All categories

3 years ago

How do I do one of these types of problems?

Mathematics

1 answer:

otez555 [7]3 years ago

3 0

Answer:

x = 5

Step-by-step explanation:

I'm assuming the question is asking to solve for variable x.

The two red arrows signify the two lines are congruent, meaning they equal the same measurement.

In other words:

23x - 5 = 21x + 5

Isolate the variable x

Add 5 to both sides to cancel 5 from the left side

23x - 5 + 5 = 21x + 5 + 5

23x + 0 = 21x + 10

23x = 21x + 10

Subtract 21x from both sides to cancel 21x from the right side

23x - 21x = 21x - 21x + 10

2x = 0x + 10

2x = 10

Divide both sides by 2 to isolate the variable x

2x ÷ 2 = 10 ÷ 2

x = 5

When solving for these types of problems, always isolate the variable. Also, keep in mind that whatever you do to one side, you must do to the other (such as how I added 5 to <em>both</em> sides).

You might be interested in

Someone help me please what is 4.3 x 20

Mars2501 [29]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

AB has coordinates A(-5,9) and B(7,- 7). Points P, Q, and I are collinear

VMariaS [17]

Answer:

$\overline{QT}$

Step-by-step explanation:

We want to find the coordinates of a certain point C(x,y) such that C divides $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio m:n=3:2

The x-coordinate is given by:

$x=\frac{mx_2+nx_1}{m+n}$

The y-coordinate is given by:

$y=\frac{my_2+ny_1}{m+n}$

AB has coordinates A(-5,9) and B(7,- 7)

We substitute the values to get:

$x=\frac{3*7+2*-5}{3+2}$

$x=\frac{21-10}{5}$

$x=\frac{11}{5}$

and

$y=\frac{3*-7+2*9}{3+2}$

$y=\frac{-21+18}{5}$

$y=-\frac{3}{5}$

Therefore C has coordinates $(\frac{11}{5},-\frac{3}{5})$

The line segment that contains C is $\overline{QT}$

See attachment.

6 0

3 years ago

Which of the following correctly expresses sin (70) + sin(30) as a product?

JulsSmile [24]

Answer:

2sin50 cos20

Step-by-step explanation:

We need to write sin (70) + sin(30) as a product. The formula used here is :

$\sin A+\sin B=2\sin (\dfrac{A+B}{2})\cos(\dfrac{a-b}{2})$

Here, A = 70 and B = 30

So,

$\sin 70+\sin 30=2\sin (\dfrac{70+30}{2})\cos(\dfrac{70-30}{2})\\\\\sin 70+\sin 30=2\sin 50\cos20$

So, the value of sin (70) + sin(30) is 2sin50 cos20. Hence, the correct option is (c).

7 0

3 years ago

if we increased one side of a square by 5 units and decreased the other by 3 units the area of the resulting rectangle would be

zvonat [6]

Answer:

18 units

Step-by-step explanation:

So let's list out the sides.

for the first square let's just call them x

for the second square then they would be x+5 and x-3

So let's write out their areas we will cal the area of the first one z

x*x = z

(x+5)*(x-3) = z+21

since z = x^2 we can set up the second equation as a quadratic.

(x+5)*(x-3) = x^2 + 21

x^2 - 3x + 5x - 15 = x^2 + 21

But look, the x^2s cancel out

2x - 15 = 21

2x = 36

x = 18

Test it out and see if it fits the description, And if you don't understand anything just let me know so I can explain more.

3 0

3 years ago

Solve for the unknown variable. 144=-12(x+5)

dolphi86 [110]

Answer: x=-17

Step-by-step explanation:

We first distribute the -12 across to the (x+5)

144=-12(x+5)

144=-12x-60

Then, we add 60 to both sides.

204 = -12x

Lastly, we divide the equation by -12.

204/-12 = -12x/-12

-17 = x

3 0

3 years ago

Read 2 more answers