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jekas [21]
3 years ago
6

13. Suppose a car dealer sells 2 sport-utility vehicles. How many mid-sized cars must be sold to earn at least S3500?

Mathematics
2 answers:
Alekssandra [29.7K]3 years ago
6 0

Answer:

incomplete information.What is the price of 1 car ?

Step-by-step explanation:

Marrrta [24]3 years ago
5 0
Some information isn’t listed for us to help you..
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Help please I will give brainleist
Natali5045456 [20]

Answer:

The answer is 80%

8 0
2 years ago
Subtract and simplify 20 2/7 - 26/35
marissa [1.9K]
20& 10/35 - 26/35 

19& 16/35 simplified to it's fullest.
4 0
3 years ago
Particle 1 of charge q1 �� ��5.00q and particle 2 of charge q2 �� ��2.00q are fixed to an x axis. (a) as a multiple of distance
Naya [18.7K]

<span>Assuming that the particle is the 3rd particle, we know that it’s location must be beyond q2; it cannot be between q1 and q2 since both fields point the similar way in the between region (due to attraction). Choosing an arbitrary value of 1 for L, we get </span>

<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>

Rearranging to calculate for d:

<span> (d-1)^2/d^2 = -q2/q1 = 0.4 </span><span>
<span> d^2-2d+1 = 0.4d^2 </span>
0.6d^2-2d+1 = 0  
d = 2.72075922005613 
d = 0.612574113277207 </span>

<span>
We pick the value that is > q2 hence,</span>

d = 2.72075922005613*L

<span>d = 2.72*L</span>

3 0
3 years ago
11/6358 long agurithum
miss Akunina [59]

Answer:

Least common multiple:

lcm (578; 11) = 6,358 = 2 × 11 × 172;

Numbers have no common prime factors: 6,358 = 578 × 11.

Step-by-step explanation:

Approach 1. Integer numbers prime factorization:

578 = 2 × 172;

11 is a prime number, it cannot be broken down to other prime factors;

Multiply all the prime factors, by the largest exponents.

Least common multiple:

lcm (578; 11) = 2 × 11 × 172;

lcm (578; 11) = 2 × 11 × 172 = 6,358

Numbers have no common prime factors: 6,358 = 578 × 11.

Integer numbers prime factorization

Approach 2. Euclid's algorithm:

Calculate the greatest (highest) common factor (divisor), gcf, hcf, gcd:

Step 1. Divide the larger number by the smaller one:

578 ÷ 11 = 52 + 6; Step 2. Divide the smaller number by the above operation's remainder:

11 ÷ 6 = 1 + 5; Step 3. Divide the remainder from the step 1 by the remainder from the step 2:

6 ÷ 5 = 1 + 1; Step 4. Divide the remainder from the step 2 by the remainder from the step 3:

5 ÷ 1 = 5 + 0; At this step, the remainder is zero, so we stop:

1 is the number we were looking for, the last remainder that is not zero.

This is the greatest common factor (divisor).

Least common multiple, formula:

lcm (a; b) = (a × b) / gcf, hcf, gcd (a; b);

lcm (578; 11) =

(578 × 11) / gcf, hcf, gcd (578; 11) =

6,358 / 1 =

6,358;

lcm (578; 11) = 6,358 = 2 × 11 × 172;

8 0
3 years ago
B) x⁴ + x² + 1 solve it​
LekaFEV [45]

Answer: x⁶︎+1 or 2⁶︎

Step-by-step explanation: you add the exponents and because x technically has a 1 in front of it, you add 1+1 and end up with 2⁶︎

7 0
2 years ago
Read 2 more answers
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