HELP PLEASE!

Is x²-5x+3=0 is a linear equations?

Neporo4naja [7]

Answer:

No, the equation is not a linear equation.

Step-by-step explanation:

There is the x^2 thing that doesn't make it a linear equation.

8 0

3 years ago

Read 2 more answers

In the given parallelogram ABCD, find the value of x and y

Pani-rosa [81]

Answer:

see explanation

Step-by-step explanation:

(a)

The consecutive angles are supplementary, sum to 180° , then

3y + 2y - 5 = 180

5y - 5 = 180 ( add 5 to both sides )

5y = 185 ( divide both sides by 5 )

y = 37

Then ∠ A = 3y = 3 × 37 = 111

The opposite angles are congruent , then

3x + 3 = 111 ( subtract 3 from both sides )

3x = 108 ( divide both sides by 3 )

x = 36

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(b)

The diagonals bisect each other , so

x + 8 = 16 - x ( add x to both sides )

2x + 8 = 16 ( subtract 8 from both sides )

2x = 8 ( divide both sides by 2 )

x = 4

and

5y + 4 = 2y + 13 ( subtract 2y from both sides )

3y + 4 = 13 ( subtract 4 from both sides )

3y = 9 ( divide both sides by 3 )

y = 3

5 0

3 years ago

Help ASAP Ill mark brainliest

Shkiper50 [21]

The answer is letter D

6 0

3 years ago

An elementary school is in a growing suburb. The school’s principal estimates that the current enrollment of 750 students will i

HACTEHA [7]

Answer:

62.9% in ten years

Step-by-step explanation:

This type of growth corresponds to what is called an exponential growth, since in the expression for the number of students keep including the multiplication of the same factor as the years go by, Let's start by analyzing what happens the first year:

The initial enrollment of 750 is expected to grow by 5%, therefore after one year the enrollment should be:

After one year = Year_1 = 750 + 5% increase = $750+750*0.05=750(1+0.05)$

where we have used the decimal form of 5% as he factor 0.05 multiplying 750 (the initial enrollment) to give the increase.

After the second year, we consider a starting value of Year_1 enrollments that will increase another 5%, which gives: Year_2 = Year_1 + Year_1 * 0.05= Year_1 (1 + 0.05).

So replacing Year_1 by its original expression (750(1+0.05)) we notice that Year_2 = 750 * (1+0.05) * (1+0.05) = 750 * (1+0.05)^2

We can go on with this same reasoning and find that each year includes a new factor (1+0.05) to the new increasing enrollment.

At the end of 10 years, the number of enrollment will be:

$750*(1+0.05)^10=1221.67$