Find k such that the equation kx^2+x+25k=0 has a repeated solution.

A rectangular plot of land is to be enclosed by a fence. One side is along a river, and does not need to be enclosed. If the tot

Bad White [126]

Answer:

Width 150 meters.

Length = 350 meters.

Step-by-step explanation:

Let us assume the width of the fence = k meters

So, both sides = k + k = 2k meters

Also, the TOTAL fencing length = 600 m

So, the one side length of the fence = (600 - 2 k) meters

AREA = LENGTH x WIDTH

⇒ A(k) = (600 - 2k) (k)

or, A = -2k² + 600 k

The above equation is of the form: ax² +bx + C

Here: a = - 2 , b = 600 and C = 0

As a< 0, the parabola opens DOWNWARDS.

Here, x value is given as: $x = \frac{-b}{2a}$

Solving for the value of k similarly, we get:

$k = \frac{-b}{2a} = \frac{600}{2(-2)} = 150$

Thus the desired width = k = 150 meters

So, the desired dimensions of the plot is width 150 meters.

And length = 650 - 2k = 650 - 300 = 350 meters.

5 0

2 years ago

Helpppp me please!! I dont understand.

Dafna1 [17]

Answer:

X = 11

X = 7

Step-by-step explanation:

Part 1:

The measure of an inscribed angle is half the measure of an intercepted arc. So 1/2 mCE = m∠CDE

158 ÷ 2 = 79

You need to find x so you're going to set your equation of (8x - 9) = 79

Then you're going to add 9 to both sides to get 8x = 88

Divide both sides by 8 to get x = 11

Part 2:

We know that WY is 180 because it's a semicircle. Using this and the knowledge that ∠WXY is half the measure of the intercepted arc (180), we know that our equation should be (13x - 1) = 90

add 1 to both sides and our equation is (13x = 91)

Divide both sides by 13 and...

x = 7

8 0

3 years ago

Read 2 more answers

Select all ratios equivalent to 4:3.

valentina_108 [34]

Answer:

8:6 and 16:12

Step-by-step explanation:

5 0

3 years ago

The owner of a football team claims that the average attendance at games is over 523, and he is therefore justified in moving th

USPshnik [31]

Answer:

C) There is not sufficient evidence to support the claim that the mean attendance is greater than 523.

Step-by-step explanation:

Let μ be the the average attendance at games of the football team

The claim: the average attendance at games is over 523

Null and alternative hypotheses are:

$H_{0}$ : μ=523

$H_{a}$ : μ>523

The conclusion is failure to reject the null hypothesis.

This means that test statistic is lower than critical value. Therefore it is not significant, there is no significant evidence to accept the alternative hypothesis.

That is no significant evidence that the average attendance at games of the football team is greater than 523.

7 0

2 years ago

Suppose Jessica opens a savings account with $5000 that compounds interest daily. The APR at the time Jessica opens the account

IrinaK [193]

Hi there

1+0.035=(1+r/360)^360
Solve for r
R=(((1.035)^(1÷360)−1)×360)×100
R=3.44%

6 0

3 years ago