Answer:166
Step-by-step explanation:
SA=2lw+2lh+2hw so 2*4*5+2*7*5+2*7*4
so 40+70+56=166
Answer:
NUMBER 1.)
Step 1
Subtract 3y3y from both sides.
5x=10-3y5x=10−3y
Step 2
Divide both sides by 55.
\frac{5x}{5}=\frac{10-3y}{5}
5
5x
=
5
10−3y
Hint
Undo multiplication by dividing both sides by one factor.
Step 3
Dividing by 55 undoes the multiplication by 55.
x=\frac{10-3y}{5}x=
5
10−3y
Hint
Undo multiplication.
Step 4
Divide 10-3y10−3y by 55.
x=-\frac{3y}{5}+2x=−
5
3y
+2
Hint
Divide.
Solution
x=-\frac{3y}{5}+2x=−5
3y+2
Step-by-step explanation:
NUMBER 2.)
Step 1
Add 4y4y to both sides.
3x=6+4y3x=6+4y
Step 2
The equation is in standard form.
3x=4y+63x=4y+6
Step 3
Divide both sides by 33.
\frac{3x}{3}=\frac{4y+6}{3}
3
3x
=
3
4y+6
Hint
Undo multiplication by dividing both sides by one factor.
Step 4
Dividing by 33 undoes the multiplication by 33.
x=\frac{4y+6}{3}x=
3
4y+6
Hint
Undo multiplication.
Step 5
Divide 6+4y6+4y by 33.
x=\frac{4y}{3}+2x=
3
4y
+2
Hint
Divide.
Solution
x=\frac{4y}{3}+2x= 3
4y+2
Answer:
When you say that, I think directly to Triangles. SO i am going to state that.
There is acute: Less than 90 degrees
There is Obtuse: more than 90 degrees
There is Right: 90 degrees
I think it’s 2
Hope that helped sorry if it’s wrong
![\bf ln\left( \sqrt[7]{e^4} \right)\implies log_e\left( e^{\frac{4}{7}} \right)\\\\ -----------------------------\\\\ log_{{ a}}{{ a}}^x\implies x=x\impliedby \textit{let's use this cancellation rule}\\\\ -----------------------------\\\\ log_e\left( e^{\frac{4}{7}} \right)\iff \cfrac{4}{7}](https://tex.z-dn.net/?f=%5Cbf%20ln%5Cleft%28%20%5Csqrt%5B7%5D%7Be%5E4%7D%20%5Cright%29%5Cimplies%20log_e%5Cleft%28%20e%5E%7B%5Cfrac%7B4%7D%7B7%7D%7D%20%5Cright%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Alog_%7B%7B%20%20a%7D%7D%7B%7B%20%20a%7D%7D%5Ex%5Cimplies%20x%3Dx%5Cimpliedby%20%5Ctextit%7Blet%27s%20use%20this%20cancellation%20rule%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Alog_e%5Cleft%28%20e%5E%7B%5Cfrac%7B4%7D%7B7%7D%7D%20%5Cright%29%5Ciff%20%5Ccfrac%7B4%7D%7B7%7D)