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3 years ago

Solve for u:−14u+32=−6.

Mathematics

1 answer:

zubka84 [21]3 years ago

3 0

Answer:u=19/7 or 2.714286 or 2 5/7

Step-by-step explanation:

−14u+32=−6

Step 1: Subtract 32 from both sides.

-14u+32-32=-6-32

-14u=-38

Step 2: Divide both sides by -14.

-14u/-14=-38/-14

U=19/7

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16 divided by four fifth

kati45 [8]

Answer:

Step-by-step explanation:

16 divided by 4/5 = 20 :)

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3 years ago

Read 2 more answers

Today, everything at a store is on sale. The store offers a 20% discount.

kotykmax [81]

Answer:

a) $19.2

b) 0.8x

c) $17.5

Step-by-step explanation:

1. We want to find the discount price given the regular price

The discount is 20%

so the discount price will be;

24 - 20% of 24

24 - 4.8 = $19.2

b) We have this as;

x - (20% of x)

x - 0.2x = 0.8x

c) let the original price be x

x - 20% of x = 14

x - 0.2x = 14

0.8x = 14

x = 14/0.8

x = $17.5

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3 years ago

The roof of the Terminal Tower is 709 feet above ground. A window washer dropped a bucket from the edge of the roof. How many se

Masja [62]

Answer:

6.64 seconds

Step-by-step explanation:

Given that:

Distance between roof of the terminal and ground = 709 feet

A bucket was dropped from the roof.

To find:

The time taken by the bucket to hit the ground?

Solution:

Distance traveled, $s$ = 709 feet

Initial velocity of the bucket, $u$ = 0 feet/sec (because initially it was at rest)

Acceleration of the bucket will be equal to acceleration due to gravity i.e.

$g$ or 32.1741 $ft/sec^{2}$

Let the time taken is $t$ seconds.

Formula for distance is given as:

$s = ut+\dfrac{1}{2}at^2$

Putting all the values:

$709 = 0 (t) + \dfrac{1}{2}(32.1741) t^2\\\Rightarrow 1418 = 32.1741 t^2\\\Rightarrow t^2 = 44.07\\\Rightarrow t \approx \bold{6.64\ sec}$

6 0

3 years ago

Please HELP!!!!!

Stels [109]

Can you explain it more clear because my english is not very good, I want to easy to help you

3 0

3 years ago

Solve the equation on the interval [0,2π). 7 sec x-7=0, x=?

Oksanka [162]

$\huge \bf༆ Answer ༄$

Let's solve ~

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:7 \sec(x) - 7 = 0$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:7 \sec(x) = 7$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: \sec(x) = 7 \div 7$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: \sec(x) = 1$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:x = 0$

8 0

2 years ago