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3 years ago

Somone help me on please and thank you

Mathematics

2 answers:

Alexeev081 [22]3 years ago

7 0

Answer:

Capitalize Europe

Step-by-step explanation:

You need to capitalize Europe because it is a state. You have to capitalize proper nouns and Europe is a proper noun so you have to capitalize it.

If this has helped you please mark as brainliest

Slav-nsk [51]3 years ago

5 0

Answer:

You need to capitalize Europe

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I really need help >_< please and thank you

SashulF [63]

So for each of the graphs, you basically just have to fill out slope form, which is y=my+b. And to make this equation, you need to find each variable.

First, find the slope. To calculate slope, count the rise (how many units up or down) the line goes from any point to the next immediate point, then the run (how many units left or right.) This should leave you with a fraction, rise/run. For example, on number 3, from labeled points (-5, -4) to (5,2) (double check cause it’s hard to see on my phone, but i think those are the points on 1??) it rises 6 units (-4 to 2) and runs 10 units (-5 to 5). This gives you your rise/run fraction, which is 6/10, simplified to 3/5.

So the slope fills out the m part of the equation. For 3, we found that slope is 3/5, and that fits into the m variable of the slope equation.

This makes it y=3/5x+b.

The last (and considerably less confusing) step is to find b. b is the y-intercept, which is just the point in the graph where the line crosses the y-axis and x=0. On 3, this would be (0,-1) or just -1.

So fill -1 into the b slot of the equation, and you get y=3/5x-1. And thats it!!

Let me know if you still need help on any of the other problems, but I hoped this helped to clear it up!! :)

8 0

3 years ago

Consider two functions f and g on [1, 8] such that integral^8_1 f(x) dx = 9, integral^8_1 g(x) dx = 5, integral^8_5 f(x) dx = 4,

Gelneren [198K]

I'll abbreviate the definite integral with the notation,

$I(f(x),a,b)=\int_a^bf(x)\,\mathrm dx$

We're given

$I(f,1,8)=9$
$I(g,1,8)=5$
$I(f,5,8)=4$
$I(g,1,5)=3$

Recall that the definite integral is additive on the interval $[a,b]$ , meaning for some $c\in[a,b]$ we have

$I(f,a,b)=I(f,a,c)+I(f,c,b)$

The definite integral is also linear in the sense that

$I(kf+\ell g,a,b)=kIf(a,b)+\ell I(g,a,b)$

for some constant scalars $k,\ell$ .

Also, if $a\ge b$ , then

$I(f,a,b)=-I(f,b,a)$

a. $I(2f,1,5)=2I(f,1,5)=2(I(f,1,8)-I(f,5,8))=2(9-4)=\boxed{10}$

b. $I(f-g,1,8)=I(f,1,8)-I(g,1,8)=9-5=\boxed{4}$

c. $I(f-g,1,5)=I(f,1,5)-I(g,1,5)=\dfrac{I(2f,1,5)}2-I(g,1,5)=10-3=\boxed{7}$

d. $I(g-f,5,8)=I(g,5,8)-I(f,5,8)=(I(g,1,8)-I(g,1,5))-I(f,5,8)=(5-3)-4=\boxed{-2}$

e. $I(7g,5,8)=7I(g,5,8)=7(5-3)=\boxed{14}$

f. $I(3f,5,1)=3I(f,5,1)=-3I(f,1,5)=-\dfrac32I(2f,1,5)=-\dfrac32(10)=\boxed{-15}$

4 0

2 years ago

3x+2y= -19 solve by elimination of addition<br> -3x-5y= 25

tensa zangetsu [6.8K]

Answer:

x=-5, y=-2

Step-by-step explanation:

3x+2y= -19

-3x-5y= 25

Add the two equations together

3x+2y= -19

-3x-5y= 25

------------------

0x -3y = 6

Divide each side by -3

-3y/-3 = 6/-3

y = -2

Now find x

3x+2y = -19

3x +2(-2) =-19

3x-4 = -19

Add 4 to each side

3x -4+4 = -19+4

3x= -15

divide by 3

3x/3 = -15/3

x = -5

8 0

2 years ago

Read 2 more answers

10. When you add the same number to both sides of an inequality, is the inequality still true? Explain how you know your stateme

avanturin [10]

The inequality is still true! If you add a number, say 5 to both sides of the following inequality, does anything change?

3 < 6

3 + 5 < 6 + 5

8 < 11

The inequality is still true. We know the statement holds for subtracting the same number because, in a way, addition and subtraction are pretty much the same operation. If I subtract 5 from both sides, I can think of it like "I add negative 5 to both sides" or something along those lines. It's kind of backwards thinking.

6 0

3 years ago

Help please

Nikitich [7]

Answer:

35 bagels

Step-by-step explanation:

Duh

8 0

2 years ago