Hi there!

If Jeremy walked 6/8 of the way to school and ran the rest we know that Jeremy ran 2/8 of the way. We can simplify this further to get to our simplest form.

6/8 ---> 3/4 Simplified.

So your answer is Jeremy walked 3/4 of the way.

6 0

3 years ago

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Given: PQ ⊥ QR , PR=20, SR=11, QS=5 Find: The value of PS.

dlinn [17]

Answer:

The value of the side PS is 26 approx.

Step-by-step explanation:

In this question we have two right triangles. Triangle PQR and Triangle PQS.

Where S is some point on the line segment QR.

Given:

PR = 20

SR = 11

QS = 5

We know that QR = QS + SR

QR = 11 + 5

QR = 16

Now triangle PQR has one unknown side PQ which in its base.

Finding PQ:

Using Pythagoras theorem for the right angled triangle PQR.

PR² = PQ² + QR²

PQ = √(PR² - QR²)

PQ = √(20²+16²)

PQ = √656

PQ = 4√41

Now for right angled triangle PQS, PS is unknown which is actually the hypotenuse of the right angled triangle.

Finding PS:

Using Pythagoras theorem, we have:

PS² = PQ² + QS²

PS² = 656 + 25

PS² = 681

PS = 26.09

PS = 26

7 0

3 years ago

The intensity of a light source at a distance is directly proportional to the strength of the source and inversely proportional

jolli1 [7]

Answer:

x= $\frac{16}{\sqrt[3]{2}+1 }$

Step-by-step explanation:

Q= illumination

I = intensity

Q= I/d^2

Q_total = $\frac{I_1}{d_1^2}+\frac{I_2}{d_2^2}$

= $\frac{I}{x^2}+\frac{2I}{(16-x)^2}$

now Q' = 0

⇒I ${-\frac{2}{x^3}}+\frac{4}{(16-x)^3}$

x= $\frac{16}{\sqrt[3]{2}+1 }$

[/tex] $\frac{1}{x^3} = \frac{2}{(16-x)^3}$

x= $\frac{16}{\sqrt[3]{2}+1 }$

is the required point

5 0

3 years ago

PLEASE HELP ASAP! Question 12. Thank you

a_sh-v [17]

Answer:

Step-by-step explanation:

In the first column, put days at the top and pounds at the bottom.

So Day 1 = 75 pounds, Day 2=150 pounds, Day 3 = 225 pounds, etc..

8 0

3 years ago