All categories

2 years ago

Given 6(x) = (x+41, what is b(-10)?

Mathematics

1 answer:

prisoha [69]2 years ago

7 0

The value of $b(-10)$ is 31

Explanation:

The given expression is $b(x)=x+41$

We need to determine the value of $b(-10)$

The value of $b(-10)$ can be determined by substituting $x=-10$ in the expression $b(x)=x+41$

Thus, we have,

$b(-10)=-10+41$

Adding the terms, we have,

$b(-10)=31$

Thus, the value is 31.

Therefore, the simplified value of the expression by substituting $x=-10$ in the expression $b(x)=x+41$ is 31.

You might be interested in

Fill in the blanks 864+2006=--------+864 5351 +(574+799)= 574+(5351+_____)

Mama L [17]

Answer:

2006

799

Step-by-step explanation:

Given the expression:

The right hand side of the sun must be equal to the left hand side ;

Therefore ;

864+2006=--------+864

The sum of 864 and 2006 must be equal to the right hand side sum ; hence

864+2006= 2006 +864

Similarly, the left hand side and right havd side must also be equal here ;

5351 + 574 + 799 = 574 + 5351 + 799

Hence, the missing value is 799

6 0

2 years ago

What is the equation of the line that passes through the point -8, 0 and has the slope of -1 / 4

svetlana [45]

Answer:

y=-1/4x-8

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

What equation represents this sentence? 18 less than the product of a number and 9 is 14.

Sonja [21]

18 - (9 x 14) This equation represents the sentence.

hope that helps :)

5 0

2 years ago

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

2 years ago

Read 2 more answers

Simplify the expression a²+a+a²

sasho [114]

Answer:

2a^2+a

Step-by-step explanation:

you can simplify a^2 into itself twice

7 0

2 years ago