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satela [25.4K]
2 years ago
15

Solve the following equation. show work. x^2 + 10x = -25

Mathematics
1 answer:
SVEN [57.7K]2 years ago
6 0

Step-by-step explanation:

x^2 + 10x = -25

x^2 + 10x + 25 = 0

(x+5)^2=0

x=-5

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alukav5142 [94]
The answer is D ……….
8 0
1 year ago
Find the values of A B C AND D of <br> 4x^2 (2x^3+5x)= Ax^B +Cx^D
iren2701 [21]

The values are A=8, B=5, C= 20 and D=3

Explanation:

The expression is 4x^{2} (2x^{3} +5x)=Ax^{B} +Cx^{D}

Simplifying, we get,

8x^{5} +20x^3=Ax^{B} +Cx^{D}

Since, both sides of the expression are equal, we can equate the corresponding values of A, B, C and D.

Thus, we get,

8 x^{5}=A x^{B} ⇒ A=8 and B=5

Also, equating, 20 x^{3}=C x^{D}, we get,

C=20 and D=3

Thus, the values are A=8, B=5, C= 20 and D=3

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3 years ago
Item 4 Find the distance between the two points. (4, −5), (−1, 7)
Rus_ich [418]

Answer:

yes , the distance between the two points is 13 units

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
Which of the following sets of numbers could not represent the three sides of a triangle?
Zanzabum

Answer:

The last one..Number 4

Step-by-step explanation:

When u add the two sides it's supposed to be larger than the bigger side

so 13+20 is smaller than 34

4 0
2 years ago
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