Answer:
length and width=4
height=8
Step-by-step explanation:
Hello to solve this problem we must propose a system of equations of 3x3, that is to say 3 variables and 3 equations.
Ecuation 1
Leght=Width
.L=W
Ecuation 2
To raise the second equation we consider that the length and width of 4 inches less than the height of the box
H-4=W
Ecuation 3
To establish equation number 3, we find the volume of a prism that is the result of multiplying length, width, and height
LxWxH=128
From ecuation 1(w=h)
solving for H
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<em>Using ecuation 2</em>
H-4=W
Now we find the roots of the equation, 2 of them are imaginary, and only one results in 4
W=4in
L=4in
to find the height we use the ecuation 2
H-4=W
H=4+W
H=4+4=8
H=8IN
Answer:
A, D, E
Step-by-step explanation:
I hope this is correct
In the diagram, the length of segment QV is 15 units.What is the length of segment TQ? The answer is C 14
The answer is 14 units
Subtract 1111 from both sides
5{e}^{{4}^{x}}=22-115e4x=22−11
Simplify 22-1122−11 to 1111
5{e}^{{4}^{x}}=115e4x=11
Divide both sides by 55
{e}^{{4}^{x}}=\frac{11}{5}e4x=511
Use Definition of Natural Logarithm: {e}^{y}=xey=x if and only if \ln{x}=ylnx=y
{4}^{x}=\ln{\frac{11}{5}}4x=ln511
: {b}^{a}=xba=x if and only if log_b(x)=alogb(x)=a
x=\log_{4}{\ln{\frac{11}{5}}}x=log4ln511
Use Change of Base Rule: \log_{b}{x}=\frac{\log_{a}{x}}{\log_{a}{b}}logbx=logablogax
x=\frac{\log{\ln{\frac{11}{5}}}}{\log{4}}x=log4logln511
Use Power Rule: \log_{b}{{x}^{c}}=c\log_{b}{x}logbxc=clogbx
\log{4}log4 -> \log{{2}^{2}}log22 -> 2\log{2}2log2
x=\frac{\log{\ln{\frac{11}{5}}}}{2\log{2}}x=2log2
Answer= −0.171
that is, hypotenuse.
Therefore, let the length of the hypotenuse be 'x.'
625+3600=
4225=
Therefore,
=x
65=x
The length of the hypotenuse is B.65 centimetres.
