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Mnenie [13.5K]
3 years ago
11

5 less than 18 times the number x, is less than of equal to -45. write and inequality for this statement

Mathematics
1 answer:
swat323 years ago
8 0

Answer:

18x-5≤-45

Step-by-step explanation:

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How many different strings can be made from the letters in MISSISSIPPI, using all the letters
leonid [27]
Hello,

Answer is 11!/(1!*4!*4!*2!)=34650

M: 1
I:4
S:4
P:2

8 0
4 years ago
D) p + 6pq - 2pq- 129<br>g) (2x - 3y) +b"(3y - 2x) + e- (2x - 3y)​
Marizza181 [45]

d=

Let's simplify step-by-step.

p+6pq−2pq−129

=p+6pq+−2pq+−129

Combine Like Terms:

=p+6pq+−2pq+−129

=(6pq+−2pq)+(p)+(−129)

=4pq+p+−129

Answer:

=4pq+p−129

g=

Let's simplify step-by-step.

2x−3y+b(3y−2x)+2.718282−(2x−3y)

Distribute the Negative Sign:

=2x−3y+b(3y−2x)+2.718282+−1(2x−3y)

=2x+−3y+b(3y−2x)+2.718282+−1(2x)+−1(−3y)

=2x+−3y+b(3y−2x)+2.718282+−2x+3y

Distribute:

=2x+−3y+(b)(3y)+(b)(−2x)+2.718282+−2x+3y

=2x+−3y+3by+−2bx+2.718282+−2x+3y

Combine Like Terms:

=2x+−3y+3by+−2bx+2.718282+−2x+3y

=(−2bx)+(3by)+(2x+−2x)+(−3y+3y)+(2.718282)

=−2bx+3by+2.718282

Answer:

=−2bx+3by+2.718282

6 0
3 years ago
Using substitution method 4x -2y = 18 -4x+2y= -18
dedylja [7]

x= 9
y= 9

4x -2y= 18.................equation 1

-4x + 2y= -18...............equation 2

From equation 1

4x= 18 + 2y

x= 18 + 2y/4

substitute 18+2y/4 for x in equation 2

-4(18+2y/4) -2y= 18

-(18+2y)-2y= 18

-18-2y-2y= 18

-18+4y= 18

4y= 18+18

4y= 36

y= 36/4

y= 9

Substitute 9 for y in equation 1

4x-2y= 18

4x-2(9)= 18

4x-18= 18

4x= 18+18

4x= 36

x= 36/4

x= 9

Hence the value of x is 9 and the value of y is 9

Read more on substitution here

brainly.com/question/28112905?referrer=searchResults

#SPJ1

6 0
1 year ago
Find y' if x = tan(y)
Travka [436]
Use the chain rule:

x=\tan y
\dfrac{\mathrm d}{\mathrm dx}x=\dfrac{\mathrm d}{\mathrm dx}\tan y
1=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}
1=\sec^2y\,y'
y'=\dfrac1{\sec^2y}=\cos^2y

Furthermore, if you restrict the domain of y, we could write

x=\tan y\iff \tan^{-1}x=y

and so we could also have

y'=\cos^2(\tan^{-1}x)=\left(\dfrac1{\sqrt{x^2+1}}\right)^2=\dfrac1{x^2+1}
5 0
4 years ago
Find the zeros of the function g(x) = 2x2 - 5x + 2.
sertanlavr [38]
G(x) = 2x² - 5x + 2 = 2x² - 4x - x + 2 = 2x · x - 2x · 2 - 1 · x - 1 · (-2)

= 2x(x - 2) -1(x - 2) = (x - 2)(2x - 1)

g(x) = 0 ⇔ (x - 2)(2x - 1) = 0 ⇔ x - 2 = 0 or 2x - 1 = 0

x = 2 or x = 0.5
8 0
3 years ago
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