Fifth grade eureka math Problems set module 2 lesson 1

Mathematics

1 answer:

lapo4ka [179]3 years ago

5 0

Step-by-step explanation:

what is the question you are just showing us the lesson and modle

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Twelve less than a number is 50.

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Answer:

Step-by-step explanation:

another way to write this is

x - 12 = 50

the answer is 62

62 - 12 = 50

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Chan rows at a rate of 8 mph in still water. It takes him three hours to row upstream from his house to the park. He rows back h

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When x=8,y=20.find y when x=42.

tino4ka555 [31]

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Simplify the expression

kow [346]

Answer:

Step-by-step explanation:

$\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}\\\\6(x^2-3x+2)\ne0\ \iff\ x=\frac{3\pm\sqrt{9-8}}{2}\ne0\ \iff\ x\ne2\ \wedge\ x\ne1\\\\\\\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}=\dfrac{x(5x^2-8x-4)}{6(x^2-3x+2)}=\dfrac{x(5x^2-10x+2x-4)}{6(x^2-2x-x+2)}=\\\\\\=\dfrac{x[5x(x-2)+2(x-2)]}{6[x(x-2)-(x-2)]} =\dfrac{x(x-2)(5x+2)}{6(x-2)(x-1)}=\dfrac{x(5x+2)}{6(x-1)}=\dfrac{5x^2+2x}{6x-6}\\\\\\f(x)=5x^2+2x\\\\g(x)=6x-6$

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2 years ago

3/8<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B8%7D%20%28x%20-%201%29%28x%20-%209%29" id="TexFormula1" title=" \fr

Verdich [7]

The first step for solving this expression is to distribute $\frac{3}{8}$ through the first set of parenthesis.
$( \frac{3}{8}x - \frac{3}{8})$ × (x - 9)
Now use the FOIL method to multiply each term in the first parenthesis by each term in the second parenthesis. This will look like the following:
$\frac{3}{8}$ x × x - $\frac{3}{8}$ x × 9 - $\frac{3}{8}$ x - $\frac{3}{8}$ × (-9)
Remember that multiplying two negatives together equals a positive,, so the expression changes to:
$\frac{3}{8}$ x × x - $\frac{3}{8}$ x × 9 - $\frac{3}{8}$ x + $\frac{3}{8}$ × 9
Calculate the product of all of the sets of multiplication to make the expression become:
$\frac{3}{8}$ x² - $\frac{27}{8}$ x - $\frac{3}{8}$ x + $\frac{27}{8}$
Lastly,, calculate the difference of - $\frac{27}{8}$ x - $\frac{3}{8}$ x to find your final answer.
$\frac{3}{8}$ x² - $\frac{15}{4}$ x + $\frac{27}{8}$
Let me know if you have any further questions.
:)

8 0

2 years ago