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tresset_1 [31]
3 years ago
10

What is the unit rate of 5/7 : 4/7?

Mathematics
1 answer:
allochka39001 [22]3 years ago
6 0

9514 1404 393

Answer:

  a = 5/4

Step-by-step explanation:

To make the second number in this ratio be 1, you need to multiply the whole ratio by 7/4:

  (7/4)(5/7) : (7/4)(4/7) = 5/4 : 1 = a : 1

  a = 5/4

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A right triangle has a leg of 11 cm and a hypotenuse of 17 cm.
Vedmedyk [2.9K]
The answer would be the second option, 13.0 cm.

To find the missing length of the leg, you would use the pythagorean theorem and substitute your inputs in.

a^2+b^2=c^2
11^2+b^2=17^2
121+b^2=289
b^2=168
b=12.96 
b=13.0 cm (rounded up)
4 0
3 years ago
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How do I do this ? I don’t understand
juin [17]
It's just dividing out the factoring to their simplest form. 
The bold numbers are the answers that go in the circles. 

1. 16 < 8 and 2
    8 < 2 and 4
    4 < 2 and 2

2. 42 < 14 and 3
     14 < 7 and 2

3. 40 < 2 and 20
    20 < 10 and 2
    10 < 2 and 5

4. I'm not able to see the rest of the factoring tree
6 0
3 years ago
PLS HELP ASAP ILL GIVE BRAINLKEST THANKS
kompoz [17]

Answer:

d<u>></u>-7

Step-by-step explanation:

4 0
2 years ago
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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
If f ( x ) f(x) is an exponential function where f ( − 3.5 ) = 25 f(−3.5)=25 and f ( 6 ) = 33 f(6)=33, then find the value of f
gavmur [86]

Given:

f(x) is an exponential function.

f(-3.5)=25, f(6)=33

To find:

The value of f(6.5).

Solution:

Let the exponential function is

f(x)=ab^x         ...(i)

Where, a is the initial value and b is the growth factor.

We have, f(-3.5)=25. So, put x=-3.5 and f(x)=25 in (i).

25=ab^{-3.5}         ...(ii)

We have, f(6)=33. So, put x=6 and f(x)=33 in (i).

33=ab^{6}         ...(iii)

On dividing (iii) by (ii), we get

\dfrac{33}{25}=\dfrac{ab^{6}}{ab^{-3.5}}

1.32=b^{9.5}

(1.32)^{\frac{1}{9.5}}=b

1.0296556=b

b\approx 1.03

Putting b=1.03 in (iii), we get

33=a(1.03)^{6}

33=a(1.194)

\dfrac{33}{1.194}=a

a\approx 27.63

Putting a=27.63 and b=1.03 in (i), we get

f(x)=27.63(1.03)^x

Therefore, the required exponential function is f(x)=27.63(1.03)^x.

4 0
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