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Debora [2.8K]
3 years ago
10

the sum of four times a number and another number is 10. The difference of the two numbers is 35. What is the larger number? Als

o please explain how you got that answer
Mathematics
2 answers:
valkas [14]3 years ago
7 0

Answer:

The two numbers are 9 and -26.

The larger number is 9.

Step-by-step explanation:

Let the two unknown numbers be x and y.

The sum of a four times a number x and another number y is 10. Hence:

4x+y=10

And the difference between the two numbers is 35. So:

x-y=35

So, we have the system of equations:

\left\{ \begin{array} \ 4x+y=10 \\ x-y=35 \end{array}

From the second equation, we can add y to both sides:

x=y+35

Now, we can substitute this into the first equation:

4(y+35)+y=10

Distribute:

4y+140+y=10

Simplify:

5y+140=10

Subtract 140 from both sides:

5y=-130

Divide both sides by 5:

y=-26

So, the value of y is -26 .

Then it follows that the value of x is:

x=y+35

Substitute -26 for y:

x=-26+35=9

So, our two numbers are 9 and -26.

The larger number is 9.

So, our answer is 9.

Irina18 [472]3 years ago
6 0

Answer:

Yup answers 9

Step-by-step explanation: just took the test

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84

Step-by-step explanation:

6 times 7 is 42

42 times 2 is 84

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3 years ago
The least common multiple of two numbers is 60. the prime factorization of one number is 3 times 5. what is the prime factorizat
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3 years ago
Please answer this correctly
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2 years ago
One number exceeds another by 24. The sum of the numbers<br> is 58. What are the numbers?
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3 0
3 years ago
A.
Reptile [31]

Answer:

The real zeros of f(x) are x = 0.3 and x = -3.3.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this problem, we have that:

f(x) = x^{2} + 3x - 1

So

a = 1, b = 3, c = -1

\bigtriangleup = 3^{2} - 4*1*(-1) = 13

x_{1} = \frac{-3 + \sqrt{13}}{2*1} = 0.3

x_{2} = \frac{-3 - \sqrt{13}}{2*1} = -3.3

The real zeros of f(x) are x = 0.3 and x = -3.3.

7 0
3 years ago
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