By which rule are these triangles congruent ?<br> A) AAS<br> B) ASA<br> C) SAS<br> D) SSS

The value of n is both 5 times as much as the value of m and 36 more than the value of m. what are the values of n and m? explai

Cloud [144]

I'm in seventh grade. I haven't learned that yet.
try asking your teacher.

8 0

3 years ago

When the family arrived in New Delhi from Rome, the youngest son asked the pilot how fast he was flying the plane. The pilot tol

Elden [556K]

Answer:

7 hours

Step-by-step explanation:

It is given that the plane was flying at 847 km/hour. So the speed of the plane was 847 kilometers per hour.

It is also given that the distance between Rome and New Delhi is 5,929 kilometers.

We know that the distance is equal to the product of speed and time, so we have,

$d= s\times t$ , where 'd' represents the distance, 's' represents the speed, and 't' represents the time.

Putting the values we get,

$5929= 847 \times t$

Solving for 't' we get,

$t= \frac{5929}{847}=7$

So the time taken to fly from Rome to New Delhi is 7 hours.

7 0

3 years ago

PLEASE ANSWER FAST!!!

Ilya [14]

S= 11/12
15%-p=x
35-$18=n

5 0

3 years ago

Read 2 more answers

Which graph shows the solution to the system of linear inequalities?

Aleksandr [31]

Answer:

its d

Step-by-step explanation:

edge

3 0

3 years ago

Read 2 more answers

Log(x-2)+log2=2logy<br>log(x-3y+3)=0

In-s [12.5K]

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}\log(x-2)+\log2=2\log y&(1)\\\log(x-3y+3)=0&(2)\end{array}\right$

$===========================\\\text{DOMAIN:}\\\\x-2>0\to x>2\\y>0\\x-3y+3>0\to x-3y>-3\\=====================\\\\\text{We know:}\ \log_ab=c\iff a^c=b,\\\\\text{therefore}\ \log_ab=0\iff a^0=b\to b=1\\\\\text{From this we have}\\\\\log(x-3y+3)=0\iff x-3y+3=1\qquad\text{add}\ 3y\ \text{to both sides}\\\\x+3=3y+1\qquad\text{subtract 3 from both isdes}\\\\\boxed{x=3y-2}\qquad(*)\\\\\text{Substitute it to (1)}$

$\log(3y-2-2)+\log2=2\log(y)\qquad\text{use}\ \log_ab^n=n\log_ab\\\\\log(3y-4)+\log2=\log(y^2)\qquad\text{subtract}\ \log(y^2)\ \text{from both sides}\\\\\log(3y-4)+\log2-\log(y^2)=0\\\\\text{Use}\ \log_ab+\log_ac=\log_a(bc)\ \text{and}\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\log\dfrac{(3y-4)(2)}{y^2}=0\qquad\text{use}\ \log_ab=0\Rightarrow b=1\\\\\dfrac{(3y-4)(2)}{y^2}=1\iff y^2=(3y-4)(2)\\\\\text{Use the distributive property}\ a(b+c)=ab+ac$

$y^2=(3y)(2)+(-4)(2)\\\\y^2=6y-8\qquad\text{subtract}\ 6y\ \text{from both sides}\\\\y^2-6y=-8\qquad\text{add 8 to both sides}\\\\y^2-6y+8=0\\\\y^2-2y-4y+8=0\\\\y(y-2)-4(y-2)=0\\\\(y-2)(y-4)=0\iff y-2=0\ \vee\ y-4=0\\\\\boxed{y=2\ \vee\ y=4}\in D$

$\text{Put the values of}\ y\ \text{to }\ (*):\\\\\text{for}\ y=2:\\\\x=3(2)-2\\\\x=6-2\\\\\boxed{x=4}\in D\\\\\text{for}\ y=4:\\\\x=3(4)-2\\\\x=12-2\\\\\boxed{x=10}\in D$

4 0

2 years ago

By which rule are these triangles congruent ? A) AAS B) ASA C) SAS D) SSS