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2 years ago

Used to find the midpoint between two points (x1,y1) and (x2,y2) what is the formula?

Mathematics

1 answer:

umka21 [38]2 years ago

8 0

Here you go! Hope this helps

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Can i get some help please??? Thanks!!!!

Varvara68 [4.7K]

Answer:

9.6

Step-by-step explanation:

$d = \sqrt{ {8}^{2} + {2}^{2} + {5}^{2} } \\ = \sqrt{64 + 4 + 25} \\ = \sqrt{93} \\ = 9.64365076 \\ = 9.6$

8 0

3 years ago

How do you factor 33x^5y^2+21x^3y+9x^3y^3?

scZoUnD [109]

Answer:

3x³y(11x²y+3y²+7)

Step-by-step explanation:

hope that helps

7 0

3 years ago

Given the function f(x)=(x+3)^2 and g(x)=3x+13 determine the x values of the point intersection

charle [14.2K]

Answer:

x = - 4, x = 1

Step-by-step explanation:

At the point of intersection

f(x) = g(x) , that is

(x + 3)² = 3x + 13 ← expand left side using FOIL

x² + 6x + 9 = 3x + 13 ( subtract 3x + 13 from both sides )

x² + 3x - 4 = 0 ← in standard form

(x + 4)(x - 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 4 = 0 ⇒ x = - 4

x - 1 = 0 ⇒ x = 1

3 0

2 years ago

X2 - 4x + 3 = 0<br> lesser x =<br> greater x =<br> o

Levart [38]

Answer:

lesser x is 1

greater x is 3

Step-by-step explanation:

Here, we want to solve the equation so as to get the bigger and the lesser x values

We proceed as follows;

x^2 -4x + 3 = 0

x^2 - x-3x + 3 = 0

x(x-1) -3(x-1) = 0

(x-3)(x-1) = 0

x = 3

or x = 1

Greater x is 3

lesser x is 1

7 0

2 years ago

Simplify. Your answer should contain only positive exponents with no fractional exponents in the denominator.

mart [117]

Answer:

$\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$ .

Step-by-step explanation:

The given expression is

$\dfrac{3y^{\frac{1}{4}}}{4x^{-\frac{2}{3}}y^{\frac{3}{2}}\cdot 3y^{\frac{1}{2}}}$

We need to simplify the expression such that answer should contain only positive exponents with no fractional exponents in the denominator.

Using properties of exponents, we get

$\dfrac{3}{4\cdot 3}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{2}}}$ $[\because a^ma^n=a^{m+n}]$

$\dfrac{1}{4}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{2}}$

$\dfrac{1}{4}\cdot \dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{y^{2}}$ $[\because a^{-n}=\dfrac{1}{a^n}]$

$\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$

We can not simplify further because on further simplification we get negative exponents in numerator or fractional exponents in the denominator.

Therefore, the required expression is $\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$ .

5 0

3 years ago