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ikadub [295]
3 years ago
11

How can you estimate 1,506 divided 2 so that it is close to the actual answer

Mathematics
1 answer:
NeTakaya3 years ago
7 0
Think of dividing by 2 splitting it in half, if you round it you get about 1500 then split that in half. it would be around 750.
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Find the inverse of the function y = x2 - 12.
Semmy [17]

Good evening ,

Answer:

The inverse of the function y = x² - 12 is

x = √(y + 12)

Step-by-step explanation:

for y ≥ -12 :

x² - 12 = y ⇔ x² = y + 12 ⇔ x = √(y + 12).

:)

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2 years ago
In KLM, LM=7 and angle K=45. Find KL. Leave your answer in simplest radical form.
Sphinxa [80]
Answer in simplest radical form would be 7.(:
5 0
3 years ago
Arnie's car used 100 cups of gasoline during a drive.He paid $3.12 per gallon of gas.How much did the gas cost?
Elan Coil [88]
There are 16 cups in a gallon. Divide 100 into 16 to get 6.25. You now just multiply 3.12 by 6.25 to get 19.5. Your answer is $19.50. Hope this helps!
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2 years ago
Nanu borrowed a certain sum at the rate of 10 % p.a. If she paid compound interest Rs. 1,290 at the end of two years compounded
Leokris [45]

Answer:

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5 0
2 years ago
Solve the following equations by factorisation method.Only factorisation not dharacharya.​
Luba_88 [7]

Hello, please consider the following.

When x_1 and x_2 are two roots, we can factorise as

ax^2+bx+c=a(x-x_1)(x-x_2)=a(x^2-(x_1+x_2)x+x_1x_2)=0

So for the first equation, we can say that the sum of the zeros is

\dfrac{a^2+b^2}{2}=\dfrac{a^2}{2}+\dfrac{b^2}{2}

and the product is

\dfrac{a^2b^2}{4}=\dfrac{a^2}{2}\dfrac{b^2}{2}

So we can factorise as below.

4x^2-2(a^2+b^2)x+a^2b^2=(2x-a^2)(2x-b^2)=0

And the solutions are

\boxed{\sf \n\bf  \ \dfrac{a^2}{2} \ \ and \ \ \dfrac{b^2}{2}}

For the second equation, we will complete the square and put the constant on the right side and take the root.

Let's do it!

9x^2-9(a+b)x+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2-9\dfrac{(a+b)^2}{4}+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9(a+b)^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9a^2+18ab+9b^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{a^2+b^2-2ab}{4}=\dfrac{(a-b)^2}{4}\\\\(x-\dfrac{a+b}{2})^2=\dfrac{(a-b)^2}{2^2\cdot 3^2}

We take the root, and we find the two solutions

\begin{aligned}x_1&=\dfrac{a+b}{2}-\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b-a+b}{6}\\\\&=\dfrac{2a+4b}{6}\\\\&\boxed{=\dfrac{a+2b}{3}}\end{aligned}

\begin{aligned}x_1&=\dfrac{a+b}{2}+\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b+a-b}{6}\\\\&=\dfrac{4a+2b}{6}\\\\&\boxed{=\dfrac{2a+b}{3}}\end{aligned}

Thank you.

6 0
3 years ago
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