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horsena [70]
3 years ago
7

X² + 15x +44 Factor the quadratic completely

Mathematics
2 answers:
Hoochie [10]3 years ago
8 0
Factor
x
2
+
15
x
+
44
x
2
+
15
x
+
44
using the AC method.
(
x
+
4
)
(
x
+
11
)
anastassius [24]3 years ago
8 0

Answer:

(x + 4) (x + 11)

Step-by-step explanation:

f

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PLEASE PLEASE PLEASE HELP!!
irga5000 [103]
  9            a
------  =  -----
  a            5

a^2 = 45 

a = 3<span>√5
</span>
  4            b
------  =  -----
  b            5

b^2 = 20

b = 2√5

answer: 

a = 3√5
b = 2√5

hope it helps
7 0
2 years ago
Expand the following expression
masya89 [10]

Answer:

\frac{5}{4}  \times 4x +  \frac{5}{4}  \times  \frac{3}{4}  = 5x +  \frac{15}{16}

5 0
2 years ago
((Pls help need then to get my grade up))
DIA [1.3K]
It's C because the equations are similar
3 0
2 years ago
Can someone please help me
AVprozaik [17]

Answer:

Lines RQ and SP are perpendicular to SR

Step-by-step explanation:

SR are parallel to PQ so that means that RQ and SP are perpendicular to SR

6 0
3 years ago
A highway map of Ohio has a coordinate grid superimposed on top of the state. Springfield is at point (1, –4) and Columbus is at
Vedmedyk [2.9K]
The midpoint, where the rest area is, will just be the average of the coordinates of Springfield and Columbus.

Rest Area=((1+7)/2, (-4+1)/2)

Rest Area=(4, -1.5)

The distance between the two cities can be found using the Pythagorean Theorem, which used in the manner is often referred to as the "distance" formula between two points.

d^2=(x2-x1)^2+(y2-y1)

d^2=(7-1)^2+(1--4)^2

d^2=36+25

d^2=61

d=√61  and we are told that each unit is equal to 5.38mi so the distance from Springfield to Columbus is:

D=5.38√61 mi

D≈42.02mi  (to nearest hundredth of a mile)
3 0
3 years ago
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