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2 years ago

Helppppppppppppppppppp

Mathematics

1 answer:

DochEvi [55]2 years ago

7 0

Angle 2 and angle 11 are alternate exterior angles. Line G and line L are parallel lines. Think of them as train tracks. On the outside or exterior of the train tracks is angle 2 and angle 11.

Angle 2 is on the right side of the transversal line, while angle 11 is on the left side of the transversal line.

So this is why they are alternate exterior angles. Because line G and line L are parallel lines, this means the alternate exterior angles are congruent (by the alternate exterior angle theorem).

Since angle 2 is 115 degrees, angle 11 must also be 115 degrees

Answer: C) 115

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Axis of symmetry: x=4; points: (3,-2), (6,-26); opening downward

Andru [333]

Vertex form is
y=a(x-h)^2+k
vertex is (h,k)
axis of symmetry is x=4, therfor h=4
y=a(x-4)^2+k

we have some points
(3,-2) and (6,-26)
input and solve for a and k

(3,-2)
-2=a(3-4)^2+k
-2=a(-1)^2+k
-2=a(1)+k
-2=a+k

(6,-26)
-26=a(6-4)^2+k
-26=a(2)^2+k
-26=a(4)+k
-26=4a+k

we have

-2=a+k
-26=4a+k
multiply first equation by -1 and add to second

2=-a-k
<u>-26=4a+k +</u>
-24=3a+0k

-24=3a
divide both sides by 3
-8=a

-2=a+k
-2=-8+k
add 8 to both sides
6=k

the equation is
$y=-8(x-4)^2+6$

3 0

3 years ago

I need help please please

kotegsom [21]

Answer:

k, a, and i I think!

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Is this correct??? Will give Brainlyist

masya89 [10]

It.s not correct, but it’s not hard to learn. All you really have to do is multiply both the numerator and the denominator by each of the numerators that are listed in the problem. If you need help just ask, but I’m not going to give you the problem because you can do it! Just ask if you don’t understand.

7 0

3 years ago

If Rib-eye steaks are on sale for $8.52 per pound. Ms. Markum bought a 0.9-pound steak. How much did the steak cost?

Pepsi [2]

Answer: The steak costs $ 7.67.

Step-by-step explanation:

Given: Cost per pound of Rib-eye steaks = $8.52

Ms. Markum bought a 0.9-pound steak.

To find : cost of 0.9 pound steak =

Then cost of 0.9 pound steak = 0.9 x (Cost per pound of Rib-eye steaks )

= $ (0.9 x 8.52)

= $7.668 ≈ $ 7.67

Hence, the steak costs $ 7.67.

7 0

2 years ago

How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?

inna [77]

A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
$\text{Case 1: } 4 \cdot 4!$

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
$\text{Case 2: } 3 \cdot 4!$

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
$\text{Case 3: } 2 \cdot 4!$

$\text{Case 4: } 1 \cdot 4!$

$\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240$

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

$\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36$

7 0

2 years ago