All categories

2 years ago

What is 110% of 250?

Mathematics

2 answers:

Likurg_2 [28]2 years ago

4 0

Answer:

Your answer 275

romanna [79]2 years ago

3 0

Answer:

275

Step-by-step explanation:

You might be interested in

What is square root of 49 over square root of 1250 simplified

Elan Coil [88]

im pretty sure its 0.19 because the sqyare root of 49 is 7. the square root of 1250 is 35.35

7/35.35=0.19

7 0

2 years ago

Read 2 more answers

Select the correct answer.

Sonja [21]

Answer:

B) 5 + 17i is your answer.

Step-by-step explanation:

In order to write this equation as a complex number in standard form, you must first simplify each term.

Apple the distributive property.

21 - 4i - 1 * 16 - (7i) + 28i

Multiply -1 by 16.

21 - 4i -16 - (7i) + 28i

Multiply 7 by -1.

21 - 4i - 16 - 7i + 28i

Now simplify by adding terms :)

Subtract 16 from 21.

5 - 4i - 7i + 28i

Subtract 7i from -4i.

5 - 11i + 28i

Add -11i and 28i.

= 5 + 17i

5 0

3 years ago

Inventory sheet A shows 256 ounces of ketchup. Inventory sheet B shows two gallons of ketchup. You tell your employee to use oun

jeka94

Answer:

18 ounces

Step-by-step explanation:

3 0

2 years ago

What is four twentyoneths simplified

dedylja [7]

4/21 cannot be reduced any more than it is

4 0

2 years ago

Read 2 more answers

Find the multiplicative inverse of 6 + 2i

Marina CMI [18]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2774989

________________

Find the multiplicative inverse of

$\mathsf{z=6+2i}$

________

The inverse multiplicative of $\mathsf{z=a+bi}$ is

$\mathsf{\dfrac{1}{z}}\\\\\\
=\mathsf{\dfrac{1}{a+bi}\qquad\quad(a\ne 0~~and~~b\ne 0)}\\\\\\
=\mathsf{\dfrac{1}{a+bi}\cdot \dfrac{a-bi}{a-bi}}\\\\\\
=\mathsf{\dfrac{1\cdot (a-bi)}{(a+bi)\cdot (a-bi)}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2-\,\diagup\hspace{-10}abi+\,\diagup\hspace{-10}abi-(bi)^2}}$

$=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot i^2}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot (-1)}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2+b^2}}\\\\\\\\
\therefore~~\mathsf{\dfrac{1}{a+bi}=\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}\,i\qquad\quad\checkmark}$

________

For this question,

$\mathsf{z=6+2i}$

So,

$\mathsf{\dfrac{1}{z}}\\\\\\
=\mathsf{\dfrac{1}{6+2i}}\\\\\\
=\mathsf{\dfrac{6}{6^2+2^2}-\dfrac{2}{6^2+2^2}\,i}\\\\\\
=\mathsf{\dfrac{6}{36+4}-\dfrac{2}{36+4}\,i}\\\\\\
=\mathsf{\dfrac{6}{40}-\dfrac{2}{40}\,i}$

$\therefore~~\mathsf{\dfrac{1}{z}=\dfrac{3}{20}-\dfrac{1}{20}\,i}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

6 0

3 years ago

Read 2 more answers