Answer:
Step-by-step explanation:
<u><em>The correct question is</em></u>
What value of b will cause the system to have an infinite number of solutions?
y = 6x – b
–3x + 1/2y = –3
we have
----->equation A
-----> equation B
we know that
If the system has infinite number of solutions then, the equation A must be equal to the equation B
so
isolate variable y in the equation B
Multiply by 
both sides
-------> new equation B
To find out the value of b, equate equation A and equation B
Answer:
- 
Step-by-step explanation:
Step 1: Add the numerator fractions.
2/5 + 3/10
Here we have to find LCD, the LCD of 5 and 10 is 10
(2*2) + 3 4 + 3
Therefore, 2/5 + 3/10 = --------------- = ------------------
10 10
= 7/10
Step 2: Now substitute 2/5 + 3/10 = 7/10 in the given fraction, we get
7/10
= ----------
-7/9
If we have fraction over fraction, we have to find the reciprocal of denominator fraction and multiply.
The reciprocal of -7/9 is -9/7
Step 3: Now multiply 13/10 and -9/7
= (7/10) x (-9/7)
= -9/10
=-9/10
The answer is -
Answer:
I am not sure .but i will try
i think so i will give correct answer
perpendicular of triangle is 22-10 cm =12 cm bcz both sides of triangle are equal =22
area of rectangle l×w= 22×17=374
area of triangle =h×b/2 Now we r not having h
lets find it
use pythagorus theorum (h2=P2+b2....2 here is square)
h=22.5cm
22.5×19=427(area of triangle )
374+427=801cm sq
i tried my best dont know its righr of wrong
pls mark me as brainlieast
2 and 3
negative numbers won't work, nor will 0
1 and 2 won't work because 1*8=2*4
2 and 3 is the answer
Answer:
SSS is the congruence theorem that can be used to prove Δ LON is congruent to Δ LMN ⇒ 1st answer
Step-by-step explanation:
Let us revise the cases of congruence
- SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
- AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd Δ
- HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ
In triangles LON and LMN
∵ LO ≅ LM ⇒ given
∵ NO ≅ NM ⇒ given
∵ LN is a common side in the two triangles
- That means the 3 sides of Δ LON are congruent to the 3 sides
of Δ LMN
∴ Δ LON ≅ LMN ⇒ by using SSS theorem of congruence
SSS is the congruence theorem that can be used to prove Δ LON is congruent to Δ LMN
