All categories

3 years ago

De acordo com o estudo da equação do 1o grau, determine o valor de x que satisfaz a equação abaixo ?

Mathematics

1 answer:

andreyandreev [35.5K]3 years ago

5 0

Answer:

11/5

Step-by-step explanation:

2x+12+4-12x+6

-10x+22

x=22/10

x= 11/5

22/5+12+4=123/5-6

102/5=123/5-30/5

102/5=97/5

102/5-97/5

5/5

You might be interested in

Find the number that makes the ratio equivalent to 12:1.<br> ?:6

Naily [24]

Right side was multiplied by 6 so you multiply the left side by 6 so 72:6

5 0

3 years ago

EMERGENCY!!PLEASE HELP

Ronch [10]

The more plot point the better but you must have at least three points, a labeled X-axis and Y-axis, and a table for the data to be organized into.

8 0

4 years ago

Read 2 more answers

I really need your help, but I don't get any.

STatiana [176]

Answer:

Step-by-step explanation:

x < -12, so x+12 < 0

| x-(-12) | = | x+12 | = -x-12

5 0

3 years ago

Q1.The cost of $16.50 for the music was reduction of 12% on the original cost.

NNADVOKAT [17]

Q1) £16.50 x 1.12 = £18.48
Q2)

2+9+1= 12
3h 33m = 213m
213/12 = 17.75
2x17.75 = 35.5 (music)
9x17.75 = 159.75 (films)
1x17.75 = 17.75 (books)

So it’s 159.75 minutes for the films to download

7 0

3 years ago

Given: PRST is a square

xxTIMURxx [149]

Answer:

$(1-\sqrt{2})a^2$

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

$PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a$

Thus,

$MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a$

Consider isosceles triangle MSC. In this triangle

$MS=MC=(\sqrt{2}-1)a.$

The area of this triangle is

$A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}$

Consider right triangle PTS. The area of this triangle is

$A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}$

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

$A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2$

5 0

3 years ago