B = 4
A line with a slope of zero is a perfectly horizontal line, which means the y value (the second value) stays the same for every single point on the line.
So, if y equals 4 in the first point, then y still equals 4 in the second point, which means b is equal to 4.
Answer:
(
2
x
−
6
)
2
+
4
(
2
x
−
6
)
+
3
=
0
Simplify the left side.
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(
2
x
−
6
)
2
+
8
x
−
21
=
0
Use the quadratic formula to find the solutions.
−
b
±
√
b
2
−
4
(
a
c
)
2
a
Substitute the values
a
=
4
,
b
=
−
16
, and
c
=
15
into the quadratic formula and solve for
x
.
16
±
√
(
−
16
)
2
−
4
⋅
(
4
⋅
15
)
2
⋅
4
Simplify.
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x
=
4
±
1
2
The final answer is the combination of both solutions.
x
=
5
2
,
3
2
Step-by-step explanation:
Answer:
−9⋅(5j+k) = -45j -9k.
Step-by-step explanation:
Given : −9⋅(5j+k)
To find : distributive property.
Solution : −9⋅(5j+k)
By distributive property a( b +c ) = a*b + a*c.
Then −9⋅(5j+k) = -9 *5j + (-9 *k)
−9⋅(5j+k) = -45j -9k.
Therefore, −9⋅(5j+k) = -45j -9k.
Answer:
So Philip made 5 bracelets and 4 necklaces.
Step-by-step explanation:
Let x = number of bracelets and y = number of necklaces.
Since we have a total of 9 bracelets and necklaces,
x + y = 9 (1)
Also, we have 8 inches of cord for each bracelet and 20 inches of cord for each necklace, then the total length for the bracelet is 8x and that for the necklace is 20y.
So, the total length for both is 8x + 20y. Since the total length of cord used is 120 inches,
8x + 20y = 120 (2)
Simplifying it we have
2x + 5y = 30 (3).
Writing equations (1) and (3) in matrix form, we have
![\left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}9\\30\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C2%265%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%5C%5C30%5Cend%7Barray%7D%5Cright%5D)
Using Cramer's rule to solve for x and y,
![x = det \left[\begin{array}{ccc}9&1\\30&5\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\](https://tex.z-dn.net/?f=x%20%3D%20det%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%261%5C%5C30%265%5Cend%7Barray%7D%5Cright%5D%20%2Fdet%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C2%265%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
x = (9 × 5 - 30 × 1) ÷ (1 × 5 - 1 × 2)
x = (45 - 30) ÷ (5 - 2)
x = 15 ÷ 3
x = 5
![y = det \left[\begin{array}{ccc}1&9\\2&30\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\](https://tex.z-dn.net/?f=y%20%3D%20det%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%269%5C%5C2%2630%5Cend%7Barray%7D%5Cright%5D%20%2Fdet%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C2%265%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
y = (30 × 1 - 9 × 2) ÷ (1 × 5 - 1 × 2)
y = (30 - 18) ÷ (5 - 2)
y = 12 ÷ 3
y = 4
So Philip made 5 bracelets and 4 necklaces.
Answer:
482.5 ft^2.
Step-by-step explanation:
The figure can be regarded as a rectangle with a triangle cut out of one corner.
The base of the triangle = 25-20 = 5 in and the height = 20 - 13 = 7 in.
The required area = area of rectangle - area of triangle
= 25*20 - 1/2 * 5 * 7
= 500 - 17.5
= 482.5 ft^2
