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2 years ago

Guys please help me I’ll give u 7 points and mark brainliest need asap

Mathematics

1 answer:

enot [183]2 years ago

8 0

Answer:

a. y = 3x + 4.5 & y = 7x + 2

b. x = 0.625 (year)

c. 6.375 feet

Explanation:

Type A starts at 4.5 and Type B starts at 2 ,each year can be marked as x

type A grow 3 inches per year = 3x

type B grows 7 inches per year = 7x

y = the total height

Since they both equal to y, you can set them equal to each other

3x + 4.5 = 7x +2

After solve you get answer for b

Work: 3x + 4.5 = 7x + 2

4.5 = 4x + 2

2.5 = 4x

0.625 = x

Substitute the value of x you got from part b into each equation then you will get 6.375 as your y value which is also your height

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Dmitriy789 [7]

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2 years ago

Divide (x^3-20x+16)/(x-4)

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$\frac{( x^{3}-20x+16) }{(x-4)}$

Use the rational root theorem:
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The dividers of $a_{o}$ : 1,2,4,8,16
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$\frac{4}{1}$ is a root of the expression, so factor out x-4

Compute $\frac{ x^{3}-20x+16 }{x-4}$ to get the rest of the equation.

$\frac{(x-4)( x^{2} +4x-4)}{x-4}$

Cancel the common factor:
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2 years ago

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Your current schedule only lets you take two 1-credit courses every six months. You need 12 credits to get the degree you want.

lianna [129]

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2 years ago

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sergeinik [125]

A partir de la definición de razón y la teoría de semejanza entre triángulos, la razón del área del triángulo AMN y el área del cuadrilátero BMNC es equivalente a 1/3.

<h3>¿Cómo determinar la medida de un lado de un triángulo desconocido?</h3>

En este problema tenemos un sistema formado por dos triángulos similares, la semejanza entre los dos triángulos se debe a la colinealidad entre los segmentos de línea AP' (triángulo pequeño) y AP'' (triángulo grande), así como de los lados AM y AB, así como los lados AN y AC, así como los mismos ángulos en la misma distribución. (Semejanza Lado - Ángulo - Lado)

En consecuencia, obtenemos las siguientes proporciones:

AP'/AP'' = MN/BC = 1/2 (1)

Finalmente, la proporción entre el triángulo AMN y el cuadrilátero BMNC es:

$\frac{AMN}{ABC - AMN} = \frac{\frac{1}{2}\cdot a \cdot \left(\frac{1}{2}\cdot h \right)}{\frac{1}{2}\cdot (2\cdot a) \cdot h - \frac{1}{2}\cdot a \cdot \left(\frac{1}{2}\cdot h \right)} = \frac{\frac{1}{4}\cdot a\cdot h }{a\cdot h - \frac{1}{4}\cdot a \cdot h }$

$\frac{AMN}{ABC - AMN} = \frac{\frac{1}{4} }{\frac{3}{4} } = \frac{1}{3}$

Para aprender sobre triángulos semejantes: brainly.com/question/21730013

#SPJ1

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1 year ago