All categories

2 years ago

What is the solution of this equation -12x-7=53

Mathematics

2 answers:

umka21 [38]2 years ago

8 0

Answer:

<em>x </em>= -5

Step-by-step explanation:

You have to work backwards.

53 + 7 = 60

60 ÷ 12 = 5

Add the negative »»» -5

Taya2010 [7]2 years ago

5 0

X= -5

Explanation: Move the constant to the right-hand side and change its sign, Add the numbers, Divide both sides of the equation by -12.

You might be interested in

The Skubic family and the Shaw family went on a safari tour in which they drove their own vehicles through the park to view the

VMariaS [17]

Answer:

$12.75

Step-by-step explanation:

Skubic Family-

22.75 = 5.25 + 7x

17.5 = 7x

2.5 = x

Therefore the cost for each person is equal to $2.5

Shaw Family-

2.5(3) + 5.25

7.5 + 5.25 = $12.75

8 0

3 years ago

Read 2 more answers

F(x) = (1)<br> What is the x intercept

Vinvika [58]

Answer:

There is no x-intercept, only a y-intercept, which would be 1

8 0

3 years ago

A small metal bar, whose initial temperature was 10° C, is dropped into a large container of boiling water. How long will it tak

IgorC [24]

Answer

According to newton law of cooling

$\dfrac{d T}{dt} = k (T - T_a)$

and

$T = Ce^{kt} + T_a$

now At Ta = 100

T₀ = 10 °C

T₁ = 12° C

$T(0) = Ce^{0} + T_a$

10 = C + 100

C = -90

now,

$12 = -90e^k + 100$

$e^k = \dfrac{88}{90}$

$k = ln{\dfrac{88}{90}}$

at time equal to t

T(t) = 70

$T(t) = -90 (\dfrac{88}{90})^t + 100$

$70 = -90(\dfrac{88}{90})^t + 100$

$(\dfrac{88}{90})^t = \dfrac{1}{3}$

$t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{3})$

t = 48.88 s

hence, after 48.88 s the temperature of the body will be 70°C

b) time taken to reach 98°C

$T(t) = -90 (\dfrac{88}{90})^t + 100$

$98 = -90(\dfrac{88}{90})^t + 100$

$(\dfrac{88}{90})^t = \dfrac{1}{45}$

$t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{45})$

t = 390 s

hence, after 390 s the temperature of the body will be 98°C

5 0

3 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.

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