Answer:
Step-by-step explanation:
In ΔABC, we have,
∠B=2∠C or ∠B=2y where ∠C=y
AD is the bisector of ∠BAC. So, Let ∠BAD=∠CAD=x
Let BP be the bisector of ∠ABC.
In ΔBPC, we have
∠CBP = ∠BCP = y ⇒ BP = PC ... (1)
Now, in ΔABP and ΔDCP, we have
∠ABP = ∠DCP = y
AB = DC [Given]
and, BP = PC [Using (1)]
So, by SAS congruence criterion, we have
Δ ABP ≅ Δ DCP
Therefore
∠BAP = ∠ CDP = 2x and AP = DP ,
So in Δ APD, AP=DP
=> ∠ADP = ∠DAP = x
In ΔABD, we have
∠ADC = ∠ABD + BAD ⇒ 3x= 2y + x
⇒ x = y
In ΔABC, we have
∠A + ∠B + ∠C = 180°
⇒ 2x + 2y + y = 180°
⇒ 5x = 180°
⇒ x = 36°
Hence, ∠BAC = 2x = 72°
Answer:
a = 7
Step-by-step explanation:
3(1 - 2a) = -3a - 18
3 - 6a = - 3a - 18
Add 18 to each side:
3 + 18 - 6a = - 3a - 18 + 18
21 - 6a = - 3a
Add 6a to each side:
21 - 6a + 6a = - 3a + 6a
21 = 3a
3a = 21
Divide each side by 3:
3a ÷ 3 = 21 ÷ 3
a = 7
Answer:
x = ⅔ hr
V = 4 km³
Step-by-step explanation:
The volume of the first cloud is given as:
V1 = 4^(3x - 1)
The volume of the second cloud is given as:
V2 = 8^x
When the two clouds have the same volume, V1 = V2:
=> 4^(3x - 1) = 8^x
We can rewrite this as:
2^[2(3x - 1)] = 2^[3(x)]
=> 2^(6x - 2) = 2^(3x)
According to the law of indices, we can equate the powers because their bases are equal (2 is the base).
=> 6x - 2 = 3x
6x - 3x = 2
3x = 2
x = ⅔ hr
Hence, after 2/3 hr, their volumes will be equal.
To find the volume after 2/3 hr, we can use the equation of either cloud but let us use the equation for the second cloud for simplicity sake:
V2 = 8^(⅔)
V2 = 4 km³ or 4 cubic kilometers
Answer:
oh expressions are numbers that represent variables
Ok so standard form is a bit tricky, but I’ll help explain
Here’s and example:
If you have 2x+5y=10 here is how you would solve it
First you divide the x value by what the number is equal to (in our example, 10), and then we get 10/2 is 5. So now we know that
x is equal to five
Keep that in mind. Now we just have to do the same thing for the y value. 5/10 is 2, so you know that
y is equal to two
NOW you put those two together, and graph on the intercepts (the positions on the y and x axis where the lines intercept)
You would put a line on the x axis at 5, and on the y axis at 2
If you have any questions, you can ask