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3 years ago

Using the quadratic formula to solve x2 = 5-x, what are the values of x?

Mathematics

2 answers:

Alexus [3.1K]3 years ago

8 0

Answer:

A) (-1 ± √21)/ 2

Step-by-step explanation:

Move all terms onto one side

$x^{2}$ = 5-x

$x^{2}$ + x - 5 = 0

Insert values into quadratic formula

a = 1 b= 1 c= - 5

[-b ± √( $b^{2}$ - 4ac)]/2a

[-1 ± √( $1^{2}$ - 4 (1(-5)]/ 2(1)

(-1 ± √21)/ 2

kodGreya [7K]3 years ago

5 0

Answer:

its the first choice, negative 1 plus or minus square root of 21 over 2

Step-by-step explanation:

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What is the value of x

nordsb [41]

Median is half of add the sum ofthe base and first side
17+5x-14=3+5x
divide it by 2
3+5x ÷2=median
(3+5x)/2=x+12
3+5x=2x+24
3x=21
x=7

Answer is D=7

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3 years ago

Find the area and perimeter of the algebra titles don't forget your units

Rufina [12.5K]

The perimeter of the large square is 24 and the area is 36 units
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2 years ago

ABC Car Share Inc. offers car sharing for a $199 yearly fee plus $14/hr for the use of its car share. What equation models the t

Katarina [22]

Total cost equals the yearly fee price plus $14 times number of hours.

T= total cost
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Equation:
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Hope this helps! :)

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3 years ago

Read 2 more answers

9. (12 pts) daniel is paying $600 for his auto insurance, and he is wondering if he is overpaying compared to his friends. he se

Karolina [17]

Answer:

Step-by-step explanation:

1) The sample and the population of this study is the friends who replied his email which includes in his contact list. then, the number of the replied to his email are 9 friends.

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3 years ago

3. Let U and V be subspaces of a vector space W. Prove that their intersection UnV is also a subspace of W

kenny6666 [7]

Answer: The proof is done below.

Step-by-step explanation: Given that U and V are subspaces of a vector space W.

We are to prove that the intersection U ∩ V is also a subspace of W.

(a) Since U and V are subspaces of the vector space W, so we must have

0 ∈ U and 0 ∈ V.

Then, 0 ∈ U ∩ V.

That is, zero vector is in the intersection of U and V.

(b) Now, let x, y ∈ U ∩ V.

This implies that x ∈ U, x ∈ V, y ∈ U and y ∈ V.

Since U and V are subspaces of U and V, so we get

x + y ∈ U and x + y ∈ V.

This implies that x + y ∈ U ∩ V.

(c) Also, for a ∈ R (a real number), we have

ax ∈ U and ax ∈ V (since U and V are subspaces of W).

So, ax ∈ U∩ V.

Therefore, 0 ∈ U ∩ V and for x, y ∈ U ∩ V, a ∈ R, we have

x + y and ax ∈ U ∩ V.

Thus, U ∩ V is also a subspace of W.

Hence proved.

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3 years ago