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3 years ago

Find the measure of

Mathematics

1 answer:

Pavel [41]3 years ago

3 0

Answer:

Step-by-step explanation:

180 - (180 - 102) - (180 - 123) = 180 - 78 - 57 = 45°

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define a variable write an equation and solve the problem. the length of a rectangle is twice the width. an equation that models

liberstina [14]

6w=36
w=6

width=6
length=12

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4 years ago

An infant is 32.625 inches (in) long. Write this as a common fraction.

Romashka-Z-Leto [24]

The fraction will be 261/8

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3 years ago

Read 2 more answers

Factor the following equation to find its zeros. y = x² - 36

alukav5142 [94]

Answer:

x = 6, x = - 6

Step-by-step explanation:

Given

y = x² - 36

To find the zeros let y = 0, that is

x² - 36 = 0 ← x² - 36 is a difference of squares and factors in general as

a² - b² = (a - b)(a + b), thus

x² - 36 = 0

x² - 6² = 0

(x - 6)(x + 6) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 6 = 0 ⇒ x = 6

x + 6 = 0 ⇒ x = - 6

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3 years ago

Mr. Ramirez orders drumsticks and sheet music for a group of students. One pair of drumsticks costs $1. 50, and the sheet music

diamong [38]

Answer:

Step-by-step explanation:

It expresses how much money he had to pay in total for the sheet music and drumsticks altogether.

Have an amazing day!

Please mark brainliest!

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3 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

3 years ago