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3 years ago

Please help meeeeeee and show your work please

Mathematics

1 answer:

Luda [366]3 years ago

6 0

It is 6.5 days actually. 3.6/0.2= 13. Since it goes by a half of a day, divide the 13 and 0.5. 6.5 should’ve been that students answer.

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Which of the following items has the lowest unit price? 4 for $5.00 $1.22 each 6 for $7.44 3 for $3.60

aev [14]

3 for 3.60 the unit price come out to 1.20

3 0

3 years ago

Can you help me please?

9966 [12]

First, change the two mixed numbers in the expression into an improper fraction: 25/6 + 5/3.
Then, find the Lowest Common Denominator of both fractions, which is 6, and set both denominators equal to that. Remember, whatever you do on one side you must do to the other: 25/6 + 10/6
Add the two together: 25/6 +10/6 = 35/6.
To make it a mixed number again, find how many times 6 goes into 35, which is 5 times, with a remainder of 5. Your answer is 5 5/6

6 0

3 years ago

Simplify (-4 + 5i) + 2(1 + 3i). Enter your answer in the form a + bi.

luda_lava [24]

(-4 + 5i) + 2(1 + 3i)

Distribute 2 inside the parentheses.

(-4 + 5i) + 2 + 6i

Combine like terms.

is your answer.

5 0

3 years ago

Evaluate

SCORPION-xisa [38]

Value of the expression will be 7

<u><em>Explanation</em></u>

Given value of the variables are : $p= \frac{1}{2}$ and $q=7$

Given expression: $8p+3q-18$

So, <u>plugging the values</u> of $p$ and $q$ into the above expression, we will get...

$8p+3q-18\\ \\ = 8(\frac{1}{2})+3(7)-18\\ \\ =4 + 21-18\\ \\ =25-18\\ \\ =7$

So, the value of the expression will be 7.

5 0

3 years ago

Read 2 more answers

I need help solving this problem. <img src="https://tex.z-dn.net/?f=4x%20cos%20%5E%7B-1%7D%20%282x%2B4%29-%20%5Csqrt%7B3-3%20x%5

Gnesinka [82]

Given:
$f(x)=4xcos^{-1}(2x+4)-\sqrt{3-3x^2}$

Using
$\frac{d}{dx}cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}$
we derive
$\frac{d}{dx}4xcos^{-1}(2x+4)$
$=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}$

Similarly, using
$\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$
we derive
$\frac{d}{dx}(-\sqrt{3-3x^2})$
$=\frac{3x}{\sqrt{3-3x^2}}$

Therefore, the derivative is
$f'(x)=\frac{d}{dx}(4xcos^{-1}(2x+4)-\sqrt{3-3x^2})$
$=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}+\frac{3x}{\sqrt{3-3x^2}}$

3 0

3 years ago

Read 2 more answers