Answer:
Amanda's box is in this case, the bigger one. Then subtract the Amanda's by Mary's. The Larger box is 70cm^3 larger then the smaller box.
Step-by-step explanation:
Amanda's box: 13.5*10*10= 1350cm^3
Mary's box: 20*8*8= 1280cm^3
1350-1280=70
The larger box is 70cm^3 larger then the smaller box.
If you have any questions regarding my answer tell me in the comments, i will come answer them. Have a good day.
Step-by-step explanation:
- y=3/4x+(-5)
- -4x+4(3/4x+-5)=-20
-4x+3x-20=-20
-x=-20+20
-x=0
x=0
y=3/4×{0}-5
y=-5
Using our congruence statement
![ABCD \cong LMNO](https://tex.z-dn.net/?f=ABCD%20%5Ccong%20LMNO)
, we know that corresponding angles match, so that means that
![\angle M \cong \angle B](https://tex.z-dn.net/?f=%5Cangle%20M%20%5Ccong%20%5Cangle%20B)
. We know that
![m \angle B=131](https://tex.z-dn.net/?f=m%20%5Cangle%20B%3D131)
, so that means that
![m \angle M=131](https://tex.z-dn.net/?f=m%20%5Cangle%20M%3D131)
. This gives us the equation
<em /><em /><em>m</em> + 45 = 131
Subtract 45 from both sides to cancel it and we have
<em>m</em> = 86.
The height attained by an objects thrown from ground level is given by the equation:
![v^2=u^2-2gs](https://tex.z-dn.net/?f=v%5E2%3Du%5E2-2gs)
At maximum height, v = 0, we assume
![g = 32 ft/s^2](https://tex.z-dn.net/?f=g%20%3D%2032%20ft%2Fs%5E2)
then we have:
![0=u^2-2(32)(49)\\ \\u^2=3,136\\ \\ u= \sqrt{3,136} =56](https://tex.z-dn.net/?f=0%3Du%5E2-2%2832%29%2849%29%5C%5C%20%5C%5Cu%5E2%3D3%2C136%5C%5C%20%5C%5C%20u%3D%20%5Csqrt%7B3%2C136%7D%20%3D56)
Therefore, the <span>minimum speed that the stone should be thrown so as to reach a height of 49 feet</span> is 56 feet per second.
Answer:
Answer is given below.
Step-by-step explanation:
3. Option a is correct
we ran a univariate ANOVA for 2 independent variables and 1 dependent variable.
for the bragger condition (truthful or non truthful), we got the F (1,116)=68.646 and p value = 0.000 < 0.01 that necessarily implies that there is no significant main effect of the bragger condition on the likeability of Edward.