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3 years ago

Can somebody help me solve for x?

Mathematics

1 answer:

Dahasolnce [82]3 years ago

7 0

Let y = the distance between the top left corner and the bottom right corner

y^2 = 24^2 + 15^2

y^2 = 801

y = 3 * sqrt(89)

Now we can find x.

12^2 + x^2 = (3 * sqrt(89))^2

x^2 = 657

x = 3 * sqrt(73) or 25.63

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if the equation x^2 +(k+2)x+2k=0 has equal roots,then the value of k is .....a.2 b.-2 c 1/2 d.none

blsea [12.9K]

Answer:

k=2

Problem:

if the equation x^2 +(k+2)x+2k=0 has equal roots,then the value of k is ..

Step-by-step explanation:

Since the coefficient of x^2 is 1, we can use this identity to aid us: x^2+bx+(b/2)^2=(x+b/2)^2.

So we want the following:

[(k+2)/2]^2=2k

Apply the power on the left:

(k+2)^2/4=2k

Multiply both sides by 4:

(k+2)^2=8k

Expand left side:

k^2+4k+4=8k *I used identity (x+c)^2=x^2+2xc+c^2

Subtract 8k on both sides:

k^2-4k+4=0

Factor using the identity mentioned a couple lines above:

(k-2)^2=0

Since zero squared is zero, we want k-2=0.

Adding both sides by 2 gives k=2.

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Read 2 more answers

While traveling to and from a certain destination, you realized increasing your speed by 40 mph saved 4 hours on your return. If

Luden [163]

Answer:

The speed driven while returning is 88 mph.

Step-by-step explanation:

We are given that while traveling to and from a certain destination, you realized increasing your speed by 40 mph saved 4 hours on your return.

Also, the total distance of the roundtrip was 420 miles.

Let the speed driven while returning be 'x mph' which means that the speed driven while going was '(x - 40) mph' because it has been given that while returning we have increased the speed by 40 mph.

As we know that the Speed-Distance-Time formula is given by;

$\text {Speed} = \frac{\text{Distance}}{\text{Time}}$ or $\text {Time} = \frac{\text{Distance}}{\text{Speed}}$