Answer:
a.70. b.50
Step-by-step explanation:
- If Cartie takes $10 from her account for 7 days she would have taken $70 out.
![7 \times 10 = 70](https://tex.z-dn.net/?f=7%20%5Ctimes%2010%20%3D%2070)
- Next we look at the net problem it is asking us how much is left we already know from the equation above that we need to subtract 70 from 120
![70 - 120 = 50](https://tex.z-dn.net/?f=70%20-%20120%20%3D%2050%20)
- We now have both answers A. being that she would have taken $70 in total out of her account and B. being that she would have $50 left in her account.
additive inverse.
Step-by-step explanation:
Due to my beliefs considering -8 + 0 = -8 that is considerd additive inverse.
Answer:
True Statement are
![\sin E =\dfrac{11}{\sqrt{185}}](https://tex.z-dn.net/?f=%5Csin%20E%20%3D%5Cdfrac%7B11%7D%7B%5Csqrt%7B185%7D%7D)
![\sin D =\dfrac{8}{\sqrt{185}}](https://tex.z-dn.net/?f=%5Csin%20D%20%3D%5Cdfrac%7B8%7D%7B%5Csqrt%7B185%7D%7D)
Step-by-step explanation:
Given:
DF = 11
EF = 8
![DE = \sqrt{185}=Hypotenuse](https://tex.z-dn.net/?f=DE%20%3D%20%5Csqrt%7B185%7D%3DHypotenuse)
In Right Angle Triangle DEF, the Sine Identity is
![\sin D = \dfrac{\textrm{side opposite to angle D}}{Hypotenuse}\\](https://tex.z-dn.net/?f=%5Csin%20D%20%3D%20%5Cdfrac%7B%5Ctextrm%7Bside%20opposite%20to%20angle%20D%7D%7D%7BHypotenuse%7D%5C%5C)
Substituting the values we get
![\sin D = \dfrac{EF}{DE}=\dfrac{8}{\sqrt{185}}](https://tex.z-dn.net/?f=%5Csin%20D%20%3D%20%5Cdfrac%7BEF%7D%7BDE%7D%3D%5Cdfrac%7B8%7D%7B%5Csqrt%7B185%7D%7D)
....True
And also,
![\sin E = \dfrac{\textrm{side opposite to angle E}}{Hypotenuse}\\](https://tex.z-dn.net/?f=%5Csin%20E%20%3D%20%5Cdfrac%7B%5Ctextrm%7Bside%20opposite%20to%20angle%20E%7D%7D%7BHypotenuse%7D%5C%5C)
Substituting the values we get
![\sin E = \dfrac{DF}{DE}=\dfrac{11}{\sqrt{185}}](https://tex.z-dn.net/?f=%5Csin%20E%20%3D%20%5Cdfrac%7BDF%7D%7BDE%7D%3D%5Cdfrac%7B11%7D%7B%5Csqrt%7B185%7D%7D)
....True
Answer:
The question
after change the mixed number to improper & using KCF will become : ![\mathbf{=\frac{7}{2}\times \frac{3}{1}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%5Cfrac%7B7%7D%7B2%7D%5Ctimes%20%5Cfrac%7B3%7D%7B1%7D%7D)
Option B is correct option.
Step-by-step explanation:
We need to change the mixed number to improper & use KCF to rewrite the question.
The expression is: ![3\frac{1}{2}\div\frac{1}{3}](https://tex.z-dn.net/?f=3%5Cfrac%7B1%7D%7B2%7D%5Cdiv%5Cfrac%7B1%7D%7B3%7D)
First converting mixed fraction into improper fraction
Multiply whole number with the denominator i,e (3*2=6) now add the numerator i.e (6+1) = 7/2
Solving:
![3\frac{1}{2}\div\frac{1}{3}\\=\frac{7}{2}\div\frac{1}{3}](https://tex.z-dn.net/?f=3%5Cfrac%7B1%7D%7B2%7D%5Cdiv%5Cfrac%7B1%7D%7B3%7D%5C%5C%3D%5Cfrac%7B7%7D%7B2%7D%5Cdiv%5Cfrac%7B1%7D%7B3%7D)
Using KCF
KCF stands fro Keep it, Change it, Flip it.
It is used when division sign is converted to multiplication the term (1/3) is reversed to (3/1)
So, our expression will be
![=\frac{7}{2}\times \frac{3}{1}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B7%7D%7B2%7D%5Ctimes%20%5Cfrac%7B3%7D%7B1%7D)
So, The question
after change the mixed number to improper & using KCF will become : ![\mathbf{=\frac{7}{2}\times \frac{3}{1}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%5Cfrac%7B7%7D%7B2%7D%5Ctimes%20%5Cfrac%7B3%7D%7B1%7D%7D)
Option B is correct option.
Answer:
Option C is correct, i.e. 16x² + 24xy + 9y² = (4x + 3y)²
Step-by-step explanation:
Perfect square trinomial is an algebraic expression of three terms which can be combined into a squared binomial like (a+b)^2 = a^2 + 2ab + b^2.
From the given options, we can eliminate first two options because they have only two terms (they are not trinomials).
Checking 3rd option:- 16x² + 24xy + 9y²
16x² + 24xy + 9y² = (4x)² + 2(4x)(3y) + (3y)²
16x² + 24xy + 9y² = (4x + 3y)²
Checking 4th option:- 49x² - 70xy + 10y²
49x² - 70xy + 10y² = (7x)² - 2(7x)(5y) + 10y²
Hence, option C is correct, i.e. 16x² + 24xy + 9y² = (4x + 3y)²