Answer:
The value of tan P = 1.875
Step-by-step explanation:
Given:
PR=17
RQ=15
PQ =8
To find:
tan P = ?
Solution:
In trigonometric ratio
tan =
Now on substituting the given values(refer the figure)
tan (P) =
tan (P) = 1.875
I’m thinking it’s the third option but not 100% sure, sowy
I think the answer would be rhombus
Answer:
Trina is 13
Kieth is 18
Step-by-step explanation:
x + 5 = Kieth's age
Divide by 2 to separate the ages:
62 ÷ 2 = 31
Subtract the difference in ages:
31 - 5 = 26
Add the 5 years to Kieth's age:
31 + 5 = 36
Divide by 2 again, to eliminate the twice of their ages
26 ÷ 2 = 13
36 ÷ 2 = 18
Trina is 13
Kieth is 18
This is what i got. Sorry if i'm wrong, lmk if i am!
Hope this helps! Brainliest if i'm correct!?
Answer:
x - 1\x + 3
Step-by-step explanation:
Factoring quadratic expressions with a Leading Coefficient of 1 → In the first equation, you have to find two numbers that when differed to 5, they also multiply to 6, and those numbers are 1 and 6. Now, the tough part for you might be figuring out which term gets which sign. Well, if you look at your MIDST term [5], you would know that the negative symbol goes to 1 [-1] and the positive symbol goes to 6, so your numerator is <em>[</em><em>x</em><em> </em><em>-</em><em> </em><em>1</em><em>]</em><em>[</em><em>x</em><em> </em><em>+</em><em> </em><em>6</em><em>]</em><em>.</em><em> </em>Now, for the second equation, it is applied the same way, but in this case, we need two numbers that when added to 9, they also multiply to 18, and those numbers are 3 and 6, and automatically receive positive symbols, so your denominator is <em>[</em><em>x</em><em> </em><em>+</em><em> </em><em>3</em><em>]</em><em>[</em><em>x</em><em> </em><em>+</em><em> </em><em>6</em><em>]</em><em>.</em><em> </em>Now that we have our denominator and numerator, we now set it up: [x - 1][x + 6]\[x + 6][x + 3]. What do you see that is... MAGIC--AL? That is correct! The factors <em>x</em><em> </em><em>-</em><em> </em><em>6</em><em> </em>neutralize each other and are left with <em>x - 1\x + </em><em>3</em><em>.</em>
To be honest, if you had posted quadratic expressions with Leading Coefficients greater than 1, that would be a little bit more tough for you, meaning taking extra steps further, but if you post one in the future, it will be there to assist you because as always...,
I am joyous to assist anyone at any time.