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Nikolay [14]
3 years ago
6

5x^2-7x=4x^2-10 solve

Mathematics
1 answer:
Luda [366]3 years ago
5 0

Answer:

x = 0, x = - 3

Step-by-step explanation:

Given

5x² - 7x = 4x² - 10 ← subtract 4x² - 10x from both sides

x² + 3x = 0 ← in standard form

Factor out x from each term

x(x + 3) = 0

Equate each factor to zero and solve for x

x = 0

x + 3 = 0 ⇒ x = - 3

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John has a ribbon that was 1 1/2 meters long. He used 2 pieces that were each 1/3 of that length. How much ribbon did John use?
Bond [772]

Answer:

1 meter.

Step-by-step explanation:

1 1/2 * 1/3

= 3/2 * 1/3

= 3/6

= 1/2.

So John used 2*1/2 = 1 meter of the ribbon.

5 0
4 years ago
Soon-Jin sent out 48 invitations, this was 4/5 of all her invitations. How many invitations will she send out in total?
stepladder [879]
If 4/5 of the total is 48, the total can be found by dividing it by 4 and multiplying it by 5:
48/4 = 12
12*5 = 60.
3 0
3 years ago
Read 2 more answers
How many 1/4 gram doses can be obtained from 10 1/2 gram vial of medication?
alexdok [17]

Answer:

2.02× 10²= 202

I hope it help

5 0
3 years ago
Priya measured the length of lizard to be 11 cm. The actual length is 12 cm. Find the % error.
Aleksandr [31]

Answer:

8.33%

Step-by-step explanation:

Formula for percent error:

|\frac{observed value - expected value}{expected value} | * 100%

Using this formula, we get:

|\frac{11-12}{12} |* 100 = \frac{100}{12} = 8.33

If you found this answer helpful please consider leaving 5 stars, giving it a like, or if you asked this question, mark this answer as Brainliest. Thanks!

8 0
2 years ago
A) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s.
Fed [463]

Answer:

A) a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

B) a_{0} = a_{1} = 0

C)   for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

Step-by-step explanation:

A) A recurrence relation for the number of bit strings of length n that contain a  pair of consecutive Os can be represented below

if a string (n ) ends with 00 for n-2 positions there are a pair of  consecutive Os therefore there will be : 2^{n-2} strings

therefore for n ≥ 2

The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

b ) The initial conditions

The initial conditions are : a_{0} = a_{1} = 0

C) The number of bit strings of length seven containing two consecutive 0s

here we apply the re occurrence relation and the initial conditions

a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}

for n = 2

  a_{2} = 1

for n = 3

 a_{3} = 3

for n = 4

a_{4} = 8

for n = 5

a_{5} = 19

7 0
3 years ago
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