 7

It's also the slope of the line parallel to it.

We can substitute 7 as m into the formula y=mx+b for our new line:

y=7x+b

Now we need to find b

Since the equation will pass through the point (4, 31), we can use it to solve for b; substitute 31 as y and 4 as x

31=7(4)+b

Multiply

31=28+b

Subtract 28 from both sides

3=b

Substitute 3 as b

y=7x+3

Hope this helps!

6 0
2 years ago
Lim x → 0 sin3x)/5x^3 -4
JulsSmile [24]
\bf \lim\limits_{x\to 0}\ \cfrac{sin^3(x)}{5x^3}-4&#10;\\ \quad \\\\ \cfrac{sin^3(x)}{5x^3}-4\implies \cfrac{[sin(x)]^3}{5x^3}-4&#10;\\ \quad \\&#10; &#10;&#10;\cfrac{[sin(x)]^3}{x^3}\cdot \cfrac{1}{5}-4&#10;\\ \quad \\&#10;thus&#10;\\ \quad \\&#10;\lim\limits_{x\to 0}\cfrac{[sin(x)]^3}{x^3}\cdot \lim\limits_{x\to 0}\cfrac{1}{5}-4\qquad \boxed{recall \qquad \lim\limits_{x\to 0} \cfrac{sin(x)}{x}\implies 1}&#10;\\ \quad \\&#10;1\cdot \cfrac{1}{5}-4
8 0
3 years ago
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