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lapo4ka [179]
2 years ago
13

There are 52 weeks in a year. Ben has lived in london for two fifths of year. How many weeks has ben lived in london?

Mathematics
2 answers:
Soloha48 [4]2 years ago
6 0

\text{We're trying to find how many weeks Ben has lived in London}\\\\\text{We know that there are 52 weeks in a year}\\\\\text{We also know that he's been in London for 2/5 of a year, or 40\%}\\\\\text{To find your answer, you could multiply 52 by 0.40}\\\\52*0.40= 20.8\\\\\boxed{\text{Ben has been in London for 20.8 weeks}}

masha68 [24]2 years ago
5 0

Answer:

\frac{104}{5} weeks or 20\frac{4}{5} weeks

Step-by-step explanation:

To do this we can set up a proportion

\frac{52}{1} *\frac{2}{5}

This will tell us how many weeks are in 2/5 of a year

Let's evaluate this

\frac{104}{5}

We cannot simplifiy this anymore, so we will leave it as an improper fraction, or a mixed number as shown below

20\frac{4}{5} weeks

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X^2+ 3 has a value of 28. What term in the sequence has a value of 28?
VikaD [51]

Answer:

5th term

Step-by-step explanation:

x^2 +3 = 28

Subtract 3 from each side

x^2+3-3 =28-3

x^2 = 25

Take the square root of each side

sqrt(x^2) = sqrt(25)

x = 5

(x could be -5, but there are not usually negative terms in a sequence)

5 0
2 years ago
Read 2 more answers
For the geometric sequence of a1=2 and r=2 find a5
Brrunno [24]

Answer:

a_5 = 32

Step-by-step explanation:

The nth term for the geometric sequence is given by:

a_n = a_1 \cdot r^{n-1}

where,

a_1 is the first term

r is the common ratio

n is the number of terms.

As per the statement:

For the geometric sequence of a_1=2 and r=2

We have to find a_5

for n = 5;

a_5=a_1 \cdot r^{n-1}

Substitute the given values we have;

a_5 = 2 \cdot 2^4 = 2 \cdot 16

⇒a_5 = 32

Therefore, the value of a_5 is, 32

3 0
3 years ago
Type the correct answer in the box,
HACTEHA [7]

<u>We'll assume the quadratic equation has real coefficients</u>

Answer:

<em>The other solution is x=1-8</em><em>i</em><em>.</em>

Step-by-step explanation:

<u>The Complex Conjugate Root Theorem</u>

if P(x) is a polynomial in x with <em>real coefficients</em>, and a + bi is a root of P(x) with a and b real numbers, then its complex conjugate a − bi is also a root of P(x).

The question does not specify if the quadratic equation has real coefficients, but we will assume that.

Given x=1+8i is one solution of the equation, the complex conjugate root theorem guarantees that the other solution must be x=1-8i.

3 0
2 years ago
Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Tju [1.3M]

Answer:

-10 because you do parenthesis first

4 0
3 years ago
Read 2 more answers
Uually Cxpressions
Oduvanchick [21]
4(x+1) is the only one equal to 0
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2 years ago
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