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shusha [124]
3 years ago
14

Find the value of X

Mathematics
2 answers:
emmasim [6.3K]3 years ago
7 0

Answer: 20 i think srry if its wrong

Step-by-step explanation:

HACTEHA [7]3 years ago
6 0

Answer:

x = -1

Step-by-step explanation:

<u>To solve this type of mathematical problem, you need to set up an equation.</u>

Since AB = 5 and BC = 2x + 6, AC must = 2x + 6 + 5.

AC is also given another equivalent value, x + 10. That means an equation is given to solve for x:

2x + 6 + 5 = x + 10

<u>Time for solving!</u>

2x + 6 + 5 = x + 10     First, bring x to one side by subtracting x on both sides.

x + 6 + 5 = 10             Now, simplify the equation to make it more readable.

x + 11 = 10                   Lastly, isolate x by subtracting 11 from both sides.

x = -1

The value of x is -1

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Original: $30.59<br> Discount:25%<br> New: ?
Scorpion4ik [409]

<em>Hello!</em>

The new value is $ 22,9425 or just $ 23 in a more simplified size.

We need to calculate 25% of 30.59.

We will do this by turning 25 into tenths:

  • 0,25 × 30,59
  • = 7,6475

Now, we subtract that from 30,59.

30,59 - 7,6475

= 22,9425

So, the new value is $ 22,9425 or just $ 23 in a more simplified size.

_______________

<em>Att. Isaiasdesign03</em>

<em>Moderator in Brainly BR</em>

<em>Sorry my English, I'm Brazilian. </em>☺

3 0
3 years ago
Sharon has some one-dollar bills and some five-dollar bills. She has 14 bills. The value of the bills is $30. Solve a system of
Scorpion4ik [409]

Answer:

10 one-dollar  and 4 five-dollar bills.

Step-by-step explanation:

Let x = the number of one-dollar bills and y = number of five-dollar bills.

Then:

x + y = 14

x + 5y = 30      (using the values of the bills)

Subtracting the 2 equations:-

-4y = -16

y = 4

so x + 4 = 14

x = 10.

There are 10  one-dollar bills and 4  five-dollar bills.

8 0
3 years ago
Find the surface area of the rectangular prism.<br> 4 in<br> 10 in<br> 3 in
yKpoI14uk [10]

Answer: 120 in

Step-by-step explanation: You have to multiply all the numbers. Remember length.width.height is how you get your answer.    .=multiply

3 0
3 years ago
BRAINLY FOR THE FASTEST WITH GOOD EXPLANATIONS
Neporo4naja [7]

So, when we're tasked with things like this, rewriting everything in terms of sine and cosine and combining fractions often trivializes things, so doing just that gives us:

\frac{\frac{1}{\cos{x}}-\frac{1}{\sin{x}}}{\frac{\cos{x}}{\sin{x}}-1}= \\\frac{\frac{=(\cos{x}-\sin{x})}{\cos{x}\sin{x}}}{\frac{\cos{x}}{\sin{x}}-\frac{\sin{x}}{\sin{x}}}=\\\frac{-(\cos{x}-\sin{x})}{\cos{x}\sin{x}}*\frac{\sin{x}}{\cos{x}-\sin{x}}=\\-\frac{1}{\cos{x}}

So out expression is 1/cos(x).

4 0
3 years ago
Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p
GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2

you replace t1 from the first equation in the second equation:

(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}

Then, for t2 = 0 and t2=1/2 you obtain for t1:

t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths  intersect are:

r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)

B) You have the following two trajectories of two independent particles:

r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t

From the first equation you have:

t=1+2t\\\\t=-1

This values does not have a physical meaning, then, the particle do not collide

answer: DNE

5 0
3 years ago
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