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2 years ago

What are the terms, like terms,coefficients, and constants of the problem 4e + 7e + 5

Mathematics

1 answer:

mote1985 [20]2 years ago

8 0

Answer:

terms: 3 terms, 4e,7e, and 5

like terms: 4e and 7e

coefficients:4e and 7e

constants:5

Step-by-step explanation:

terms are the numbers and or variables being added

like terms consist of the same variables

coefficients are the numbers being multiplied by a variable

constants are any numbers that aren't being multiplied by a variable

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Graph a right triangle with the two points forming the hypotenuse. Using the sides, find the distance between the two points, to

VashaNatasha [74]

Answer: 69, -69

Step-by-step explanation:

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3 years ago

Olivia and 4 friends order 1 Chocco-Brocco bar to share equally, what fraction is the WHOLE Chocco-Brocco bardoes each of the fr

Olin [163]

Answer: 1/4

Step-by-step explanation:

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3 years ago

Express as a trinomial. (3x+3)(2x−4)

Artyom0805 [142]

Use the FOIL method: First, Outside, Inside, Last

3x(2x) = 6x²

3x(-4) = -12x

3(2x) = 6x

3(-4) = -12

6x² -12x + 6x-12

Combine like terms

-12x + 6x = -6x

6x² - 6x - 12 is your answer

hope this helps

5 0

3 years ago

Select the pair of equations whose graphs are perpendicular.

viktelen [127]

$k:\ y=m_1x+b_1\ \ \ \ and\ \ \ \ l:\ y=m_2x+b_2\\ \\the\ line\k\ is\ perpendicular\ to\ the\ line\ l\ \ \ \Leftrightarrow\ \ \ m_1\cdot m_2=-1\\----------------------------\\\\A.\\2y=-3x+5\ \ \Rightarrow\ \ \ y=- \frac{3}{2} x+2.5\ \ \Rightarrow\ \ m_1=- \frac{3}{2} \\\\2x+3y=4\ \ \Rightarrow\ \ 3y=-2x+4\ \ \Rightarrow\ \ y=- \frac{2}{3}x+1 \frac{1}{3} \ \ \Rightarrow\ \ m_2=- \frac{2}{3} \\\\m_1\cdot m_2=- \frac{3}{2} \cdot (- \frac{2}{3})=1 \neq -1$

$B.\\5x-8y=9\ \ \Rightarrow\ \ -8y=-5x+9\ \ \Rightarrow\ \ \ y= \frac{5}{8} x-1 \frac{1}{8} \ \Rightarrow\ \ m_1= \frac{5}{8} \\\\12x-5y=7\ \Rightarrow\ \ -5y=-12x+7\ \ \Rightarrow\ \ y= \frac{12}{5}x-1 \frac{2}{5} \ \ \Rightarrow\ \ m_2= \frac{12}{5} \\\\m_1\cdot m_2= \frac{5}{8} \cdot \frac{12}{5}= \frac{3}{2} \neq -1$

$C.\\y=2x-7\ \ \Rightarrow\ \ m_1=2 \\\\x+2y=3\ \Rightarrow\ \ 2y=-x+3\ \ \Rightarrow\ \ y= -\frac{1}{2}x+1 \frac{1}{2}\ \ \Rightarrow\ \ m_2=- \frac{1}{2} \\\\m_1\cdot m_2= 2 \cdot (- \frac{1}{2})= -1\ \ \Rightarrow\ \ y=2x-7\ \ \ \ \perp\ \ \ \ x+2y=3$

$
D.\\x+6y=8\ \ \Rightarrow\ \ 6y=-x+8\ \ \Rightarrow\ \ \ y=- \frac{1}{6} x+1 \frac{1}{3}\ \ \Rightarrow\ \ m_1= -\frac{1}{6} \\\\y=2x-8\ \Rightarrow\ \ m_2= 2 \\\\m_1\cdot m_2= -\frac{1}{6} \cdot2=- \frac{1}{3} \neq -1$

7 0

3 years ago

In this figure, AB || CD and mZ3 = 120°.<br> What is mZ6?<br> (Look at picture”

Lerok [7]

Answer:

60°

Step-by-step explanation:

m3 = m1 = m5 = 120°

m5 + m6 = 180°

120° + m6 = 180°

m6 = 180° - 120°

= 60°

4 0

1 year ago