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damaskus [11]
2 years ago
9

A painter uses 1.2 liters of the paint to cover 1m2 of the wall.

Mathematics
1 answer:
xz_007 [3.2K]2 years ago
7 0

Answer:

10

Step-by-step explanation:

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Plz help ill give you brainlist
Daniel [21]
Consider this option:
1. formula of perimeter is: P=2(a+b), where a & b - the sides of rectangle.
2. according to the condition 2(x+(x+4))<120; ⇒ x<28
answer: D. x<28
5 0
3 years ago
Mandy's square quilt measures 99 in on each edge. how many yd of trim does she need to buy to go around the entire quilt
Masteriza [31]
4x99=396in
1 yard=36 in
so there for mandy needs 11 yards of trim
5 0
3 years ago
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Simplify (x^2y)^3. <br> 1)X^2y^3<br> 2)x^5y^3<br> 3)x^6y^3
dsp73

Answer:

the answer is 2

Step-by-step explanation:

3 0
3 years ago
What is 1/6x=4 in solving equations with rational coefficients
Debora [2.8K]
To solve an equation with a rational (fraction) coefficient you need to multiply both sides of the equation by the reciprocal (invert the fraction - flip it).

\frac{1}{6}x =4 The reciprocal of 1/6 is 6/1 or 6
6( \frac{1}{6} x)= 6(4)  The product of reciprocals equals 1 
1x = 24
x = 24


7 0
3 years ago
Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai
marta [7]

f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1

Lets check with every option

(a) [-4,-3]

We plug in -4  for x  and -3 for x

f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3  for x  and -2 for x

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.  

(c) [-2,-1]

We plug in -2  for x  and -1 for x

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1  for x  and 0 for x

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0  for x  and 1 for x

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.  

(f) [1,2]

We plug in 1  for x  and 2 for x

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1],  (d) [-1,0], (f) [1,2]

8 0
3 years ago
Read 2 more answers
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