-5.5 + 3.7 can u please help will mark branliest

Mathematics

2 answers:

Scrat [10]3 years ago

6 0

Answer:

1.8 es pero que te ayude amiga

Nadusha1986 [10]3 years ago

4 0

Answer:

-1.8

Step-by-step explanation:

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Simple interest. 2700=2200(1+r)^5

kykrilka [37]

Solve your equation step-by-step.

2700=2200(1+r)^5

Flip the equation.

2200(r+1)^5 = 2700

Divide both sides by 2200.

2200(r+1)^5 / 2200 = 2700 / 2200

(r+1)^5=27/22

Solve Exponent.

(r+1)^5=27/22

((r+1)5) ^1/5 = (27/22)^1/5 (Raise both sides to power)

r+1=1.041809

r+1−1=1.041809−1(Subtract 1 from both sides)

r = 0.041809

Check answers. (Plug them in to make sure they work.)

r = 0.041809 (Works in original equation)

5 0

4 years ago

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A friend is having trouble finding the sum of -84 and 28. A) What is the solution to this problem and B) what explanation would

sergejj [24]

Answer:

A) -56

Step-by-step explanation:

B) First, you have to ignore the plus sign, and recognize that the second negative number means you are subtracting that number.

If this is correct, please like the comment so others can know, too!

4 0

4 years ago

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I don’t understand, HELPPP

goldenfox [79]

$\frac{-6xy^{3}z^5}{-36x^{2}y^{2}z^4}$

First cross out the -6 and -36. Six will go into 6 one time and 6 will go into -36 six times so we have the following after crossing out.

$\frac{-6xy^{3}z^5}{-36x^{2}y^{2}z^4}$
$\frac{xy^{3}z^5}{6x^{2}y^{2}z^4}$ <---After crossing out

Next cross out the x / x^2 which will give us the following:
$\frac{xy^{3}z^5}{6x^{2}y^{2}z^4}$
$\frac{y^{3}z^5}{6xy^{2}z^4}$ <---After crossing out

Now cross out the y^2 / y^3
$\frac{y^{3}z^5}{6xy^{2}z^4}$
$\frac{yz^5}{6xz^4}$ <----After crossing out

All that is left is to cross out the z^5 / z^4 and we are done.
$\frac{yz^5}{6xz^4}$
$\frac{yz}{6x}$ <----After crossing out and we are done.

$\frac{-6xy^{3}z^5}{-36x^{2}y^{2}z^4} = \frac{yz}{6x}$

8 0

3 years ago

Is x less than or equal to 5

frez [133]

Answer:

good evening

Step-by-step explanation:

i dont think

its not equal

4 0

3 years ago

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