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yuradex [85]
3 years ago
7

13. The outboard engine on Steve's boat uses a fuel-to-oil mixture in a 50:1 ratio. If Steve puts 2.5 gallons of fuel in the tan

k, how many fluid ounces of oil does he need to add?
Mathematics
1 answer:
Nataly [62]3 years ago
7 0
You would put 2.5 over 50 and x over 1. Then would cross multiply then divide by 2.5 and should get the answer.

I hope you do well
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(a) Simplify the expression: (4x - 6) - (x + 5)
Liono4ka [1.6K]

Answer:

a) 3x - 11

b) 10x + 4

Step-by-step explanation:

a) The negative sign before the (x+5) means to subtract those values. Reverse the signs. Then combine like terms:

4x –6 –x –5. 4x–x–6–5

3x –11

b) Put all the x terms together.

Put all the integers together.

Add. Adding a negative is the same as subtracting.

4x+x+5x = 10x

+6-2 = 4

Combine them:

10x + 4

8 0
3 years ago
Factor the polynomials
photoshop1234 [79]
C. -X to the 3rd power, 4x to the second power and 5x all share the common factor or X, therefore you put it outside the parenthesis. What you have left is -X to the second power minus 4x minus 5.
4 0
3 years ago
Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
3 years ago
What are the roots of the equation x2 - 5x + 6 = 0?
Alenkasestr [34]

The answer is

B. -2 and 3

Step-by-step explanation :

Given the expression

x²- 5x + 6 = 0

Firstly we need to find two numbers that when multiplied will give us the constant term 6, and when added will give us the 2nd term 5

These numbers are 3 and 2

Substituting 2x +3x for 5x

x²- (2x+3x)+ 6 = 0

x²-2x-3x+6=0

(x²-2x)-(3x+6)=0

Factoring we have

x(x-2)-3(x-2)=0

x-3=0, x-2=0

x=3, x= 2

7 0
3 years ago
1. When do you need to state a domain in your final answer?
Ket [755]

<em>Answer:</em>

<em>Divide them all</em>

<em>Hope you like it!</em>

5 0
3 years ago
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