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3 years ago

13

Don, Shuan and Monica have 5 hallways to decorate for the student council if they share the work equally how much will easch stu

dent decorate

1 answer:

Romashka-Z-Leto [24]3 years ago

8 0

Each student will have a little more then a hallway and a half to decorate

If you were to be exact each student would have 1.6 (repeating) hallways which would make sense in the problem situation <span />

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IM CONFUSED: A used car dealer has the following information for 2 months of sales:

Anuta_ua [19.1K]

Answer:

increase because 324 >>>>> 239

percentage: 324/239 = 135%

8 0

2 years ago

Read 2 more answers

Find the complete factorization of the expression.<br><br> 16xz + 20yz

Andreyy89

To completely factor you must from the Greatest common factor

In this case it is 4z

So the complete factorization is

4z(4x + 5y)

8 0

4 years ago

Evaluate the expression when y=6 and x = 35.<br> x-3y

vredina [299]

Answer:

17

Step-by-step explanation:

Given: y = 6; x + 35

Step 1: Equation

x - 3y

Step 2: Substitution

35 - 3(6)

Step 3: Solving

35 - 18 = 17

Answer:

17

Hope This Helps :)

4 0

3 years ago

Read 2 more answers

Perform the indicated operations. Write the answer in standard form, a+bi.<br> 5-3i / -2-9i

Vsevolod [243]

$\huge \boxed{\mathfrak{Answer} \downarrow}$

$\large \bf\frac { 5 - 3 i } { - 2 - 9 i } \\$

Multiply both numerator and denominator of $\sf \frac{5-3i}{-2-9i} \\$ by the complex conjugate of the denominator, -2+9i.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)}) \\$

Multiplication can be transformed into difference of squares using the rule: $\sf\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$ .

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}}) \\$

By definition, i² is -1. Calculate the denominator.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85}) \\$

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

$\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85}) \\$

Do the multiplications in $\sf5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9\left(-1\right)$ .

$\large \bf \: Re(\frac{-10+45i+6i+27}{85}) \\$

Combine the real and imaginary parts in -10+45i+6i+27.

$\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85}) \\$

Do the additions in $\sf-10+27+\left(45+6\right)i$ .

$\large \bf Re(\frac{17+51i}{85}) \\$

Divide 17+51i by 85 to get $\sf\frac{1}{5}+\frac{3}{5}i \\$ .

$\large \bf \: Re(\frac{1}{5}+\frac{3}{5}i) \\$

The real part of $\sf \frac{1}{5}+\frac{3}{5}i \\$ is $\sf \frac{1}{5} \\$ .

$\large \boxed{\bf\frac{1}{5} = 0.2} \\$

3 0

3 years ago

Simplify the difference quotient f ( x + h ) â f ( x ) h for the function f ( x ) = 6 x â x 2

ladessa [460]

Answer:

final answer is 6

Step-by-step explanation:

Given $f(x)=6x-2$

To find the difference quotient use formula

$\frac{f(x+h)-f(x)}{h}$

$f(x)=6x-2$

$f(x+h)=6(x+h)-2=6x+6h-2$

Now plug it in the formula

$\frac{f(x+h)-f(x)}{h}$

$\frac{6x+6h-2-(6x-2)}{h}$

$\frac{6x+6h-2-6x+2)}{h}$

$\frac{6h}{h}$

6

5 0

3 years ago