6z^2-48x-120/3z^2+6z • 14z^5+140z^4/4z^2-400 =7z^3 Which values of s make the resulting expression undefined?

Can someone help me please please

solong [7]

Answer:

With? I dont see an image if you attached one

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the missing length for the right triangle described below. B = 21 ft, c=27 ft. Find side a. Round to the nearest whole numb

maksim [4K]

Answer:

16.97 ft

Step-by-step explanation:

According to the scenario, computation of the given data are as follows,

Side B = 21 ft

Side c = 27 ft

As we know, formula for right triangle is as follows,

$a^{2} + b^{2} = c^{2}$

$a^{2} = c^{2} - b^{2}$

So, by putting the value in the formula, we get,

$a^{2} = 27^{2} - 21^{2}$

$a^{2}$ = 729 - 441

$a^{2}$ = 288

a = 16.97 ft

Hence, the length of the side for the right triangle is 16.97ft.

6 0

2 years ago

ben earns a base salary of $400.00 every week with an additional 11% commission on everything he sells. if ben sold $6050.00 wor

RUDIKE [14]

If you would like to know what was his total pay last week, you can calculate this using the following steps:

a base salary ... $400.00
an additional 11% commission on everything Ben sells
11% of $6050.00 = 11% * 6050 = 11/100 * 6050 = $665.5

$400 + $665.5 = $1065.5

Result: His total pay was $1065.5.

4 0

3 years ago

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Find the modulus of the complex number 6-2i

Papessa [141]

Answer:

The modulus of the complex number 6-2i is:

$|z|\:=2\sqrt{10}$

Step-by-step explanation:

Given the number

$6-2i$

We know that

$z = x + iy$

where x and y are real and $\sqrt{-1}=i$

The modulus or absolute value of z is:

$|z|\:=\sqrt{x^2+y^2}$

Therefore, the modulus of $6-2i$ will be:

$z=6-2i$

$z=6+(-2)i$

$|z|\:=\sqrt{x^2+y^2}$

$|z|\:=\sqrt{6^2+\left(-2\right)^2}$

$=\sqrt{6^2+2^2}$

$=\sqrt{36+4}$

$=\sqrt{40}$

$=\sqrt{2^2}\sqrt{2\cdot \:5}$

$=2\sqrt{2\cdot \:5}$

$=2\sqrt{10}$

Therefore, the modulus of the complex number 6-2i is:

$|z|\:=2\sqrt{10}$

3 0

2 years ago

What is the answer please.... need help

mina [271]

Answer:

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

$f\left(x\right)=x^3-6x^2+3x+10$

As the highest power of the x-variable is 3 with the leading coefficients of 1.

So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

$f\left(x\right)=x^3-6x^2+3x+10$

$0=x^3-6x^2+3x+10$ ∵ $f(x)=0$

as

$Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

so

$\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

if $ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x=-1,\:x=2,\:x=5$

Therefore, the zeros of the function are:

$x=-1,\:x=2,\:x=5$

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.

8 0

2 years ago