7-15 please!!! I would really appreciate it!!!

Mathematics

1 answer:

vichka [17]3 years ago

7 0

7. x=45 The angle is a right angle, which is 90 degrees. 45+45=90
8. x=45 3x+x=180>> 3(45)+45=180>> 135+45=180
9. x=30 2x+x=90>> 2(30)+30=90>> 60+30=90
10. x=50 2x+(x+30)=180>> 2(50)+(50+80)=180>> 100+(80)=180
11. x=30 (x+20)+(x+10)=90>> (30+20)+(30+10)=90>> 50+40=90
12. x=25 (x+20)+(5x+10)=180>> (25+20)+(5(25)+10)=180>> 45+(125+10)=180>> 45+135=180
13. x=9 6x+4x=90>> 6(9)+4(9)=90>> 54+36=90
14. x=34 (2x-10)+(3x+20)=180>> (2(34)-10)+(3(34)+20)=180>> (68-10)+(102+20)=180>> 58+122+180
15. x=18 (2x+15)+(3x-15)=90>> (2(18)+15)+(3(18)-15)=90>> (36+15)+(54-15)=90>> 51+39=90

You might be interested in

Arrange the following from least to greatest 0.08,0.1, 1.8, 0.049, 0.103

8_murik_8 [283]

0.049, 0.08, 0.1, 0.103, 1.8.

3 0

3 years ago

Each side of a square is increasing at a rate of 5 cm/s. At what rate (in cm2/s) is the area of the square increasing when the a

Naddika [18.5K]

The area of the square increasing at 30 $\frac{cm^{2} }{sec}$

The rate of change function is defined as the rate at which one quantity is changing with respect to another quantity. In simple terms, in the rate of change, the amount of change in one item is divided by the corresponding amount of change in another

Let the side of a square be x.

So,

Area of square = $x^{2}$

A = $x^{2}$

Differentiating with respect to time we get,

dA/dt = 2x dx/dt

When area = 9 $cm^{2}$ the side becomes 3 cm

At x = 3cm and dx/dt = 5 cm/s (Given)

dA/dt = 2.3.5 = 30 $\frac{cm^{2} }{sec}$

Thus the area of the square increasing at 30 $\frac{cm^{2} }{sec}$

Learn more about rate of change here :

brainly.com/question/12786410

#SPJ4

4 0

2 years ago

The graph of f(x) and g(x) are shown below Which of the following is the graph of (g-f)(x)

anyanavicka [17]

Answer: (g-f)(x)=-4x

Explanation: According to the graph, line created by function f(x) passes through the points (1,-3) and (0,0) and similarly, line created by the function passes through the points (1,1) and (0,0).

Thus, we can find the equation of the lines with help of formula $y-y_1=\frac{y_2-y_1}{x_2-x_1}$ × $(x-x_1)$