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docker41 [41]
3 years ago
6

V² + 2v-8=0 this question accords using the quadratic formula as equation​

Mathematics
2 answers:
yaroslaw [1]3 years ago
8 0
Hope this helps with your answer.

kirill [66]3 years ago
6 0
That’s how to solve it
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Plz Answer this question ASAP!
nydimaria [60]

Answer:

564 tiles

Step-by-step explanation:

there are many ways to solve this soo...

1. convert to centimetres (there are 100 centimetres in 1 metre)

2. the square tiles area = 20*20=400cm

2. find the large area (4.8m x 3.8m or 480cm x 380cm)

480 *380=182400cm

2. find the two smaller areas (1.2m x 1.8m or 120cm x 180cm)

2(120*180)=4.3200cm

3. add to find total area

182400 +43200=225600cm

4. find how many tiles are needed

\frac{225600}{400}= 564 tiles

8 0
3 years ago
Question is in picture<br> IS MNL=QNL? Why or why not?
True [87]

Answer:

B) Yes, they are both right angles.

Step-by-step explanation:

3 0
2 years ago
Write the following in slope intercept form:
BabaBlast [244]
To change it into slope intercept (y=mx+b) you just replace the inequality with an equality and isolate y

x<=2:
x=2=y-> y=2 or y=2+0x

2x-3y>=4:
2x-3y=4
2x-4=3y
(2/3)x-(4/3))=y
y=(2/3)x-(4/3)

x+y>0:
x+y=0
y=-x or y=-x+0
7 0
3 years ago
Please help me out with this
Naily [24]

Answer:

y = -3x - 2

Step-by-step explanation:

Equation of a line passing through (x_1,y_1) and slope m is,

y-y_1=m(x-x_1)

Slope of a line passing through (x_1,y_1) and (x_2,y_2) is,

Slope = \frac{y_2-y_1}{x_2-x_1}

Slope of the line passing through (0, -2) and (1, -5) will be,

Slope = \frac{-2+5}{0-1}

m = - 3

Therefore, equation of the line passing through (0, -2) and slope = -3 will be,

y + 2 = -3(x - 0)

y + 2 = -3x

y = -3x - 2

8 0
3 years ago
Find the following Euler Totients using Euler’s Theorem, as explained on p.409 of the text (10 points each): a.ϕ(13) b.ϕ(81) c.ϕ
ad-work [718]

(a) \varphi(13)=12 since 13 is prime.

(b) 81=3^4, and there are 81/3 = 27 multiples of 3 between 1 and 81, which leaves 81 - 27 = 54 numbers between 1 and 81 that are coprime to 81, so \varphi(81)=54.

(c) 100=2^2\cdot5^2; there are 50 multiples of 2, and 20 multiples of 5, between 1 and 100; 10 of these are counted twice (the multiples of 2*5=10), so a total of 50 + 20 - 10 = 60 distinct numbers not coprime to 100, leaving us with \varphi(100)=100-60=40.

(d) 102=2\cdot3\cdot17; there are 51 multiples of 2, 34 multiples of 3, and 6 multiples of 17, between 1 and 102. Among these, we double-count 17 multiples of 2*3=6, 3 multiples of 2*17=34, and 2 multiples of 3*17=51; we also triple-count 1 number, 2*3*17=102. There are then 51 + 34 + 6 - (17 + 3 + 2) + 1 = 70 numbers between 1 and 102 that are not coprime to 102, and so \varphi(102)=102-70=32.

4 0
4 years ago
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