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3 years ago

Who is more deserving of a statue helen keller or jackie robison

Mathematics

2 answers:

PtichkaEL [24]3 years ago

8 0

Hellen Keller is more deserving of a statue.

Tems11 [23]3 years ago

6 0

A statue will not help either of them in their present condition.

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Find the area of the shape shown below

PSYCHO15rus [73]

Add the top and bottom together, divide by 2, then multiply by the height:

Area = (7.5 + 2.5)/2 x 2.5

Area = 10/2 x 2.5

Area = 5 x 2.5

Area = 12.5 square units.

7 0

4 years ago

Mr Doyle mowed 2/7 of his lawn. His son mode 1/3 of it. Who made most of the lawn? How much of the lawn still needs to be mowed?

CaHeK987 [17]

Answer:

His son mowed most of it and 8/21 of it still needs to be done.

Step-by-step explanation:

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Rate 5 stars mark as brainliest and press thancks plz!

4 0

3 years ago

A jar of marbles has five blue marbles, three red marbles, and four green marbles. What is the possibility of drawing a marble t

solniwko [45]

Answer:

5/36

Step-by-step explanation:

The chance of drawing a blue marble is 5/12. The chance of drawing a green marble is 4/12. You multiply the fractions to get the probability.

5/12 * 4/12 =20/144 = 5/36 simplified.

8 0

3 years ago

Read 2 more answers

What additional information is necessary to prove that SILY is a parallelogram

kirza4 [7]

Answer:

Are the sides parallel?

Step-by-step explanation:

8 0

4 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