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swat32
3 years ago
6

Jake is y years old. His grandma is 6 times his age. His cousin is 5 years older than him. a. Write an expression for each of th

eir ages.
Mathematics
1 answer:
gtnhenbr [62]3 years ago
7 0

Answer:

y+5=c

6y=g

Step-by-step explanation:

C for cousin

G for grandma

You might be interested in
Points
liq [111]

Answer:

[x-h]^2 + [y-k]^2 =r^2

Step-by-step explanation:

Let x and y be an arbitrary point along the circle

The equation of the circle is given as:

[x-h]^2 + [y-k]^2 =r^2

3 0
3 years ago
EACH PAIR OF FIGURES IS SIMILAR. FIND THE MISSING SIDE!!!!
inessss [21]

Answer:

58.1 and 17

Step-by-step explanation:

For the first triangles the similarity ratio is 1:7 so x is 8.3 × 7 = 58.1

For the second triangles the similarity ratio is 1:5 so x is 3.4 × 5 = 17

8 0
3 years ago
URGENT WORTH 20 POINTS
Triss [41]
If we were to draw a best fit line, the y-interest would be near 15, and the slope would be negative.

In the options given, the slope is either -3 or -13. If we imagine a line through the plotted points, the rise would be about 15 while the run would be about 5. This means the slope is about -3.

The best answer is y = -3x + 15





5 0
3 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
3 years ago
I need this done rn plz help
My name is Ann [436]

Answer:

answer for yes total, is 5, the answer for no total is 20, the total for junior is 17, and the total for senior is 8.

Step-by-step explanation:

What you do is you add the 2 values. for example, to find the total for the juniors, you would add 1 to 16.

8 0
3 years ago
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