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3 years ago

90 units needed 8per case

1 answer:

6 0

Are you asking how much cases is needed to to reach the 90 units? If yes then 90/8 which would give you 11.25. But since it’s by case then round up to get 12 cases. I think I answered it wrong but

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Pls help i will give 5 starts and brainliest question is down below 9:22

OleMash [197]

4) -0.3 = -1/3 < -1/7 since the bigger a negative number is the less positive it is. 1/8 is positive so it is clear that it should be placed last. & thus do all options this way.

5) B

Area of a triangle = ½ bh.

where b = base &

h = height.

Area of outer shape = (½ x 16 x 12) in^2.

= 96 in^2.

Area of unshaded inner shape = (½ x 3 x 4) in^2.

= 6 in^2.

Area of shaded region = (96 – 6) in^2.

= 90 in^2.

4 0

2 years ago

heights of 200 adults were recorded and divided into two categories which two way freuency table correctly shows the marginal fr

Vsevolod [243]

Answer:

SO THE 1309 IS THE NEW 1 1(239) THE ANSWER OF THAT IS IN THE LINK MP MODULE .COM

7 0

3 years ago

2 Here are two equations:

MatroZZZ [7]

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Step-by-step explanation:

a. Decide whether each neither of the equations,

i (3,4)

ii. (4,2.5)

ill. (5,5)

iv. (3,2)

To decide whether each point is solution to equations or not we will put the point in the equations

Equations are:

Equation 1: 6x + 4y = 34

Equation 2: 5x – 2y = 15

i (3,4)

Putting in Equation 1:

$6(3) + 4(4) = 34\\18+16=34\\34=34\\$

Putting in Equation 2:

$5(3) - 2(4) = 15\\15-8 = 15\\7\neq 15$

ii. (4,2.5)

Putting in Equation 1:

$6(4) + 4(2.5) = 34\\24+10=34\\34=34\\$

Putting in Equation 2:

$5(4) - 2(2.5) = 15\\20-5 = 15\\15=15$

ill. (5,5)

$6(5) + 4(5) = 34\\30+20=34\\50\neq 34$

Putting in Equation 2:

$5(5) - 2(5) = 15\\25-10 = 15\\15=15$

iv. (3,2)

$6(3) + 4(2) = 34\\18+8=34\\26\neq 34$

Putting in Equation 2:

$5(3) - 2(2) = 15\\15-4 = 15\\11\neq 15$

Hence,

(3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. Is it possible to have more than one (x, y) pair that is a solution to both

equations?

The simultaneous linear equations' solution is the point on which the lines intersect. Two lines can intersect only on one point. So a linear system cannot have more than one point as a solution

So,

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Keywords: Linear equations, Ordered pairs

Learn more about linear equations at:

#LearnwithBrainly

8 0

3 years ago

What is the area of this face?