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2 years ago

1+1 =? P.S free give away

Mathematics

2 answers:

Neporo4naja [7]2 years ago

7 0

2 and the answer is 2 because 1+1=2 :)

Lina20 [59]2 years ago

4 0

Answer:

Step-by-step explanation:

The question is asking us to simply evaluate 1 + 1. 1 + 1 equals 2.

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What can i do to be brainiest

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3 years ago

Read 2 more answers

25) Which graph represents the equation below? ) = w+ 2 A ܠܐ 5 4 (2, 4) 3 2 1 (-2,0) -5 -4 -3 -2 -1 0 -1 x 1 2 3 4 5 -2 -3 -4 -5

victus00 [196]

Answer:

Can you attach a pic please ?

Step-by-step explanation:

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2 years ago

100 points! Mhanifa can you please help? Look at the picture attached. I will mark brainliest!

dimulka [17.4K]

Answer:

1 and 2.

Midpoints calculated, plotted and connected to make the triangle DEF, see the attached.

D= (-2, 2), E = (-1, -2), F = (-4, -1)

As per definition, midsegment is parallel to a side.

Parallel lines have same slope.

Find slopes of FD and CB and compare.

m(FD) = (2 - (-1))/(-2 -(-4)) = 3/2
m(CB) = (1 - (-5))/(1 - (-3)) = 6/4 = 3/2
As we see the slopes are same

Find the slopes of FE and AB and compare.

m(FE) = (-2 - (- 1))/(-1 - (-4)) = -1/3
m(AB) = (1 - 3)/(1 - (-5)) = -2/6 = -1/3
Slopes are same

Find the slopes of DE and AC and compare.

m(DE) = (-2 - 2)/(-1 - (-2)) = -4/1 = -4
m(AC) = (-5 - 3)/(-3 - (-5)) = -8/2 = -4
Slopes are same

As per definition, midsegment is half the parallel side.

We'll show that FD = 1/2CB

FD = $\sqrt{(2+1)^2+(-2+4)^2}$ = $\sqrt{3^2+2^2}$ = $\sqrt{13}$
CB = $\sqrt{(1 + 5)^2+(1+3)^2}$ = $\sqrt{6^2+4^2}$ = 2 $\sqrt{13}$
As we see FD = 1/2CB

FE = 1/2AB

FE = $\sqrt{(-4+1)^2+(-1+2)^2}$ = $\sqrt{3^2+1^2}$ = $\sqrt{10}$
AB = $\sqrt{(-5 -1)^2+(3-1)^2}$ = $\sqrt{6^2+2^2}$ = 2 $\sqrt{10}$
As we see FE = 1/2AB

DE = 1/2AC

DE = $\sqrt{(-2+1)^2+(2+2)^2}$ = $\sqrt{1^2+4^2}$ = $\sqrt{17}$
AC = $\sqrt{(-5 +3)^2+(3+5)^2}$ = $\sqrt{2^2+8^2}$ = 2 $\sqrt{17}$
As we see DE = 1/2AC

3 0

3 years ago

Read 2 more answers

Rationalize the denominator of sqrt -49 over (7 - 2i) - (4 + 9i)

zubka84 [21]

$\sqrt{ \frac{-49}{(7-2i)-(4+9i) } } 
$

This one is quite the deal, but we can begin by distributing the negative on the denominator and getting rid of the parenthesis:

$\frac{ \sqrt{-49}}{7-2i-4-9i}$

See how the denominator now is more a simplification of like terms, with this I mean that you operate the numbers with an "i" together and the ones that do not have an "i" together as well. Namely, the 7 and the -4, the -2i with the -9i.
Therefore having the result:

$\frac{ \sqrt{-49} }{3-11i}$

Now, the $\sqrt{-49}$ must be respresented as an imaginary number, and using the multiplication of radicals, we can simplify it to $\sqrt{49} \sqrt{-1}$
This means that we get the result 7i for the numerator.

$\frac{7i}{3-11i}$

In order to rationalize this fraction even further, we have to remember an identity from the previous algebra classes, namely: $x^2 - y^2 =(x+y)(x-y)$
The difference of squares allows us to remove the imaginary part of this fraction, leaving us with a real number, hopefully, on the denominator.

$\frac{7i (3+11i)}{(3-11i)(3+11i)}$

See, all I did there was multiply both numerator and denominator with (3+11i) so I could complete the difference of squares.
See how $(3-11i)(3+11i)= 3^2 -(11i)^2$ therefore, we can finally write:

$\frac{7i(3+11i)}{3^2 - (11i)^2 }$

I'll let you take it from here, all you have to do is simplify it further.
The simplification is quite straightforward, the numerator distributed the 7i. Namely the product $7i(3+11i) = 21i+77i^2$ .
You should know from your classes that i^2 = -1, thefore the numerator simplifies to $-77+21i$
You can do it as a curious thing, but simplifying yields the result:
$\frac{-77+21i}{130}$

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3 years ago

Can I get some help ???

Nikitich [7]

Answer is D. 288 . i am glad to help

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2 years ago