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3 years ago

X+2y=9 and x=2y+1helppppppppppppppppp meeeeereeer

Mathematics

2 answers:

Anastasy [175]3 years ago

6 0

Answer:

(5, 2)

Step-by-step explanation:

Given system:

x + 2y = 9
x = 2y + 1

Solve for y by substitution:

(2y + 1) + 2y = 9
4y + 1 = 9
4y = 8
y = 2

Find the value of x:

x = 2*2 + 1
x = 5

skelet666 [1.2K]3 years ago

5 0

Answer: y = 2

Step-by-step explanation:

we need to use systems of equations, basically since we know what x equals, we plug that into first equation so:

since x=2y+1,

we can say that:

(2y+1)+2y=9

then we solve this equation,

first we expand the brackets- 2y+1+2y=9,

then we combine like terms- 4y+1=9

then we take 1 away from both sides- 4y=8

then we divided both sides by 4

y=2

Hope this helps xx

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What are the zeros of f(x)=x^2+3x-10

vekshin1

You must remember that a polynomial is written like so...

ax^2 + bx + c

In this case...

a = 1

b = 3

c = -10

To factor you must find two numbers who both add up to b (3) AND multiply to c (-10)

-2 + 5 = 3

-2 * 5 = -10

so...

(x - 2)(x + 5)

To find the zero you must set each factor equal to zero and solve for for x like so...

x - 2 = 0

x = 2

x + 5 = 0

x = -5

Hope this helped!

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4 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

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coldgirl [10]

Answer:

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Step-by-step explanation:

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let

to make things easier to see

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−

See it now?

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(

−

)

(

)

Now substitute back in

for

(

−

)

(

)

Since

(

−

)

is a difference of two squares,

(

−

)

(

)

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(

−

)

(

)

(

)

x can be 3, -3 for the first two parts

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−

Taking the positive and negative root means

√

−

Thus

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3 years ago

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Find the x -intercept of the line whose equation is 10 x + 3 y = 1. 1/3 1/10 1

prisoha [69]

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3 years ago

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X+2y=9 and x=2y+1helppppppppppppppppp meeeeereeer​

X+2y=9 and x=2y+1helppppppppppppppppp meeeeereeer