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Nostrana [21]
3 years ago
7

A3 + 27 what sign is the first binomial

Mathematics
1 answer:
lakkis [162]3 years ago
3 0

Answer:

a3+27

a +BC

=a 30

30+a

=30ⁿ

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<img src="https://tex.z-dn.net/?f=%28v%2B6%29%5E%7B2%7D%3D2v%5E%7B2%7D%2B14v%2B12" id="TexFormula1" title="(v+6)^{2}=2v^{2}+14v+
timofeeve [1]

Answer:

v=-6 or 4

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
What is the point slope form of Y=1/2X-2
vagabundo [1.1K]
Point-slope form of a line: We need a point (x₀,y₀) and the slope "m";
y-y₀=m(x-x₀)

We have the next equation of line:
y=1/2 x-2  (slope-intercept form y=mx+b)

the slope of this line is 1/2  (m=1/2)
And any one point could be:
if x=0; then y=1/2 (0)-2=-2  (0,-2)

Therefore, we already have the point (0-,2) and the slope (m=1/2)
y-y₀=m(x-x₀)
y+2=1/2(x-0)

Answer: the point slope form of y=1/2 x-2; would be:
y+2=1/2(x-0)
8 0
3 years ago
ANSWER ASAP<br> THANKS<br> ...............
mylen [45]

Answer:

250

Step-by-step explanation:

5*5=25

25*10=250

+$++$$+$

6 0
3 years ago
Solve x plssssss :)))))
blondinia [14]

Answer:

x = 43

Step-by-step explanation:

180 - 65 = 115

115 = x + 72

<u>-72        -72 </u>

43 = x

x = 43

7 0
2 years ago
Verify a(b-c)=ab-ac for a=1.6;b=1/-2;&amp; c=-5/-7​
harina [27]

Given:

a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}

To verify:

a(b-c)=ab-ac for the given values.

Solution:

We have,

a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}

We need to verify a(b-c)=ab-ac.

Taking left hand side, we get

a(b-c)=1.6\left(\dfrac{1}{-2}-\dfrac{-5}{-7}\right)

a(b-c)=1.6\left(-\dfrac{1}{2}-\dfrac{5}{7}\right)

Taking LCM, we get

a(b-c)=1.6\left(\dfrac{-7-10}{14}\right)

a(b-c)=\dfrac{16}{10}\left(\dfrac{-17}{14}\right)

a(b-c)=\dfrac{8}{5}\left(\dfrac{-17}{14}\right)

a(b-c)=-\dfrac{68}{35}\right)

Taking right hand side, we get

ab-ac=1.6\times \dfrac{1}{-2}-1.6\times \dfrac{-5}{-7}

ab-ac=-\dfrac{16}{20}-\dfrac{8}{7}

ab-ac=-\dfrac{4}{5}-\dfrac{8}{7}

Taking LCM, we get

ab-ac=\dfrac{-28-40}{35}

ab-ac=\dfrac{-68}{35}

Now,

LHS=RHS

Hence proved.

7 0
2 years ago
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