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3 years ago

The width of a rectangle is "w" inches long. If the length is 4

Mathematics

1 answer:

umka2103 [35]3 years ago

6 0

Answer:

Step-by-step explanation:

Area of rectangle=length(l)×width(w)

In this case,length= w+4

so Area=(w+4)×w

Therefore Area=w^2+4w

You might be interested in

What is 4 over 10 plus 88 over 100

denpristay [2]

Answer:

1.28

Step-by-step explanation:

4/10 + 88/100

so this is unlike denominators so

4 times 10 and 10 times 10 equal to 40/100

so 40/100 + 88/100 equal to 128/100 so we simplify 128÷2 and 100÷2 equal to 64/50 and 64÷2 and 50÷2 equals 32/25 we do long division then it is equal to 1.28

6 0

2 years ago

Solve by the addition method

Sladkaya [172]

Answer is no solution I believe

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

Please awnser this asap

Illusion [34]

Answer:

Base area = 25.2 in^2

Step-by-step explanation:

Base area = 63/2.5 = 25.2 in^2

8 0

3 years ago

Find sin(α) in the triangle.

cluponka [151]

$\frac{20}{29}$ Answer:

sin α = $\frac{20}{29}$

Step-by-step explanation:

sin α = $\frac{opposite}{hypotenuse}$ = $\frac{BC}{AB}$ = $\frac{20}{29}$

6 0

2 years ago