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3 years ago

Solve. 4x + 10= 2x – 14

Mathematics

2 answers:

hammer [34]3 years ago

8 0

Answer:

x = -12

Step-by-step explanation:

Step 1: Subtract 2x from both sides

2x + 10 = -14

Step 2: Subtract 10 from both sides

2x = -24

Step 3: Divide both sides by 2

x = -12

dalvyx [7]3 years ago

4 0

Answer:

x = -12

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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Cara has 25$ to buy dry pet food and treats for the animal shelter. A pound of dog food cost $2 and treats are $1 apiece. If she

alina1380 [7]

Let x be the number of treats bought
Let y be the pounds of food bought

25= 2y+x

Given that she has buys 9 pounds of dog food:

25= 2(9)+x
25=18 +x
25-18 =x
7=x

With the money she has left over, she can buy 7 treats.

Hope I helped :)

5 0

3 years ago

Twenty years ago, you began investing $3,000 a year. Because your investments earned an average of 8 percent a year, your invest

Ulleksa [173]

Answer:

$32,000

Step-by-step explanation:

Yearly investment = $3,000

Number of years = 20

Investments earned an average of 8 percent a year.

Total invested amount = $3,000 × 20 = $60,000

Current value of investment = $92,000.

We need to find the total earnings on the investment.

Earnings = Current value of investment - Total invested amount

Earnings = $92,000 - $60,000

Earnings = $32,000

Therefore, the total earnings is $32,000.

6 0

2 years ago

PLEASE! Someone help me answer this and explain it

xenn [34]

Answer:

12.5

Step-by-step explanation:i did that problem before

3 0

2 years ago

For integers a, b, and c, consider the linear Diophantine equation ax C by D c: Suppose integers x0 and y0 satisfy the equation;

Dmitrij [34]

Answer:

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

b. x = -8 and y = 4

Step-by-step explanation:

This question is incomplete. I will type the complete question below before giving my solution.

For integers a, b, c, consider the linear Diophantine equation

$ax+by=c$

Suppose integers x0 and yo satisfy the equation; that is,

$ax_0+by_0 = c$

what other values

$x = x_0+h$ and $y=y_0+k$

also satisfy ax + by = c? Formulate a conjecture that answers this question.

Devise some numerical examples to ground your exploration. For example, 6(-3) + 15*2 = 12.

Can you find other integers x and y such that 6x + 15y = 12?

How many other pairs of integers x and y can you find ?

Can you find infinitely many other solutions?

From the Extended Euclidean Algorithm, given any integers a and b, integers s and t can be found such that

$as+bt=gcd(a,b)$

the numbers s and t are not unique, but you only need one pair. Once s and t are found, since we are assuming that gcd(a,b) divides c, there exists an integer k such that gcd(a,b)k = c.

Multiplying as + bt = gcd(a,b) through by k you get

$a(sk) + b(tk) = gcd(a,b)k = c$

So this gives one solution, with x = sk and y = tk.

Now assuming that ax1 + by1 = c is a solution, and ax + by = c is some other solution. Taking the difference between the two, we get

$a(x_1-x) + b(y_1-y)=0$

Therefore,

$a(x_1-x) = b(y-y_1)$

This means that a divides b(y−y1), and therefore a/gcd(a,b) divides y−y1. Hence,

$y = y_1+r(\frac{a}{gcd(a, b)})$ for some integer r. Substituting into the equation

$a(x_1-x)=rb(\frac{a}{gcd(a, b)} )\\gcd(a, b)*a(x_1-x)=rba$

$x = x_1-r(\frac{b}{gcd(a, b)} )$

Thus if ax1 + by1 = c is any solution, then all solutions are of the form

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

In order to find all integer solutions to 6x + 15y = 12

we first use the Euclidean algorithm to find gcd(15,6); the parenthetical equation is how we will use this equality after we complete the computation.

$15 = 6*2+3\\6=3*2+0$