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cricket20 [7]
3 years ago
15

Multiplying binomials: (a+7) * (a-4)

Mathematics
2 answers:
navik [9.2K]3 years ago
6 0
The equation when multiplying binomials is a^2 + 2ab + b^2. Therefore when you use FOIL (First, Outer, Inner, Last), you will multiply a times a and a times -4, which gives you a^2 - 4a. Now you take the second term and do 7 times a and 7 times -4, which gives you 7a - 28. Combine them and you have a^2 - 4a + 7a - 28. Then combine the middle two terms and that gives you 3a. You will always have to combine like terms and it will always be the two middle terms. So your answer is a^2 + 3a - 28. Hope this helps. Please rate, leave a thanks, and mark a brainiest answer (Not necessarily mine). Thanks, it really helps! :D

oee [108]3 years ago
4 0
<span>(a+7) * (a-4)

a^2 - 4a + 7a - 28

a^2 + 3a - 28</span>
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* WILL GIVE BRAINIEST*
k0ka [10]

Answer:

B: 280

Step-by-step explanation:

The regression line predicts that when x equals 5:

log_{10}(y) = 2.447

In order to find the value for y, one must simply apply the following logarithmic property:

if : log_{b}(a) = c

then: b^c = a

Applying it to this particular problem:

log_{10}(y) = 2.2447\\10^{2.447}= y\\y=280

Therefore, the regression line predicts y will equal 280 when x equals 5.

3 0
3 years ago
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sammy [17]

Answer:

Not clear enough

Step-by-step explanation:

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8 0
3 years ago
What is the midpoint of the segment shown below?
Burka [1]

Answer:

choice. c.) (5, 1/2)

Step-by-step explanation:

(5, 4) and (5, -3)

Use midpoint formula  ( (a + x)/2 , (b + y)/2)  for (a,b), (x,y)

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5 0
3 years ago
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bezimeni [28]
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7 0
3 years ago
Read 2 more answers
Una recta que pasa por los puntos A (2,1) y B (6,3) y otra recta pasa por A y por el punto (0,y) ¿Cuánto vale y, si ambas rectas
ASHA 777 [7]

The translation of the question given is

A line that passes through the points A (2,1) and B (6,3) and another line passes through A and through the point (0, y). What is y worth, if both lines are perpendicular?

Answer:

y = 5

Step-by-step explanation:

Line 1 that passes through A (2,1) and B (6,3)  

Slope (m1) = 3-1/6-2 = 2/4 = 1/2

y - 1 = \frac{1}{2} ( x -2)

2y - 2 = x- 2

y = \frac{x}{2}

Line 2 passes through A (2,1) and (0,y)

slope (m2) =\frac{y-1}{-2}

Line 1 and Line 2  are perpendicular

m1*m2 = -1

\frac{1}{2}  * \frac{y-1}{-2} = -1

y-1 = 4

y = 5

slope = -2

Equation of Line 2

Y-1 = -2(x-2)

y -1 = -2x +4

2x +y = 5

7 0
2 years ago
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