1) Answer: The part of the quadratic formula tells us whether the quadratic equation can be solved by factoring is b^2-4ac
2) 4x^2+6x+2=0
ax^2+bx+c=0; a=4, b=6, c=2
b^2-4ac=(6)^2-4(4)(2)=36-32
b^2-4ac=4
Answer: b^2-4ac=4
1 Face has 0 face painted
Step-by-step explanation:
As per the question a painted cube that is cut into 27 equal-size smaller cubes has been cut into a 3 * 3 * 3 arrangement.
There is 1 cube in the very center
On each of the 6 sides of the cube, there is a central smaller cube that is painted once.
Of the 27 (3*3*3) smaller cubes formed by the cutting, 1 gets no paint, 6 are painted on 1 face, 12 are painted on 2 faces, and 8 are painted on 3 faces (total 27 cubes).
1 (no paint) + 6 (painted 1) + 12 (painted 2) + 8 (painted 3) = 27 total
The sum of the painted faces (of the small cubes) is 54 (which is 6*9).
Answer:
C.) 2x+3 = 12
Step-by-step explanation:
x+x=2x
2x-(-3) is 2x+3
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Answer:
m∠A = 50°
m∠E = 50°
Step-by-step explanation:
7x - 17 and 2x + 8 are same side exterior angles. They have a sum of 180°.
7x - 17 + 2x + 8 = 180
9x - 9 = 180
9x = 189
x = 21
∠A and 7x - 17 are supplementary. ∠E and 2x + 8 are vertical angles.
m∠A + 7x - 17 = 180
m∠A + 7(21) - 17 = 180
m∠A + 147 - 17 = 180
m∠A + 130 = 180
m∠A = 50
m∠E = 2x + 8
m∠E = 2(21) + 8
m∠E = 42 + 8
m∠E = 50
firstly let's convert the mixed fraction to improper fraction, then hmmm let's see we have two denominators, 5 and 3, and their LCD will simply be 15, so we'll multiply both sides by that LCD to do away with the denominators, let's proceed,
![\bf \stackrel{mixed}{2\frac{1}{3}}\implies \cfrac{2\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{7}{3}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{z}{5}-4=\cfrac{7}{3}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{15}}{15\left( \cfrac{z}{5}-4 \right)=15\left( \cfrac{7}{3} \right)}\implies 3z-60=35 \\\\\\ 3z=95\implies z=\cfrac{95}{3}\implies z = 31\frac{2}{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B1%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%203%2B1%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B7%7D%7B3%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7Bz%7D%7B5%7D-4%3D%5Ccfrac%7B7%7D%7B3%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B15%7D%7D%7B15%5Cleft%28%20%5Ccfrac%7Bz%7D%7B5%7D-4%20%5Cright%29%3D15%5Cleft%28%20%5Ccfrac%7B7%7D%7B3%7D%20%5Cright%29%7D%5Cimplies%203z-60%3D35%20%5C%5C%5C%5C%5C%5C%203z%3D95%5Cimplies%20z%3D%5Ccfrac%7B95%7D%7B3%7D%5Cimplies%20z%20%3D%2031%5Cfrac%7B2%7D%7B3%7D)
