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3 years ago

Please help me please

Mathematics

1 answer:

Lyrx [107]3 years ago

8 0

Answer:

1. Paralleogram

2. Trapeziod

3. Square

4. Rhombus

5. Rectangle

6. I don't know

7. Rectangle

8. can't see it

9. Trapeziod

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Last June only 0.17 inches of rain fell all month. What is the difference between the average rainfall and the actual rainfall f

spin [16.1K]

Answer:

Part 1) 0.50 inches

Part 2) 1.14 inches

Part 3)

a) 1.25 inches

b) -0.32 inches

Step-by-step explanation:

The complete question in the attached figure

Part 1) Last June only 0.17 inches of rain fell all month. What is the difference between the average rainfall and the actual rainfall for the last June

Let

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

The difference between the average rainfall and the actual rainfall for the last June is given by the expression

$(x-y)$

we have

$x=0.67\ in\\y=0.17\ in$

substitute

$0.67-0.17=0.50\ in$

Part 2) The departure from the average rainfall last July was -0.36 inches. How much rain fell last July?

we know that

The departure from the average is the difference between the actual amount of rain and the average amount of rain for a given month

Let

D -----> the departure from the average

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

$D=y-x$

we have

$D=-0.36\ in\\x=1.5\ in$

substitute the given values

$-0.36=y-1.5$

solve for y

$y=1.5-0.36\\y=1.14\ in$

Part 3)

<u>Part a)</u> How much rain would have to fall in August so that the total amount of rain equals the average rainfall for these three months?

<u>Part b</u>) What would the departure from the average be in August in that

situation?

Part a) How much rain would have to fall in August so that the total amount of rain equals the average rainfall for these three months?

Find the average rainfall for these three months

Adds the average rainfall and divide by 3

$(0.67+1.5+1.57)/3=1.25\ in$

Part b) What would the departure from the average be in August in that

situation?

we know that

The departure from the average is the difference between the actual amount of rain and the average amount of rain for a given month

Let

D -----> the departure from the average

x ----> the average rainfall for the last June

y ----> the actual rainfall for the last June

$D=y-x$

we have

$x=1.57\ in\\y=1.25\ in$

substitute the given values

$D=1.25-1.57=-0.32\ in$

7 0

4 years ago

If dy/dx=2xy, then d^2y=dx^2=

Anastasy [175]

$\bf \cfrac{dy}{dx}=2xy\implies \cfrac{d^2y}{dx^2}=2\stackrel{product~rule}{\left( 1\cdot y+x\cfrac{dy}{dx} \right)}\implies \cfrac{d^2y}{dx^2}=2y+2x\cfrac{dy}{dx}$

4 0

3 years ago

I need help on #10

Ede4ka [16]

D = sq.root of((x2 - x1)^2 + (y2 - y1)^2)
sq.root of 13 = sq.root of((x - (-2))^2 + (1 - 3)^2)
= sq.root of((x + 2)^2 + (-2)^2)
= sq.root of (x^2 + 4x + 4 + 4)
= sq.root of (x^2 + 4x + 8)

Now if we square both sides:
x^2 + 4x + 8 = 13
x^2 + 4x - 5 = 0
(x + 5)(x - 1) = 0
x = -5 or x = 1

6 0

3 years ago

If i'm in 9th grade right now what years was my 6th grade year ?

creativ13 [48]

Answer:

2017-2018

Step-by-step explanation:

5 0

3 years ago

Can someone help me please

Misha Larkins [42]

Answer:

x= 13, angle 1= 37, and angle 2= 53

Step-by-step explanation:

So you have 2x+11+1+4x=90 degrees

If you simplify it, it becomes 6x+12=90 degrees

Then, you rearrange; 6x=90-12 or 6x= 78

So then, x=13

For angle 1 its 2x+11= 37 degrees

And angle 2 is 1+4x= 53 degrees

7 0

3 years ago