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3 years ago

6

12 + (-4) + 5j + -2j + 5

1 answer:

Nonamiya [84]3 years ago

3 0

Answer:

12+(-4)+5j+(-2j)+5

=12-4+5j-2j+5

=8+3j+5

=13+3j

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Which polynomial represents the sum below? (14x2 – 14) + (–10x2 – 10x + 10) A. 4x2 – 10x + 24 B. 4x2 – 10x – 4 C. 24x2 – 24x – 1

Gennadij [26K]

Answer:

answer : B

Step-by-step explanation:

(14x2 – 14) + (–10x2 – 10x + 10) = (14x² - 10x²) -10x + (-14+10)

(14x2 – 14) + (–10x2 – 10x + 10) = = 4x² - 10x - 4

answer : B

3 0

4 years ago

Read 2 more answers

Use the diagram to solve for x<br> 15x + 5<br><br> 16x+2

kondor19780726 [428]

Answer:

3

Step-by-step explanation:

I think you set them equal to each other.so you would do 16x+2=15x+5 and you would get x=3. I hope this helps out! ☺️

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3 years ago

Read 2 more answers

I need help right away

Archy [21]

Here is what the graph looks like

3 0

3 years ago

At what time is the cake done

svp [43]

<h2>Explanation:</h2>

Here we know that at 3:17 pm you started to bake the cake. The total revolution the minute hand on the clock makes is:

$\frac{5}{8}+\frac{13}{24}=\frac{7}{6} \ revolutions$

We know that:

$1 \ revolution ----------- 60min \\ \\ 7/6 \ revolutions -----------x$

Solving for x:

$x=\frac{7}{6}\times 60=70 minutes$

So it takes 70 minutes after you put the cake in the oven. But:

$70minutes=1h \ 10min$

Adding this to the initial time:

$3:17pm+1h \ 10min=4:27pm$

Finally, the cake is done at 4:27 pm, option C.

5 0

3 years ago

A person on tour has dollar 360 for his daily expenses. If he extends his tour for 4 days, he has to cut down his daily expenses

mash [69]

Answer:

The original duration of the tour = 20 days

Step-by-step explanation:

Solution:

Total expenses for the tour = $360

Let the original tour duration be for $x$ days.

So, for $x$ days the total expense = $360

<em>Thus the daily expense in dollars can be given by</em> = $\frac{360}{x}$

Tour extension and effect on daily expenses.

The tour is extended by 4 days.

<em>Tour duration now</em> = $(x+4)$ days

On extension, his daily expense is cut by $3

<em>New daily expense in dollars </em>= $(\frac{360}{x}-3)$

Total expense in dollars can now be given as: $(x+4)(\frac{360}{x}-3)$

Simplifying by using distribution (FOIL).

$(x.\frac{360}{x})+(x(-3)+(4.\frac{360}{x})+(4(-3))$

$360-3x+\frac{1440}{x}-12$

$348-3x+\frac{1440}{x}$

We know total expense remains the same which is = $360.

So, we have the equation as:

$348-3x+\frac{1440}{x}=360$

Multiplying each term with $x$ to remove fractions.

$348x-3x^2+1440=360x$

Subtracting $348x$ both sides

$348x-348x-3x^2+1440=360x-348x$

$-3x^2+1440=12x$

Dividing each term with -3.

$\frac{-3x^2}{-3}+\frac{1440}{-3}=\frac{12x}{-3}$

$x^2-480=-4x$

Adding $4x$ both sides.

$x^2+4x-480=-4x+4x$

$x^2+4x-480=0$

Solving using quadratic formula.

For a quadratic equation: $ax^2+bx+c=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plugging in values from the equation we got.

$x=\frac{-4\pm\sqrt{(4)^2-4(1)(-480)}}{2(1)}$

$x=\frac{-4\pm\sqrt{16+1920}}{2}$

$x=\frac{-4\pm\sqrt{1936}}{2}$

$x=\frac{-4\pm44}{2}$

So, we have

$x=\frac{-4+44}{2}$ and $x=\frac{-4-44}{2}$

$x=\frac{40}{2}$ and $x=\frac{-48}{2}$

∴ $x=20$ and $x=-24$

Since number of days cannot be negative, so we take $x=20$ as the solution for the equation.

Thus, the original duration of the tour = 20 days

6 0

3 years ago