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3 years ago

PLEASE HELP NEED ANSWER NOW!!!!!!

1 answer:

7 0

Slope intercept form is y = mx + b
You would have to reorder.
For this you need to isolate y.
First subtract 5x from both sides.
This would result in: -y = -4 - 5x
Then multiply both sides by negative one to make the y positive.
This would result in: y = 5x + 4
Hope I helped:)

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Solve for A in the equation AB + CD + EFG = Z.

Semenov [28]

Answer:

A = (Z-CD - EFG)/B

Step-by-step explanation:

AB + CD + EFG = Z.

Subtract CD and EFG from each side

AB + CD + EFG-CD -EFG = Z-CD - EFG

AB = Z-CD - EFG

Divide each side by B

AB /B = (Z-CD - EFG)/B

A = (Z-CD - EFG)/B

7 0

3 years ago

What is 2/3 plus 11/12 equal in a fraction

dlinn [17]

19/12

welcome :):):):)

6 0

3 years ago

Read 2 more answers

Will give brainliest to whoever answers first and correct

aliina [53]

Answer:

A is f(t)=-(t-4)^2+50

B is 50. the top of the graph is at 50

C is at 4. that is the x value when y=50

If you need the graph put the answer from question A into Desmos

4 0

2 years ago

my brother and i walk the same route every day. my beother takes 40minutes to get to school and i take 30minutes to get to schoo

Goryan [66]

In 1 minute, I walk = 1/30

In 1 minute, My brother walk = 1/40

In T minutes, I walk = T*1/30

My brother walk 8 minute before = T + 8

In 1 minute, My brother walk = (T + 8)*1/40

When these two distances are same, then I will catch him.

T*(1/30) = (T + 8)*(1/40)

30*(T + 8) = 40*T

30*T + 240 = 40*T

240 = 10*T

T = 24 minutes

hence, I will catch him in 24 minutes.

Take 5 instead of 8.

My brother walk 5 minute before = T + 5

In 1 minute, My brother walk = (T + 5)*1/40

When these two distances are same, then I will catch him.

T*(1/30) = (T + 5)*(1/40)

30*(T + 5) = 40*T

30*T + 150 = 40*T

150 = 10*T

T = 15 minutes

Hence, I will catch him in 15 minutes.

6 0

3 years ago

A clock was reading the time accurately on Friday at noon. On Monday at 6pm the clock was running late by 468 seconds. On averag

Setler [38]

The clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

The clock was still accurate by Friday noon. The clock was late by 468 seconds by Monday, 6 pm.

To solve the problem, we must:

Know how many 30-minutes have passed during the time period.

1 day = 24 hours

1 hour = 60 minutes = 2 × (30 minutes)

1 day = 24 hours × 2 × (30 minutes)

1 day = 48 × (30 minutes)

Thus, there are 48, 30-minutes in a day. On Friday, however, we start counting at noon, which is half of the day. Moreover, on Monday, the mark is only up to 6 pm, which is three-fourths of the day.

Friday = 48 × $\frac{1}{2}$ = 24

Saturday = 48

Sunday = 48

Monday = 48 × $\frac{3}{4}$ = 36

TOTAL = 24 + 48 + 48 + 36 = 156

Therefore, the total number of 30-minutes that have passed is 156. There were 156, 30-minutes that passed during the time period.

Divide the number of total seconds late by the number of 30-minutes passed.

That is, the number of total seconds late= 468 seconds ÷ 156

= 3 seconds

Therefore, the clock was skipping 3 seconds every 30 minutes from Friday noon to Monday 6 pm.

To learn more about clock problems visit:

brainly.com/question/27122093.

#SPJ1

3 0

2 years ago