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OLEGan [10]
3 years ago
5

Can you please help asap

Mathematics
1 answer:
dybincka [34]3 years ago
7 0
For question 38 it would be C. -38 because -38 - 7 is -45 and -45 divided by 9 is -5.

Question 35 would be B. 4/5 because 80% converted into a decimal is 4/5.

Hope these helped!! :)

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18. Let z = a + bi represent a general complex number. As noted in the lesson, the conjugate of z, abbreviated conj(z) or conj(a
AnnyKZ [126]

Question (18):

Answer: (A) All of the following

(I am using z^* in place of conj(z)

z*z^*=(a+ib)(a-ib)=a^2-(ib)^2= a^2+b^2=\mbox{modulus(z)}^2

z+z^*=a+ib+a-ib=2a\\z-z^*=a+ib-a+ib=2bi


Question (19):

Answer: (B) -7

modulus of -7 is sqrt(49)=7

the other choices are all < 7


5 0
3 years ago
Determinati numarul b astfel incat : (4x8)^15 : 4^5:2^6=2^b si 3^4x3^5x3^17=3^b
Serhud [2]
The answer is 3 becasue 

3 0
3 years ago
Read 2 more answers
Can I get lil help on theses please
pav-90 [236]

these are the answers is order:

10, 22, 57, 36, 42, 60, and 16,000.

Hope these helped! :D


5 0
3 years ago
Read 2 more answers
What is the expression in radical form?<br><br> (4x3y2)310
sertanlavr [38]

Given:

Consider the given expression is

(4x^3y^2)^{\frac{3}{10}}

To find:

The radical form of given expression.

Solution:

We have,

(4x^3y^2)^{\frac{3}{10}}=(2^2)^{\frac{3}{10}}(x^3)^{\frac{3}{10}}(y^2)^{\frac{3}{10}}

(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{6}{10}}(x)^{\frac{9}{10}}(y)^{\frac{6}{10}}

(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{3}{5}}(x)^{\frac{9}{10}}(y)^{\frac{3}{5}}

(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{2^3}\sqrt[10]{x^9}\sqrt[5]{y^3}       [\because x^{\frac{1}{n}}=\sqrt[n]{x}]

(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{8y^3}\sqrt[10]{x^9}       [\because x^{\frac{1}{n}}=\sqrt[n]{x}]

Therefore, the required radical form is \sqrt[5]{8y^3}\sqrt[10]{x^9}.

8 0
4 years ago
Write a unit rate word problem. Please include the solution.
yanalaym [24]

Answer:

bob goes 28 miles in 2 hours while traveling to the store. how many miles does he drive per hour. the answer is 14 miles per hour

3 0
3 years ago
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